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Constructing Triangles Worksheet Grade 6
Constructing a Triangle when 2 sides, and 1 angle between them is know from
Web the corbettmaths practice questions on constructing triangles. Web in worksheet on construction of triangles we will solve 10 different types of questions. Web the focus is on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no.
Source: helpingwithmath.com
Web this detailed constructing triangles worksheet includes 12 questions, giving your pupils plenty of chances to practise. A triangle whose each angle is less than 90° is known as _____ triangle.
Once completed, use the answer sheet to go over each. Web triangles worksheets and online activities.
Some Of The Worksheets Displayed Are Mathematical C.
Find the measure of the third angle. Features 12 questions with answers, so you can easily gauge pupils' progress. Web showing 8 worksheets for constructing triangles grade 6.
Web Showing 8 Worksheets For Constructing Triangles Grade 6.
A triangle whose each angle is less than 90° is known as _____ triangle. Web this detailed constructing triangles worksheet includes 12 questions, giving your pupils plenty of chances to practise. Web triangles worksheets and online activities.
A Triangle Whose One Angle Is Equal To _____ Is.
The answers are actual size, so can be printed on transparencies and used by pupils for self marking. Free interactive exercises to practice online or download as pdf to print. Web a worksheet on constructing asa and sas triangles.
Once completed, use the answer sheet to go over each. List conditions of a triangle activity | constructing triangles error analysis created by maneuvering the middle this conditions of a. Web the focus is on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no. | 677.169 | 1 |
10 Real-life Examples Of A Cone To Understand It Better
Geometry is not something that is limited to the domains of theory on paper. Examples of shapes and geometrical properties are in abundance in our surroundings, after all, it is by observing our environments that these figures took shape on paper. However, it is challenging to teach about different shapes and their properties to students.
One such crucial yet complex figure is that of a cone. When it comes to studying this 3-dimensional figure, it might not be a piece of cake for many children. It is very easy to encounter the figure of a cone in our daily life without even realizing it. So, why not connect theoretical knowledge to real-life applications and help students to learn and remember by building associations? Summarized below are some examples of cones from our day-to-day life that can ease out the process of identifying and familiarizing the shape.
Interesting real-life examples of a cone
Realistic examples of geometric shapes, including cones, help students understand the properties and elements better. With such examples, students learn to identify and differentiate between various geometrical shapes.
1. Eat An Ice-cream Cone!
Who doesn't love ice cream? We all scream and run for ice cream! Surely you must have eaten ice cream in a cup or of course, a cone. Who would have thought our favorite dessert would be served in a geometrical shape and that too a cone? An ice cream cone has a circular face with one vertex only. Such properties of a shape make it a cone. So, the next time you have an ice cream make sure you observe the properties of a cone!
2. Wear your Hat!
So who all are Harry Potter fans over here? Who remembers the talking hat which granted Harry's wish and entitled him to Gryffindor? Sure you all would like to wear that hat, but for now how about putting on that standard birthday party cap? Surprisingly, both hats depict a cone shape as such only have one vertex with a circular base and a continuous curve.
3. Decorate the Christmas Tree!
Ho, Ho, Ho, Merry Christmas, Santa. The Christmas tree is actually one big large cone in itself! If you observe carefully a Christmas tree, it has only one apex/vertex. If you observe mathematically, it has a circular base making it a perfect shape of a cone. A cone is a three-dimensional solid geometric shape and a Christmas tree is just the same!
4. Let's Eat Carrots!
Bugs Bunny keeps eating carrots all day long, even the rabbit and the hare keep munching on the carrot. As you keep seeing carrots in your everyday life, know that even carrots are a good example of a cone. Just like a cone, carrots have a circular base with only one vertex. It also uses two line segments that connect to a common point just like in a carrot.
5. Pass it through a Funnel
How many times have the teacher asked you to use a funnel in the science lab? It always has been fun to use the funnel, with how much ease transferring and sorting happen through the use of a funnel! A funnel is also a cone that is used in everyday life. Just like a cone, a funnel also has one circular base which is basically an open face of it along with one vertex point.
6. Watch the Traffic Cone
A traffic cone is a piece of plastic with a pointed top that is placed on the road for systematic parking. As you walk through, you must have observed a traffic cone. A traffic cone is also a geometric shape of a cone with only one vertex. Moreover, it has a circular base making it a cone.
7. Play with Playing Top
You must have seen a spinning wheel but it is indeed fun to watch a spinning top. Did you know a playing top is a perfect example of a cone? The playing top has a circular base and just one vertex joined together with two line segments. So, the next time you see a playing top, know that it is a cone.
8. Rest in a Tent
Have you seen tents shaped in different ways? If you observe carefully, you can see various tents built in a cone shape. A tent has only one point where two line segments meet from a circular face making it a perfect geometrical cone. So, the next time you think of buying a tent, buy a cone-shaped one!
9. Hear through a Megaphone
Sure you have heard so many announcements in school and public areas. Carefully observe a Megaphone and see how it depicts a cone shape. A cone shape follows a circular base just like the one a megaphone has. A megaphone also has two line segments that are connected to a single point. Just like a cone, a megaphone also has just one vertex.
10. Look at the Pencil Tip
Something that you use every day might also be a part of geometrical shapes. Have you ever observed the tip of a pencil? There you go, look at it carefully now, how this conical tip helps you write such long essays and solve sums. The tip of a pencil, when sharpened fully, depicts a cone shape with one vertex which is the tip, and the circular base which is the part we hold while writing.
Wrapping up
Real-life examples of geometrical figures can help students in remembering not only the shapes better but also some of their properties. While students learn a lot through drawing and watching shapes on a chart or in class, examples make it different. As students get an opportunity to view these shapes in reality, they can form a better association with them. It also helps them understand how some shapes like squares and triangles are present in everyday life along with their uses. So, teachers and educators can build a comprehensive learning environment with realistic examples | 677.169 | 1 |
Dentro del libro
Resultados 1-5 de 36
Página 7 ... joining two opposite angular points . The Complement of an angle ( complementum , that which fills up ) , is what is wanted to make an acute angle equal to a right angle , or to 90 ° . The Supplement of an angle ( supplementum , a ...
Página 11 ... joining them lies wholly in that surface . -HERON . * " A plane surface is that which lies evenly , or equally , with the straight lines in it . " - EUCLID . " A plane surface is one whose extremities hide all the intermediate parts ...
Página 12 ... join that mark to the angular point C by a st . line , the two strait lines will form the angle required . The Semicircle may be employed with a plummet suspended from D , or 90 degrees , to ascertain both the horizontal and vertical ...
Página 13 ... joining some one angular point and the other angular points . 22. Quadrilateral figures ( quadrilaterus , four sided ) , are bounded by four straight lines . The Diagonals of a quadrilateral are the lines joining the opposite angles ...
Página 15 ... joining two opposite angles : the parallelograms about the diagonals AEKG and KHDF , are those through which the diagonal passes ; and the Complements to fill up the figure , C are the two parallelograms , ECHK and GKFB , through which ...
Pasajes populares
Página 36 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Página 17 5. If equals be taken from unequals, the remainders are unequal. 6. Things which are double of the same are equal to one another.
Página 22 - Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other extremity.
Página 12 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it. | 677.169 | 1 |
Quant Question Of The Day: 182
Geometry
Rajat has two congruent sheets of paper in triangular shape. He built a parallelogram with them by three different methods. It is known that the perimeters of the three parallelograms are 32, 41 and 43. Find the perimeter of the triangular piece of paper. | 677.169 | 1 |
...the parts. Give a geometrical illustration of the identity (a — 6)2 = a2 + b* - 2ab, and show how to divide a given straight line into two parts such that the sum of the squares on the two parts may be the least possible. (20) 22. Given two points A and B, show...
...triangle, and any rectilineal 'figure. 3. (a) Describe a square equal to the sum of n squares. (6) To divide a given straight line into two parts such that the difference of the squares on the parts may be equal to a given square. (c) Point out the three cases...
...meet in E. Prove that } = {EALD}=X{KCYD}, and that XYZ is a straight line. APPENDIX. THE PENTAGON. To divide a given straight line into two parts such that the square on the greater part may be equal to the rectangle contained by the whole line and the smaller...
...area would be doubled by increasing its radius 1 in. (Take 3} for тг.) 30. Divide a line 20 in. long into two parts, such that the rectangle contained by the whole line and one part may be equal to the square on the other part. Exercise 150. Review of Chapters I-XVII Solve the...
...area would be doubled by increasing its radius 1 in. (Take 3} for тт.) 30. Divide a line 20 in. long into two parts, such that the rectangle contained by the whole line and one part may be equal to the square on the other part. Exercise 215. Review of Chapters I-XVIII Solve the...
...Describe a square that shall be equal to a given rectilineal I figure. 6. Divide a given straight line so that the rectangle contained by the whole line and one of its parts may be equal to the square on the other part. 7. I Given the base, altitude, and the radius of the...
...the mean proportional between two lines or two numbers. He also seems to have been able to divide a line into two parts such that the rectangle contained by the whole line and one of the parts is equal to the square on the other part. Expressed algebraically, if a is the length of...
...this can be done by employing the properties of similar triangles. CONSTRUCTION OF REGULAR PENTAGON. To divide a given straight line into two parts such that the square on the greater part may be equal to the rectangle contained by the whole line and the smaller...
...external segment of the base made by a straight line bisecting the external angle at the vertex 1 4. Divide a given straight line into two parts, such that the rectangle under the whole and one of them may have a given ratio to the square of the other. 5. Show that, if...
...rigorously that if two squares are of equal area, their sides are equal and conversely. 3. Divide a straight line into two parts such that the rectangle contained by the whole line and one of the parts shall be equal to the square on the other part. Divide a given straight line into two parts... | 677.169 | 1 |
Table of Contents The Perpendicular Distance of a Point from a Line What is Perpendicular Distance? Calculating Perpendicular Distance Method 1: Using the Formula Method 2: Using Vector Calculus Real-World Applications 1. Architecture and Construction 2. Physics and Engineering 3. Computer Graphics When it comes to geometry, understanding the relationship between points and lines is… | 677.169 | 1 |
Circles Circular Logic and choose the correct answers to the following questions on circles logic. And learn more about this maths parts quiz. All the best!
Questions and Answers
1.
What do you call a line that connects 2 points on the circumference, that is not a diameter?
A.
Diameter
B.
Tangency line
C.
Chord
Correct Answer C. Chord
Explanation A chord is a line that connects two points on the circumference of a circle, but it is not a diameter. Unlike a diameter, which passes through the center of the circle, a chord can be located anywhere on the circumference. Therefore, a chord is the correct term for a line that connects two points on the circumference of a circle, excluding the diameter.
Rate this question:
3
2
2.
What do you call the point where the tangency line meets the circumference?
A.
Point of tangency
B.
Tangency point
C.
None of the above
Correct Answer A. Point of tangency
Explanation The point where the tangency line meets the circumference is referred to as the point of tangency. This is the exact location where the line touches the circumference of the circle. It is called the point of tangency because it is the point at which the line is tangent to the circle, meaning it touches the circle at only one point without intersecting it.
Rate this question:
1
3
3.
If the radius of a circle is 10 what is the diameter?
A.
21
B.
20
C.
5
Correct Answer B. 20
Explanation The diameter of a circle is twice the length of its radius. Therefore, if the radius of a circle is 10, the diameter would be 10 multiplied by 2, which equals 20.
Rate this question:
2
2
4.
If the diameter of a circle is 80 what is the radius?
A.
40
B.
160
C.
180
Correct Answer A. 40
Explanation The diameter of a circle is twice the length of its radius. Therefore, if the diameter is 80, the radius would be half of that, which is 40.
Rate this question:
2
1
5.
There can be 2 points of tangency
A.
True
B.
False
Correct Answer B. False
Explanation The statement "There can be 2 points of tangency" is false. In geometry, a point of tangency is a point where a line or curve touches a circle or another curve at a single point. Two points of tangency would imply that the line or curve intersects the circle or curve at two different points, which contradicts the definition of tangency. Therefore, there can only be one point of tangency between a line or curve and a circle or another curve.
Rate this question:
1
2
6.
There can be more then one chord
A.
True
B.
False
Correct Answer A. True
Explanation The statement suggests that it is possible to have more than one chord. A chord is a line segment that connects two points on a curve. Since a curve can have multiple points, it is logical to conclude that there can be more than one chord. Therefore, the answer is true.
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2
0
7.
A diameter is a chord
A.
True
B.
False
Correct Answer A. True
Explanation A diameter is a chord because a chord is a line segment that connects two points on a circle. A diameter is a special type of chord that passes through the center of the circle, dividing it into two equal halves. Therefore, every diameter is a chord, but not every chord is a diameter.
Rate this question:
2
1
8.
If a radius is 8 the diameter is 4
A.
True
B.
False
Correct Answer B. False
Explanation The statement "If a radius is 8 the diameter is 4" is incorrect. The diameter of a circle is always twice the length of its radius. Therefore, if the radius is 8, the diameter would be 16, not 4. | 677.169 | 1 |
Plane Geometry
From inside the book
Results 1-5 of 30
Page 6 ... point , forming equal angles , find one angle ( a ) in degrees , ( b ) in right angles , and ( c ) in straight angles . 13. What kind of an angle is less than its supplement ? equal to its supplement ? greater than its supplement ? 14 ...
Page 26 ... point without a line there can be only one perpendicular to that line . 99. COR . 6. Each angle of an equiangular ... Find each angle of a triangle if the second equals twice the first , and the third equals three times the first . Ex . 85. | 677.169 | 1 |
right triangle
How To Use right triangle In A Sentence
The virtue of this conception is that it explains how two or more people can be said to be thinking of the same abstract object, such as the number ˜two,™ and how various properties, such as the Pythagorean Theorem, can be said to follow logically from the idea of a right triangle.
Malebranche's Theory of Ideas and Vision in God
A Pythagoras tiling covers the plane with periodic copies of the squares on the sides of the right triangle.
He sees the hypotenuse of a right triangle made by the deck, the mast and the front edge of the beam moving down to the deck.
He imagined how a diagonal line cutting across a square would divide the square into two right triangles.
When my brain was filled with verb tenses, right triangles and pulmonary veins, there was no room for personal thoughts.
The base of the wedge is an isosceles right triangle in a vertical plane.
Of indeterminate age - at least 150 but maybe much older, say, dating back to Pythagoras and his right triangles - the 7 tangram pieces (credited by some sources to the Chinese) can form countless different designs.
Drag to add a right triangle with an adjustable hypotenuse and numerically adjustable angle.
Consider, for example, the Pythagorean theorem that the square on the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides. | 677.169 | 1 |
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In an elegant half page paper, Burk (1985) proves the following order of sample means: harmonic mean<geometricmean-<arithmetic mean<root mean square (see appendix for definition of these sam-ple means).
We show that our definition is the only one in the literature.
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For example: For the given two numbers: 20 and 25.
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1 Geometric Mean Name _____ 1) If an altitude is drawn to the hypotenuse of triangle BAN below, then name and redraw the 3 similar triangles created. | 677.169 | 1 |
Balbharati Solutions Class 5 Mathematics Angles
Welcome to NCTB Solutions. Here with this post we are going to help 5th class students for the Solutions of Balbharati Class 5 Math Book, Problem Set 24, 25, 26 and 27, Angles. Here students can easily find step by step solutions of all the problems for Angles. Also our Expert Mathematic Teacher's solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Angles solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.
Angles all Question Solutions :
Problem Set 24 :
Question no – (1)
Solution :
Diagram – (i)
In this diagram two lines PQ and QR are arms of the angle
In PQ and QR the common point is Q is called the vertex
Write the common point in middle in angle name
The name of angle is ∠PQR Or ∠RQP
Here, ∠ represents the angle
Diagram – (ii)
In this diagram two lines LM and MN are arms of the angle
In LM and MN the common point is M is called the vertex
Write the common point in middle in angle name
The name of angle is ∠LMN Or ∠NLM
Here, ∠ represents the angle
Diagram – (iii)
In this diagram two lines SU and UT are arms of the angle
In SU and UT the common point is M is called the vertex
Write the common point in middle in angle name
The name of angle is ∠SUT Or ∠TUS
Here, ∠ represents the angle.
Problem Set 25 :
Question no – (1)
Solution :
Figure – (i no at that point is angle between this two arms
Figure – (iiFigure – (iii
Figure – (ivProblem Set 26 :
Question no – (1)
Solution :
(1) The steps to draw the angle 60 60 on protractor and mark in paper at that point
Using the ruler join the mark with vertex
Give the name of that line is QR it means that Q is vertex
In this way we draw the PQR of 60°
(2) The steps to measure the angle 90 90 on protractor and mark in paper at that point
Using the ruler join the mark with vertex
Give the name of that line is QR it means that Q is vertex
In this way we draw the PQR of 90°
(3) The steps to measure the angle 150 150 on protractor and mark in paper at that point
Using the ruler join the mark with vertex
Give the name of that line is QR it means that Q is vertex
In this way we draw the PQR of 150°
(4) The steps to measure the angle 30 30 on protractor and mark in paper at that point
Using the ruler join the mark with vertex
Give the name of that line is QR it means that Q is vertex
In this way we draw the PQR of 30°
(5) The steps to measure the angle 165 165 on protractor and mark in paper at that point
Using the ruler join the mark with vertex
Give the name of that line is QR it means that Q is vertex
In this way we draw the PQR of 165°
(6) The steps to measure the angle 45 45 on protractor and mark in paper at that point
Using the ruler join the mark with vertex
Give the name of that line is QR it means that Q is vertex
In this way we draw the PQR of 45°
Problem Set 27 :
Question no – (1)
Solution :
The example of parallel line see in environment are –
(1) We know that parallel lines do not intersect the railway track lines are not intersect they are parallel to each other
(2) We know that parallel lines do not intersect the zebra crossing on the roads are not intersect they are parallel to each other.
Question no – (2)
Solution :
The example of perpendicular line see in environment are
(1) We know that perpendicular lines are intersect each other the addition symbol lines are intersect.
(2) We know that perpendicular lines are intersect each other the two arms of the scissors are intersect each other.
Question no – (3)
Solution :
In the above picture two lines do not intersect each other
So, they are parallel to each other.
In the above picture two lines are intersect each other
Hence, they are perpendicular to each other.
In the above picture two lines do not intersect each other
So, they are parallel to each other
In the above picture two lines do not intersect each other
Thus, they are parallel to each other | 677.169 | 1 |
The ancient Egyptians knew the $3-4-5$ triangle was a right triangle, but they did not possess the Pythagorean theorem or any equivalent theory. Can it be shown that the $3-4-5$ triangle is a right triangle without using the Pythagorean theorem or any ideas related to it?
This problem was shown to me by a fellow peer tutor a while ago. I gave it some thought initially, but then I gave up trying when I realized it might not be doable. What do you think?
$\begingroup$I can think of many ways to prove that the $3,4,5$ triangle has a right angle without assuming the Pythagorean Theorem as a given, but the obvious ones could all be converted into proofs of the Pythagorean Theorem by replacing the constants $3$ and $4$ by arbitrary parameters $a$ and $b.$ The wording of the question seems to imply that this is not acceptable (supposing that any proof of the theorem would constitute "ideas related to it"). In other words, it seems you want a geometric proof, but one that can't easily be generalized to other right triangles.$\endgroup$
3 Answers
3
Take a stick length $3$ and another length $4$. Place these at right-angles to each other. Create a stick to measure the diagonal. Demonstrate that this stick plus the $3$-stick is the same length as two $4$-sticks. (You can create a $3$-stick from a $4$-stick by bisecting a $4$-stick, appending, and bisecting again).
$\begingroup$@CameronWilliams For the purposes of the ancient Egyptians, "so close that our regular measuring instruments cannot distinguish them" was the best they could hope for anyway, since they were laying out right angles with knotted cords. It seems plausible that they found this as an empirical observation, just as suggested in this answer, and then decided to use it.$\endgroup$
This is a proof (without using Pythagoras' theorem) of the following: If a right-angled triangle has a hypotenuse of length 5 and one leg of length 3, the remaining leg is of length 4, i.e. it must be a 3-4-5 triangle.
Given that the sides of a triangle uniquely determine its angles we should conclude from the above that a 3-4-5 triangle has to be right-angled.
Start by drawing a 5-5-6 rectangle with one of the length-5 sides as its base. Extend the base by a line of length 5:
Draw a circle with the three lines of length 5 as radii:
Draw an altitude of the original triangle to the side of length 6. Also, join the top of the triangle to the end of extended line of length 5:
If we call the equal angles in the isosceles 5-5-6 triangle $\theta$ then we can calculate some of the other angles. In particular the angle between the altitude drawn earlier and the base is $\pi/2 - \theta$ and this equals the angle between the line joining the top of the triangle and the extended line.
So the altitude is parallel to the line joining the top to the right extension. Also, by similar triangles, if the altitude has length $x$ then the second line has length $2x$
If we extend the other radius of length 5 to make it a diameter of the circle, we can complete a cyclic quadrilateral with the edges shown and with its diagonals both being diameters of the circle
Ptolemy's theorem says that, for a cyclic quadrilateral, the
product of the two diagonal lengths equals the sum of the products of opposite sides. In our case, this means that
On a Cartesian coordinate plane start with $O=(0,0), A=(2,0), B=(0,1)$. Construct lines parallel to $OA$ through $B$ and parallel to $OB$ through $A$, which intersect at $C=(2,1)$. Draw $OC$ and construct the perpendicular to it through $C$, which intersects line $OB$ at $D$. Triangle $OCD$ is similar to triangle $OBC$ so $|BD|/|BC|=|BC|/|OB|=2$, then $|BD|=4, |OD|=1+4=5$.
Extend $DC$ through $C$ to point $E$ such that $EC$ is congruent to $CD$. Extend $BC$ through $C$ to point $F$ such that $CF$ is congruent to $BC$. Draw $EF$ which intersects line $OA$ at $G$. Vertical angles $BCD, FCE$ are congruent, therefore triangles $BCD, FCE$ are also congruent by SAS. So $|FE|=|BD|=4$ and $\angle CFE$ is a right angle like $\angle CBD$. The latter fact makes $FG$ parallel to $BO$ and we have by earlier construction $BE$ parallel to $OA$. Quadrilateral $OACB$ is then a parallelogram with opposite sides congruent, thus
$|OG|=|BF|=|BC|+|CF|=2+2=4$
$|GF|=|OA|=1, |GE|=|FE|-|GF|=4-1=3$
Now draw the $OE$ to complete triangle $OEF$. $\angle OCE$ is supplementary to $\angle OCD$ which is a right angle, so these angles are congruent making triangles $OCD, OCE$ congruent by SAS. Then $OE$ is congruent to the corresponding side $OD$ and the hypoteneuse $OE$ of right triangle $OEG$ with legs measuring $3$ and $4$ units satisfies $|OF|=|OD|=5$. QED. | 677.169 | 1 |
Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry
05/23/2021 08/25/2022 / 7 minutes of reading
Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry
Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry are created for class 9 CBSE students to help them in doing homework and in preparation for the forthcoming exam. Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry will clear the basic concept of coordinate geometry since chapter 3 of class 9 maths is based on the introduction of the coordinate geometry which is required to solve the questions of coordinate geometry in higher classes. Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry are explained here beautifully with the help of proper diagrams.
Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry
Exercise 3.1
Q1.How will you describe the position of a table lamp on your study table to another person?
Ans. Let the table is in the shape of a rectangle.
Drawing two line of symmetry perpendicular to each other such that both the lines intersect at the centre of the table.
The horizontal line is supposed as XX' and vertical line as YY'
The position of the table lamp is 'a' unit distance from YY' and 'b' unit distance from XX',the position of the table is represented by (a,b) with respect to the position of the centre of the table (0,0)
Q2.Street plan: A city has two main roads which cross each other at the centre of the city . These two roads are along the north-south direction and east-west direction. All other streets of the city-run parallel to these roads and are 200 m apart. There are 5 streets in each direction.Using 1 cm = 200m.Draw a modal of the city on your notebook. Represent the roads/streets by single lines.
There are many cross streets in your modal. A particular cross street is made by two streets, one running in the north-south direction and another in an east-west direction. Each cross street is referred to in the following manner. If the second street running in the north-south direction and 5 th in the east-west direction meet at some crossing, then we will call this cross street. Using this convention, find
(i) How many cross streets can be referred to as (4,3)
(ii) How many cross streets can be referred to as (3,4)
Ans. Drawing one horizontal line XX' and Vertical line YY' as main roads crossing each other at the centre (0,0), thereafter drawing 5 lines parallel to XX' and YY' showing the 5 streets crossing each other.
(i) There is only one cross streets can be represented by the unique point A(4,3)
(ii) There is only one cross streets can be represented by the unique point B(3,4)
Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry
Exercise 3.2
Q1. Write the answer to each of the following questions.
(i) What is the name of the horizontal and the vertical line is drawn to determine the position of any point in the cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
Ans. (i) The name of the horizontal line is X-axis and the name of the vertical line is Y-axis
(ii) Each part of the plane formed by the horizontal line X-axis and the vertical line Y-axis are known as quadrants.
(iii) The name of the point where these two lines intersect is the origin
Q2.See the given figure and write the following:
(i) The coordinates of B
(ii) The coordinates of C
(iii)The point identified by the coordinates (-3,-5)
(iv)The point identified by the coordinates (2,-4)
(v) The abscissa of point D
(vi) The ordinate of point H
(vii) The ordinate of point L
(viii) The ordinate of point M
Ans. (i) The coordinates of B is (-5,2)
(ii) The coordinates of C is (5,-5)
(iii)The point identified by the coordinates (-3,-5) is E
(iv)The point identified by the coordinates (2,-4) is G
(v) The abscissa of point D is 6
(vi) The ordinate of point H is -3
(vii) The coordinates of point L is (5,0)
(viii) The coordinates of point M is (-3,0)
Solutions of class 9 NCERT maths chapter 3 Coordinate Geometry
Exercise 3.3
Q1.In which quadrant or on which axis do each of the points (-2,4),(3,-1),(-1,0),(1,2) and (-3,-5) lie? Verify your answer by locating them on the cartesian plane.
Ans.(-2,4) lies on second quadrant
(3,-1) lies on the fourth quadrant
(-1,0) lies on X-axis
(1,2) lies on the first quadrant
(-3,-5) lies on third quadrant
Q2.Plot the points (x,y) given in the following table on the plane,choosing suitable units of distance on the axis.
x
-2
-1
0
1
2
y
8
7
-1.25
3
-1
Ans. We are given the coordinates (-2,8),(-1,7),(0,-1.25), (1,3),(2,-1)
Drawing XX' and YY' lines perpendicular to each other and intersect at the point O known as origin.
Taking 1 unit = 1 big square
Locating the points as following
(-2,8) : lies at the distance of 2 units from the left of the origin and 8 unit above the X-axis
(-1,7): lies at the distance of 1 unit from the left of the origin and 7 unit above the X-axis
(0,-1.25): lies at the distance of 1.25 unit below the origin on the Y-axis
(1,3): lies at the distance of 1 unit left of the origin and 3 units above the X-axis | 677.169 | 1 |
Activity Time
Device
Software
TI-Nspire Version
Accessories
Transversals
Activity Overview
Students will explore corresponding, alternate interior, and same-side interior angles. This is an introductory activity where students will need to know how to change between pages and grab points.
Key Steps
Prior to this activity, students should know the definition of corresponding, alternate interior and same-side interior angles. The figure that is shown on page 1.3 will also be used on pages 2.1, 3.1, and 4.1 | 677.169 | 1 |
Thank you for reaching out for assistance with triangulation using Excel. I can definitely help you with this.
To determine the bearing and distance from the target to the new spot, you will need to use trigonometry. Specifically, you will need to use the law of sines and the law of cosines.
Here are the steps to follow:
Calculate the length of the base of the new triangle (the distance between the two observation towers). You can use the law of cosines for this. Let's call this length "b".
Calculate the angles at the target. You already know that one angle is 70 degrees. To find the other two angles, you can use the law of sines. Let's call the angle opposite the side of length "b" "A", and the angle opposite the side of length "40" "B". Then:
Formula:
sin(A)/b = sin(70)/60 sin(B)/40 = sin(70)/60
Solve for A and B.
Calculate the length of the side opposite angle A (the distance from the target to the new spot). You can use the law of sines again:
Formula:
sin(A)/x = sin(70)/40
Solve for x.
To find the bearing, you will need to use the angles you calculated in step 2. The bearing is the angle between due north and the line connecting the target to the new spot. Let's call this angle "C". Then:
Formula:
C = 90 - A
If the new spot is due north of the target, then the bearing is 360 degrees or 0 degrees.
Let me know if you have any questions or if there's anything else I can assist you with. | 677.169 | 1 |
Fill in the blank In complementary angle one angle is 48∘, then the other angle is………………….Solution in Punjabi
Video Solution
Text Solution
Verified by Experts
The correct Answer is:42∘
|
Answer
Step by step video, text & image solution for Fill in the blank In complementary angle one angle is 48^(@), then the other angle is…………………. by Maths experts to help you in doubts & scoring excellent marks in Class 7 exams. | 677.169 | 1 |
45-45-90 And 30-60-90 Triangles Worksheets
Students who study our 45-45-90 and 30-60-90 triangles worksheets will learn about special right triangles and be able to solve related problems.
30-60-90 And 45-45-90 Triangles Worksheet PDF
30-60-90 Triangles And 45-45-90 Triangles Worksheet
45-45-90 And 30-60-90 Triangles Worksheet
30 60 90 And 45 45 90 Triangle Worksheet
45 45 90 Triangle And 30 60 90 Triangle Worksheet
30-60-90 And 45-45-90 Triangles Worksheet
What is 45-45-90 triangle?
What is 30-60-90 triangle?
Meanwhile, a 30-60-90 triangle is a right triangle with angles that are 30 degrees, 60 degrees, and 90 degrees. The ratio of the sides in a 30-60-90 triangle is always 1:sqrt(3):2. This means that if the length of the shorter leg (the side that is opposite the 30 degree angle) is "x," then the length of the longer leg (the side that is opposite the 90 degree angle) is "x*sqrt(3)" and the hypotenuse is "2x".
Both of these triangles are special because of their unique side-angle relationships. This make them useful in solving geometric problems and in applied math like trigonometry. You can learn more from our 30-60-90 and 45-45-90 triangle worksheet.
45-45-90 And 30-60-90 Triangles Worksheets PDF
45-45-90 And 30-60-90 Triangles Worksheets PDF
Our 45-45-90 and 30-60-90 triangle worksheet is free to download and is designed to make the concept of special right triangles easy for your kid to understand. You can also download our 30 60 90 and 45 45 90 triangle worksheet with answers. The 45-45-90 and 30-60-90 triangles worksheet answer key will help you kids to solve problems related to special right triangles grasp geometry basics geometry lessons? An online tutor could provide the necessary help.
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Multiplying Fractions Worksheets
Many teachers agree that the process of teaching fractions is complicated and the topic may seem confusing to students. But kids can't skip it, especially the fractions multiplication part. Some schools are pushing kids to learn this topic without empowering teachers. Luckily, you can use tools like a multiplying fractions worksheet in PDF format to […]
Identify Coins Worksheets
Money is an integral element of our daily lives. Thus, it's important to introduce kids to the concept of money and teach them how to identify coins. However, this isn't always an easy feat to pull off as the sheer amount of different denominations can be confusing. This is where our values of coins | 677.169 | 1 |
θ is angle between pair of lines , then
10
20
5
15
Hint:
First find the coefficient of y square and then obtain the angle between the pair of straight lines.
The correct answer is: 10
Given That: If θ is angle between pair of lines , then >>> We first need to find the value of >>> Since, it is a pair of straight lines. Then, the determinant should be equal to zero. >>> >>> This gives = 2. >>> Therefore, now the angle between the pair of straight lines becomes: tan = tan = >> We know that | 677.169 | 1 |
Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length $24 \mathrm{km}$. When the slower ship travelled $8 \mathrm{km}$, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in $\mathrm{km}$, between the other ship and the port will be | 677.169 | 1 |
If a polytope is circumscribable, the center of its circumsphere can be said to be the polytope's center. The same is true for polytopes that can have spheres inscribed into them, and the centers of said inspheres.
For any definition of center that is preserved by rotation, reflection and scaling, the center of a polytope must be preserved by the polytope's symmetries. If there is only one point that is preserved by a polytope's symmetries (as in all isogonal polytopes), that point is the polytope's center.
In computer graphics and especially geographic information systems, a common problem concerns finding a "visual center" of a polygon that is neither regular nor convex (but has a defined interior). The centroid is unsuitable for this; a polygon with strong concave elements, such as one shaped like the letter "C," may not contain its own centroid, and a center outside the polygon is not considered visually salient.
One solution is to find a circle of largest possible radius that is completely contained in the polygon (incircle), and take the circle's center as the "visual center." Such a circle is not necessarily unique but often yields pleasing solutions.[1] | 677.169 | 1 |
Unit 3Lesson 6
Lesson 6Claims and ConjecturesSolidify Understanding
Ready
Conjectures are statements which are believed to be true based on the current data or evidence. However, they have not yet been proven or disproven. To move toward logical proof, true statements that properly build on one another, along with justifications for those statements, are needed. Based on the diagram in Figure 1, provide a justification for each statement.
a venn diagram with 4 right triangles in the centerFigure 1
1.
∠AGB,∠CGB,∠CGD, and ∠AGD are right angles.
2.
AG―≅BG―≅CG―≅DG―
3.
△ABG≅△CBG≅△ADG≅△CDG
4.
AB―≅BC―≅CD―≅DA―
5.
Sometimes when creating a proof, it is useful to add additional segments to a diagram. Amelia is conjecturing EB―≅FD―. To prove this conjecture, Amelia thinks she will need to add to the diagram in Figure 1. What would you add to the diagram? How might this addition of an auxiliary segment be used in the proof?
a venn diagram with 4 right triangles in the centerFigure 1
Set
Based on the information in the statements or diagrams it is possible to make claims about what is likely true. The claims made are considered conjectures because they have not yet been proven or disproven. For each problem, use the diagram and statements to generate at least one conjecture about the quadrilateral. Attempt to conjecture about the most precise classification for the quadrilateral.
6.
Quadrilateral TRAV with diagonal RA
Given: △TRA≅△VAR
Conjecture:
7.
Quadrilateral ABDE with Diagonals EB and AD intersecting at Point C.
Given: △ABC≅△DEC
Conjecture:
8.
Quadrilateral ARHC with diagonals RC and AH intersecting at Point E triangles.
Conjecture:
9.
Quadrilateral ABCD with diagonal BC. Line segment and BC and CD are marked with two tics. Line segment AC and AB are marked with one tic.
Given: △ABC and △DCB are isosceles triangles.
Conjecture:
10.
Quadrilateral ACRH with diagonals AH and RC that intersect at Point E. Line segments AE, EC, EH, and ER are marked with one tic. angle AEC and REH are marked with as right angles right triangles.
Conjecture:
Go
Write the equation of the line with the given slope and passing through the given point. | 677.169 | 1 |
11In one step, the existing element enlarges and a new element appears inside this element. In the next step, the outer element is lost.
Explanation: Figure 1 has a circle, figure 2 shows a triangle inside a circle and figure 3 shows a triangle
Therefore we could conclude that the circle shows the way to the next pattern which is a triangle
we could come to an option that the answer has an outer square - None of the shapes in the questions should be included in the answer
Answer (1) has q square inside the outer square,so not possible since question already has a square. Answer (2) does not have an outer square. Answer (3) has a triangle inside the outer square, not possible since question already has a triangle. Answer (4) it is correct, because it has a pentagon representing the next order, and it's not fond in the question and has an outer square.
Parthipan said:
6 years ago
It should be E only, because Triangle is having 3 sides which comprises of square (4 sides) in side it in Dth position of the question. Similarly, Square having 4 sides will be comprising of Pentagon (5 sides). So, answer is D only.
Prabh said:
6 years ago
Why it is not E?
(1)
Prabh said:
6 years ago
Why it is not E?
(1)
Ganesh said:
8 years ago
We should consider the inner figure also I feel. Option D has a pentagon but it is inverted. That's why I feel it should not be the best option to ahead with but E.
Also if we consider number of times a figure is repeated in the series then both triangle and square are repeated 3 times but the circle only two times. So E is best suitable option in this case.
Bikash said:
9 years ago
D is the correct answer because the number of sides of the inner shapes are increasing by +1 accordingly.
Ron said:
10 years ago
But if it is seen as a circular loop then E might be the write answer, Otherwise D is fine. @Jack's logic is reasonable.
Jack said:
1 decade ago
First you have a triangle (3 sides), then a square (4 sides) so the next figure should be a pentagon (5 sides).
Divya said:
1 decade ago
Hello rohit
first if we take A as the answer square inside square. if we take C triangle is repeated before, if we take E circle is repeated before so the best one is D | 677.169 | 1 |
Triangle Area - Basenji
4159
Triangle Building - Coworking Space in Denver WeWork
For three decades the gospel of kitchen design has been found in basic geometry. For maximum efficiency and ease of movement, draw an imaginary triangle from the center of
A triangle has zero diagonals. Diagonals must be created across vertices in a polygon, but the vertices must not be adjacent to one another. A triangle has A triangle has zero diagonals.
Try this Drag the orange dots on each vertex to reshape the triangle. The formula shown will recalculate the area using this method. How To Find The Area of a Triangle. View Course Next Lesson . Instructor: Malcolm M. Malcolm has a Master's Degree in education and holds four teaching certificates.
"Mi piace": 2775, commenti: 309 - Mathsmania - Pinterest
Bermuda triangle or Devil's triangle has been a mysterious area for long time.There are many questions and now we want to find out a little about it. Click here to get an answer to your question ✍️ 22 for den (23) sin( tam 's) ola. If R is the circumradius of a triangle ABC then area of its pedal triangle is. The Swedish School in the Triangle is a very active organization offering Swedish studies in the Triangle, NC area for students in 1st through 5th grade.
1)The area A of a triangle varies jointly as the lengths of its base b and height h. Triangle, is plane-sided or which looks like a pyramid, have three sides and three angles in whatever length or angle it may be, but sum-it-up to 180 degrees. What is the formula to find the calculate the area of a triangle? The formula is varied for different types of triangle, but the most common formula that was used as (Height X Base /2)
The area formulas for all the different types of triangles like an area of an equilateral triangle, right-angled triangle, and isosceles triangle are given below. Area of a Right Angled Triangle: A right-angled triangle, also called a right triangle, has one angle at 90° and the other two acute angles sums to 90°. Teknisk linje gymnasium
Click here👆to get an answer to your question ️ Each side of an equilateral triangle measures 8cm . Find(i) the area of the triangle, correct to 2 places of decimal and(ii) the height of the triangle, correct to 2 places of decimal. (Take √(3) = 1.732 )
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Area of Triangles. There are several ways to find the area of a triangle.
It is simply half of b times h. Area = 12 bh (The Triangles page explains more) The most important thing is that the base and height are at right angles. Have a play here:
Area of a triangle The formula for the area of a triangle is side x height, as shown in the graph below: There are different starting measurements from which one can solve a triangle, calculate the length of a side and height to it, and finally calculate a triangle's area. There are 4 common rules for solving a triangle, as explained below. Lars karlsson konstnär nyköping | 677.169 | 1 |
Maths - 1D Euclidean Space - Outer Product
Here we will use the definition of outer product as the values in the geometric product table such that : if gradeof(row)+gradeof(column) != gradeof(entry) then the entry is set to zero otherwise the entry is the same as the geometric product. Where a table entry is the sum of terms then this check is done on each term separately. For more information about the concept of an outer product see this page.
I can think of two ways of calculating the outer product for this null vector basis:
Generate the geometric table for the null basis (as explained on this page), then set terms to zero depending on the sum of the grades as explained above (or equivitently set term to zero if operands contain a common basis term).
Start with the outer product of e1,e2 and e3 then rotate the basis vectors to give null basis vectors.
I'm not sure if these methods are equivalent? So we will use the first method but do some tests to see if the second method is equivalent.
Meet
Two Points
In the above example if x1=x2 (and only if x1=x2) then:
scalar = 0
n0∞ = 0
n01 = 0
n∞1 = 0
So this gives us a way to test if x1 and x2 represent the same point since, if they do then,
p^q=0
This may not be specially useful in one dimensional Euclidean space but when we go to higher dimensions it will allow us to test if, a point lies on a line or a plane, or two lines intersect for examplePrerequisites | 677.169 | 1 |
You can avail free PDF of RD Sharma Solutions for Class 8 Maths Chapter 18 - Practical Geometry which is solved by expert Mathematics teachers of Vedantu, available for download on its website and mobile application. The PDFs contain all the Chapter 18 Practical Geometry exercise questions with solutions to help you revise complete syllabus and score more marks in the examination. You can also register for online coaching for IIT JEE (Mains & Advanced), NEET, Engineering and Medical entrance exams on Vedantu.
Register online for Class 8 Science tuition on Vedantu to score more marks in CBSE board examinations. EveryNCERT Solution is available on Vedantu to make your study simple and interesting. Vedantu is a leading learning virtual platform in India that provides you free reading materials like the PDF ofNCERT Solutions for Class 8 Mathssolved by expert teachers as per NCERT (CBSE) book guidelines. The PDFs contain all chapter wise questions with solutions to help you revise the complete syllabus and score more marks in your board examinations.
Class 8 RD Sharma Textbook Solutions Chapter 18 - Practical Geometry
In this Chapter 18 - Practical Geometry, several exercise questions with solutions for RD Sharma Class 8 Maths are given to help the students and understand the concepts better.
We have provided step by step solutions for all exercise questions given in the pdf of Class 8 RD Sharma Chapter 18 - Practical Geometry. All the exercise questions with solutions in Chapter 18 - Practical Geometry are given below:
At Vedantu students can also get Class 8 Maths Revision Notes, formulas and list of important questions and they can also refer to the complete syllabus for Class 8 Maths, sample papers and previous year question papers all in one place, to prepare for their exams to score higher marks.
R.D. Sharma Maths Solution for Class 8 Chapter 18 Practical Geometry
Practical Geometry is the 18th chapter in the R.D. Sharma Maths book for Class 8 and it provides in-depth knowledge of geometry to the students and makes it easier for the young minds to understand geometry in such a way that it piques their interest. Before quadrilaterals, students learned about the construction of triangles and it is just like that but with the addition of one more side and subsequent angle and diagonals.
The chapter is about the construction of quadrilaterals; how to construct them through the different elements given. To make it easier for students of Class 8, Vedantu provides solutions for each exercise and each class if the students get stuck somewhere while solving the questions. Class 8 is the stage where students start learning and adapting new things and this construction work would be interesting other than something boring as it requires them to draw but on a slight senior level.
Construction of Quadrilaterals
Four sides, four angles and two diagonals make up a quadrilateral. In the exercises afterward, the chapter teaches about how to construct them if one or more than one of these elements are missing. Various situations of constructing the quadrilaterals according to the given elements are listed below:
Constructing a quadrilateral is, anyways, not a big task but when the five measures of a quadrilateral are specified, it is fairly simple to create one.
The four sides' lengths, as well as the diagonal length, are known.
The three sides' lengths, as well as the lengths of the two diagonals, are known.
If you know the three angles and two adjacent sides, you can solve the problem.
If you have three sides and two angles, you can make a triangle.
2. When can a unique quadrilateral be constructed?
We need at least five measurements of sides and angles to draw a unique quadrilateral. However, having the dimensions of any five combinations of sides and angles does not guarantee that we will receive a unique quadrilateral.
1. If we are given the measurements of four sides and one diagonal of a quadrilateral, we can design a unique quadrilateral.
2. If we are given the measurements of two diagonals and three angles of a quadrilateral, we will not be able to design a unique quadrilateral.
3. Is it important to solve every question of the exercises?
Yes, it is very important to solve all the questions of all the exercises because questions are never the same; there might be one or more than one change(s) in the question and any of such questions can come in your exam so you must have already solved it beforehand to solve in the examination as well. Apart from this, it also enhances your problem-solving ability and the practice you will be doing of solving every question will help you perform better in your exams as practice is everything.
4. What are some key points to remember for the chapter Practical Geometry?
There are many things that one should keep in mind while constructing quadrilaterals, a few of them are listed below:
It is necessary to know at least five independent elements to draw the quadrilateral properly.
With enough data (other than five simple situations), a quadrilateral can be constructed with less than five pieces but some other relationships between them.
Drawing a basic sketch of the quadrilateral and indicating the data on it is convenient and helpful in all circumstances.
5. How to construct a rhombus?
The rhombus itself is a quadrilateral and can be constructed by following some simple steps. To construct a rhombus, the steps are given below.
Draw a perpendicular bisector of the diagonal base and a diagonal of a given length.
Cut off arcs on both sides of the perpendicular bisector with half of the supplied measurement of the second diagonal. It will yield two rhombus points.
Connect these points to the first diagonal points. It will produce the rhombus that is necessary.
6 the play store. | 677.169 | 1 |
geometry in real life conclusion
Gaussian surfaces are imaginary three-dimensional shapes generated to make it easier to calculate electric and magnetic flux through an area. turned the study of geometry into an axiomatic form at around 3rd century Roleof geometry in the daily life is the foundation of physical mathematics. The first thing you probably see is your computer screen, which is rectangular. The ancient Egyptians and Mayans made pyramids made of stone and modern art includes a pyramid at the Louvre art museum in Paris, France. Circle-- [pic]
The same equation used to calculate the volume, or surface area of a small plastic sphere is used to determine the volume and surface area of planets moons and stars. By the evidence the ancient Greeks left behind in their amazing ruins, such as the Parthenon, it's no doubt that they had a deep knowledge and understanding of the science of geometry. All other trademarks and copyrights are the property of their respective owners. To unlock this lesson you must be a Study.com Member. Technologies such as CT scans and MRIs are used both for diagnosis and surgical just create an account. Geometry is one of the oldest sciences and is concerned with questions of shape, size and relative position of figures and with properties of space. are one of the concepts concerning length and area. Geometry is considered an important field pf study because of its applications in daily life. These are circular in nature. In addition, professions such as medicine benefit from geometric imaging. Role of geometry in daily life... ...NTRODUCTION:
Create an account to start this course today. Refrigerators are cubic prisms and the elements on stoves or ranges are rings of circles. Certain electron orbitals are spherical in shape. Find the quotient and the remainder of the division of
Geo means earth and metron means measurement. help them a lot in determining the proper style (and more importantly, optimize Since atoms and molecules make up everything that physically exists let's branch out into everyday life and see what geometric applications we can find. imaginable degree, area of Thanks to the Pythagoreans,
of material (ex. What do you see?
Sometimes a picture helps form our hypothesis or conclusion. Then these hypotheses can be tested rigorously using other methods. The use of geometric shapes makes the math much more manageable when determining electric or magnetic flux, which is the amount of electric or magnetic field that penetrates an area. To learn more, visit our Earning Credit Page. These conditions define uniquely Q and R, which means that Q and Rdo not depend on the method used to compute them. Geometry is actually called as Euclidean geometry.
flashcard sets, {{courseNav.course.topics.length}} chapters | Let's take a walk to the kitchen. Log in or sign up to add this lesson to a Custom Course. In the world , Anything made use of geometrical constraints this is important application in daily life of geometry. History of geometry
Lines -- [pic]
Geometry in everyday life deals with function or appearance. A conditional statement has two parts: hypothesis (if) and conclusion (then). In real life, geometry has a lot of practical uses, from the most basic to the most advanced phenomena in life. Pyramids:- Sphere:-
Numbers are further utilized when Descartes was able to formulate Divide the first term of the dividend by the highest term of the divisor (meaning the one with the highest power of x, which in this case is x). Geometry is the mathematics of space and shape, which is the basis of all things that exist. From It was put into practice by the ancient Greeks and continues to be used throughout the world today. 1. Read this article to know more about how one can find Geometry in Daily Life. Enrolling in a course lets you earn progress by passing quizzes and exams.
Geometry is mainly divided in to two which is plane geometry and solid geometry. Description
Matthew has a Master of Arts degree in Physics Education. Receive free lesson plans, printables, and worksheets by email: Geometry is one of the classical disciplines of math. For example, space Cylinder:-
Whether you're aware of it or not, geometry quite literally shapes our lives. Geometry is the study of points in space, lines between points and three dimensional objects. measuring lengths, areas, and volumes, and is still in use up to now. As some more professions use geometry in order to do their job properly. Geometry was recognized to be not just for mathematicians. - Definition, Formula & Examples, Properties of Shapes: Rectangles, Squares and Rhombuses, OSAT Elementary Education (CEOE) (050/051): Practice & Study Guide, Biological and Biomedical
They should know about the use of geometry in our daily or real life. of your home would fit around it. credit by exam that is accepted by over 1,500 colleges and universities. and career path that can help you find the school that's right for you. pd.4 I think your example of real-life geometry is great. For instance, For example, using the concept of perimeter, you can compute the amount b) Walk a known distance along the bank and fix another pole at C. Walk another known distance to a point D. From D, walk at right angle to the bank till the point P is... ...-------------------------------------------------
Solid geometry - It is about all kinds of three dimensional shapes like
Even the very basic concept of area but numbers will soon make its way to geometry. 3 dimensional objects such as cubes, cylinders, pyramids, and spheres can
Those are some of the more basic uses of geometry, but it doesn't end there. Both geometry and astronomy were considered in the classical world to be part of the Quadrivium, a subset of the seven liberal arts considered essential for a free citizen to master. For example, computer imaging, something that is used nowadays for creating Geometry in Everyday Life Geometry in everyday life Geometry was thoroughly organized in about 300bc, when the Greek mathematician, Euclid gathered what was known at the time; added original work of his own and arranged 465 propositions into 13 books, called Elements. | 677.169 | 1 |
CCGPS Geometry
3 - Similarity & Right Triangles
3.10 – Homework
Name: ________________________________________________ Date: _______________________
Similarity Review – Homework
1) Given BD AE , find DE and CE.
C
6
4
B
D
10
A
E
2) A model of a building has a scale of 2 in to 15 ft.
If the model is 5 in tall, how tall is the actual building?
3) In the diagram, CAT DOG. Use the diagram to find each of the following.
a) Scale factor of CAT to DOG (Simplify if necessary)
Scale Factor = _______
A
y
b) Find x and y (Show Work!)
6
81˚
35˚
x = __________
y = __________
C
8
c) Find mD = __________°
T
15
O
x
d) Find mO = __________°
e) Find the perimeter of CAT = __________
81˚
D
12
G
Find the perimeter of DOG = __________
f) What is the ratio of the perimeter of CAT to the perimeter of DOG? _____________
4) A boy who is 5 ft. tall cast a shadow that is 12 ft long. At the same time, a building
nearby cast a shadow that is 72 ft long. How tall is the building? Draw a picture!
CCGPS Geometry
3 - Similarity & Right Triangles
3.10 – Homework
5) Explain why the triangles are similar and write a similarity statement.
a) ABC~__________ by ______
b) RST~__________ by ______
b) ABC~__________ by ______
6) Determine which of the triangles (Δ DEF or ΔGHJ) is similar to ΔABC
a) Complete the Similarity Statement to ΔABC ~ Δ ____________
b) Find the Scale Factor = __________
7) Determine whether the dilation from Figure A to Figure B is a reduction or an
enlargement. Then find its scale factor and simplify if possible.
a)
b)
B
A
Reduction or enlargement?
Reduction or enlargement?
scale factor = __________
scale factor = __________ | 677.169 | 1 |
1 Answer
1
Unfortunately I don't have 50 rep to leave my answer as a comment, so I'll do it here. It's probably more appropriate to do it here anyways.
Answer
I would wager to say that some, not all, of the Oblique projections are subsets of the Dimetric projection, and that at least one is a subset of the Trimetric projection.
Premise
Descriptively, per Wikipedia's projection grid graphic, a Dimetric projection (2-angle congruency) contains a third ray that bisects the explementary angle defined by two other rays. This establishes a basis and defines the 3D space, albeit in a 2D viewing plane.
In the Dimetric example therein, the primary angle is $150°$, the explementary angle is $210°$, and the two bisected angles created from the explementary angle by the third ray are equally $105°$.
For both the Military and Cavalier examples, the primary angle is $90°$ (one of the basis planes is parallel to the viewing plane, with defining rays diagonal (Military) or parallel/perpendicular (Cavalier)), the explementary angles are both $270°$, and the bisected angles are both $135°$.
The Top-Down example might not appear to be Dimetric at first glance if you start by inspecting the bottommost $90°$ angle as the primary. However, it is indeed Dimetric if you consider the rightmost $180°$ angle as the primary. The explementary angle in this case is equal to the primary angle, $180°$, and when bisected results in two congruent, right, $90°$ angles.
Wikipedia also lists Cabinet projection as a Diagonal projection in their 3D Projection entry. The requirement given for this projection is that one of the faces of the displayed object, not necessarily a basis plane, must be parallel to the viewing plane. Besides that, it appears that the ray angles are all different, meaning that Cabinet projection can be considered a subset of Trimetric projection (0-angle congruency).
Hence, because the Oblique Cabinet projection resembles a Trimetric projection, and that the Oblique Military, Cavalier, and Top-Down projections resemble a Dimetric projection, I don't think we can claim that all Oblique projections are Trimetric (or Dimetric for that matter).
Inflection
I think the crucial choice in distinguishing conventional Oblique projections from Axonometric ones is that for an Oblique Dimetric projection at least one of the angles between the basis vectors is an integer multiple of $90°$ / $π/4$: at least $90°$ / $π/4$ and at most $2 \cdot 90°$ or $180°$ / $2 \cdot π/4$ or $\pi/2$. Axonometric Dimetric projections can have a primary angle of any value (seemingly between $\pi/4$ and $\pi/2$), where the explement is still bisected to form 2 remaining congruent angles.
For Military and Cavalier projections, the primary angle choice is right, or $π/4$. For Top-Down, the choice is straight, or $π/2$.
We could theoretically conceive of other projections where the primary angle choice is acute, $<π/4$, or reflex, $>π/2$. It doesn't appear that many sources online engage with these forms of projections, probably because they screw up shapes so much as to make them appear inconsistent with reality and from an M. C. Escher-type work, or because they screw up mathematics.
This discussion doesn't even touch on scaling of the basis vectors aside from orientation/direction, so I'm sure you could add further classifications: one class for three congruent* basis vectors, one class for two congruent basis vectors and 1 of different length, and one class for three basis vectors of different lengths.
*: In the sense of congruent line segments as described in Hilbert's first Congruence axiom. I can't seem to find a term that describes vectors of equal magnitude. Parallel and antiparallel seem to describe vectors in the same or opposite direction of each other, while collinear describes vectors lying along the same line in space, necessarily being parallel or antiparallel. Congruence in other contexts may mean equality of all components of a larger abstract object, referring to magnitudes and angles, like how equilateral refers to equal magnitudes of segments constituting a shape, but not of the segment magnitudes themselves. Equi-magnitudinal could serve as the substitute, like how some sources offer co-directional as a substitute for anti/parallel, but this likely warrants another post on English or Math Stack Exchange. | 677.169 | 1 |
MATHEMATICS- GEOMETRICAL EQUATIONS
Since old times, mathematics has been an indispensable part of various societies such as physical sciences and technological works. In recent times, mathematics has gained a good amount of significance in other fields too. This significance can be elucidated by the fact that mathematics is a study of geometry, algebra, numbers, and various shapes. One of the specific aspects of this subject includes geometry.
Geometry is the learning of structure and knowledge of different shapes and equations. Knowledge of these shapes includes the area, perimeter, and several other key points about these figures. The concept of geometry also includes the topic of circles, lines, and equations. A circle as it is known is a round plane figure whose every point on the circumference is equidistant to the center fixed point. On the other hand, the line is a straight length that can be extended from either side and consists of no width. Both of these structures have abundant topics to understand. Here, we are going to learn and expertise the basics of the equation of line and equation of a circle. Now let us discuss this concept more.
Contents
MATHEMATICS- GEOMETRICAL EQUATIONS
Equations of a circle
The equation of a circle can be defined as an algebraic expression of the circle located in the XY plane. This equation helps us in recognizing the center of the circle and the measure of the radius. The Basic equation of circle is recorded as:
(x-h)2 + (y-k)2 = r2,
where h and k are the fixed center points and r is referred to as the center.
Equations of a line
Equation of line can be elaborated as a geometric equation that expresses the length and structure of line in an algebraic expression that is easily understandable to be located in an XY plane. The standard equation of a line in two variables is:
Ax + By +C = 0.
Where, A, B, and C are constants and A, B are not equal to zero. These equations can also be converted to a linear equation in a single or three variables.
There are some other equations of circles as well except the standard equation. These algebraic expressions exhibit different situations like the equation of a circle when the center is the origin, the equation when the center is not the origin. In the same way, there is some other equation of line as well. To illustrate- the standard forms of the linear equation can be differentiated into three parts. These are slope-intercept form, intercept form, and normal form. These equations are further known as straight-line formulas.
Conclusion:
Coming to an end, this was a basic commencement about the concepts of equations of lines and circles whereas, this does include plenty of content to study, understand and learn. These are vast concepts and require needful practice and effort. It results in several queries and doubts in every student.
Cuemath is an online tutoring center that provides you with highly qualified and experienced teachers. They provide one-to-one attention to every student in personal sessions to clear all the queries. This assists in the efficient, effective, and mindful growth of students. Another advantage of these classes is that they are online. Anyone from anywhere can join these classes.
It benefits everyone as you won't have to travel and can learn everything from the comfort of your home. Moreover, they also provide coding lectures which help in the enhancement and growth of students. Hence, joining these classes can result in the positive development of the student and benefit them well. | 677.169 | 1 |
Arc Length Calculator
By
Marija Kondic
Marija Kondic
A highly motivated student of mechanical engineering looking to extend her knowledge and apply her skills in a dynamic work environment. Communicative, ambitious, and a fast learner with exceptional knowledge of English and a deep understanding of mathematical calculations.
This calculator enables the easiest and fastest calculation to the arc length problem, which is very common in geometry through schooling. Simply by downloading the CalCon calculators mobile app, the calculations of the arc length will be provided to you, anytime and anywhere. CalCon has developed many calculators and explained each topic if math is your weak spot.
What is the arc length?
It is possible to calculate the length of any curved line. Our article will be based mainly on calculating the arc length of a circle, but we will also provide you with the solution for calculating the arc length of the curved line. Firstly, let's define the arc length of a circle. The arc length of a circle is a part of its circumference. It is a distance between any two points drawn onto the perimeter, which are not parallel to the centre point. Because if they are parallel, their distance equals their diameter, and if their angle equals 360 degrees, the arc length will then be equal to the circumference. However, a better insight into this definition is showed in the picture underneath.
To find an equation for the perimeter of a circle, first, we need to define the relationship between the circumference and diameter of a circle.
\Pi =perimeter/diametar
History of the number Pi
The first document to mention the number \Pi is The Rhind Mathematical Papyrus from 1650 B.C. that estimated this number to be 3.16. More than 1000 years later, the number \Pi was mentioned in the Bible, and its value was estimated to be equal to 3. Later, around 250 BC, Archimedes had an idea of determining the circumference ratio. His method consisted of drawing a polygon inscribed within a circle and circumscribed outside a circle. The more sides of the polygon he drew, the better the calculation was. His approximation of the number \Pi is 3,1418. However, the use of the \Pi was standardised by Leonhard Euler in 1736.
Further analyses by other mathematicians led to discovering more digits that this number contains. The development of computing has given more precise calculations that reach hundreds of decimals for the number \Pi . The value of this number with an adequate number of decimals for our calculations is \Pi = 3.14159.
Arc length-formula
An arc length is directly dependent upon the central angle. Therefore, the formula for the arc length could be calculated in radians and degrees, according to the unit of the central angle. To ease our data manipulation, we will use the following letters to mark the values used in equations:
L- arc length
C-circumference
r-radius
\theta -central angle
A-area of a sector of a circle
To get a formula for an arc length, we will start with the proportion among the arc length and central angle:
\frac{L}{\theta}=\frac{C}{2} \cdot \Pi
By inserting C=2*r*pi and doing the algebraic manipulation of the previously given proportion, we get the formula for an arc length:
L=(\theta\cdot \frac{\Pi}{180})\cdot r
We use the equation above when \theta is in degrees.
Arc length formula in radians
This paragraph will show you how to convert degrees into radians and vice versa. Probably, the best way to explain this is by providing you with an example. Also, CalCon calculators provided a converter for this issue.
Converting degrees into radians:
Our conversion will be based on the fact that a full circle contains 360 degrees, which is the equivalent of 2 \cdot \Pi radians. If 360\circ=2 \cdot\Pi , dividing the relation by 2, then 180\circ =\Pi . So, the 1\circ = \frac{180}{\Pi} and by acknowledging this, we got ourselves a basis for conversion. For example, let's convert 30 degrees into radians:
Area of the sector of a circle
The area of a circle sector is the amount of space enclosed with two radii and an arc. A sector always emerges from the centre of the circle. A segment of a circle is the area that is bounded by an arc and a chord. To find the area of a circle sector, we will also use the proportions. While knowing that the area of a circle equals Ac=r^{2} \cdot\Pi , the proportion will have the following setup:
\frac{r^{2}\cdot \Pi}{2\cdot \Pi}= \frac{A}{\Theta}
After fractional sorting, the formula for the area of the sector of a circle will have the following form:
A=( \frac{\Theta}{360})\cdot r^{2}\cdot \Pi – in degrees
A= \frac{1}{2}\cdot r^{2}\cdot \Theta – in radians
How to find the length of an arc (an example)?
The steps to calculate the length of an arc:
Measure the radius of your circle or write it down if you already know it.
Decide whether you will use the formula for calculating the arc length in radians or degrees. According to that, the value of the central angle has to be in the same unit, so make sure that the unit of your angle is paired with the chosen formula.
Now, input the values into the equation, do the math and get the result.
To gain a better insight into previous steps, we will show how it works on an example.
How to find the chord length?
The endpoints of the circle sector are the centre of a circle and two points on its edges. If we connect those two edge points, we create a chord. The difference between a chord and a secant is that a chord does not exceed the outer edges. First, to find chord length, we will connect the endpoints (marked with letters H and D, in the picture underneath) with a straight line. Then we will form a triangle by connecting those points with the centre (O). Now, focusing solely on this triangle and drawing a perpendicular bisector from the centre, we will get two right triangles. And the length of a chord can be determined using Pythagorean theory or using trigonometry functions.
Using Pythagorean theory to calculate the chord length:
r^{2}=b^{2}+(\frac{c}{2})^{2} c=2\cdot\sqrt{r^{2}-b^{2}}
Using trigonometry to calculate the chord length:
c=2\cdot r\cdot sin\frac{\Theta}{2}
How to find the length of any curve?
The length of a curve can be found using the following formula: L=\int_{a}^{b} ds
where a and b represent x, y, t, or θ-values as appropriate, and ds can be calculated:
1. rectangular form:
ds=\sqrt{1+(\frac{dy}{dx})^{2}} dx ,
ds=\sqrt{1+(\frac{dx}{dy})^{2}} dy
2. perimetric form:
ds=\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}} dt
3. polar form:
ds=\sqrt{r^{2}+(\frac{dr}{d\Theta})^{2}} d\Theta
FAQ
What is the perimeter of an arc?
The perimeter of an arc is equal to the sum of the three sides of the rectangle plus the half-circumference of the semi-circular portion.
How does Arc Length affect cirmumference?
The arc length is a part of the circle's circumference. For example, an arc with a measure of 30 degrees is one-twelfth of the circle's circumference, so if we want to find the arc length, it will be one-twelfth of the circle's circumference.
What do you call the longest chord?
The longest chord is called the diametar of the circle.
What is a central angle of an arc?
A central angle is an angle whose vertex is the centre of a circle and whose sides are radii.
How to find radius with the arc length?
It is possible to calculate the value of the radius if we know the arc length simply by inserting its value into the formula L=(\theta\cdot \frac{\Pi}{180})\cdot r .
How to calculate the arc length by using sector area and central agle?
The first step is to calculate the measurement of the radius by using the formula for the sector area. While knowing the values of the radius and central angle, we can proceed into the formula for the arc length. | 677.169 | 1 |
Given a path segment by three points \(A\), \(B\) and \(C\) as well as a the radius \(r\) we want to have at point \(B\). Now the vectors the vectors from \(B\) to \(A\) and \(C\) respectively are \(\mathbf{a}=A-B\) and \(\mathbf{b}=C-B\). The bisector of these two vectors is \(\mathbf{v}=\hat{\mathbf{a}}+\hat{\mathbf{b}}\). | 677.169 | 1 |
Triangles, the fundamental shapes that form the backbone of geometry, are a captivating realm within the world of mathematics. In this blog post, we embark on a fascinating journey into the intricate world of triangles, unraveling their properties, classifications, and the profound role they play in various mathematical and real-world scenarios.
Triangles are polygons with three sides and three angles, but their simplicity belies their complexity. Understanding the basics of angles, sides, and vertices sets the stage for delving deeper into the rich tapestry of triangle properties.
Classifications
Triangles come in diverse forms, each with its unique set of characteristics. From equilateral triangles, where all sides and angles are equal, to isosceles and scalene triangles, each classification opens a door to a new realm of geometric possibilities.
Properties
Exploring the interior and exterior angles, the Pythagorean theorem, and the Law of Sines and Cosines reveals the hidden symmetries and relationships within triangles. These properties not only form the foundation of geometry but also find practical applications in fields ranging from architecture to physics.
Applications
Triangles have a ubiquitous presence in our surroundings. From the stability of architectural structures to the calculations involved in navigation, triangles play a vital role in diverse fields. Unraveling these applications showcases the real-world significance of this geometric wonder.
Challenges and Puzzles
Engaging with triangles often involves solving puzzles and problems that stimulate critical thinking. From classic geometry problems to real-world conundrums, tackling these challenges not only sharpens mathematical skills but also fosters a deeper appreciation for the elegance of triangles.
Basic Proportionality Theorem
In Mathematics, the basic proportionality theorem states that "If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio."
Now, let us understand the Basic proportionality theorem with the help of a diagram.
Consider the triangle ABC, as depicted in the diagram. We draw a line PQ parallel to the side BC of ABC and intersect the sides AB and AC in P and Q, respectively.
Thus, according to the Basic proportionality theorem,
AP/PB = AQ/QC
Mathematical Expression for Similarity
In Δ ABC and Δ DEF, if
(i) ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F and
(ii)AB/DE = BC/ EF = CA/ FD, then the two triangles are similar.
To find whether the given two triangles are similar or not, it has four criteria. They are:
Side-Side-Side (SSS) Similarity Criterion – When the corresponding sides of any two triangles are in the same ratio, then their corresponding angles will be equal, and the triangle will be considered similar triangles.
Angle Angle Angle (AAA) Similarity Criterion – When the corresponding angles of any two triangles are equal, then their corresponding side will be in the same ratio, and the triangles are considered to be similar.
Angle-Angle (AA) Similarity Criterion – When two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are considered similar.
Side-Angle-Side (SAS) Similarity Criterion – When one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are said to be similar triangle?
Ans A triangle is a polygon with three sides, three angles, and three vertices.
Q2 What are the different types of triangles?
Ans Triangles can be classified into three types based on side lengths: equilateral (all sides equal), isosceles (two sides equal), and scalene (no sides equal).
Q3 What is the significance of the Pythagorean theorem in triangles?
Ans The Pythagorean theorem establishes a relationship between the sides of a right-angled triangle, stating that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Q4 How are triangles used in architecture and construction?
Ans Triangles provide stability in architectural structures. Trusses, a framework of triangles, are commonly used to distribute weight and maintain structural integrity.
Q5 What are interior and exterior angles in a triangle?
Ans Interior angles are the angles inside a triangle, while exterior angles are formed by extending one side of the triangle | 677.169 | 1 |
It depends. Strictly speaking, a semi-regular tessellation uses
two (or more) regular polygons and, since neither an isosceles
triangle nor a parallelogram is regular, it cannot be a
semi-regular tessellation. However, a less strict definition allows
non-regular components.
Wiki User
∙ 8y ago
This answer is:
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Q: Is an isosceles triangle and parallelogram a semi regular tessellation? | 677.169 | 1 |
Quadrilaterals Worksheet Grade 8
Class 8 understanding quadrilaterals test papers for all important topics covered which can come in your school exams download in pdf free. These worksheets for grade 8 understanding quadrilaterals class assignments and practice tests have been prepared as per syllabus issued by cbse and topics given in ncert book 2020 2021.
With thousands of questions available you can generate as many understanding quadrilaterals worksheets as you want. Polygons that have any portions of. Printable worksheets and tests.
All topics with exact weightage real exam experience detailed analytics comparison and rankings questions with full solutions. Start new online test. Grade 8 understanding quadrilaterals unlimited worksheets every time you click the new worksheet button you will get a brand new printable pdf worksheet on understanding | 677.169 | 1 |
Missing angle of a Quadrilateral
How to find the missing angle of a quadrilateral?
The sum of all four angles of the quadrilateral is 360°. To
find the fourth angle or the missing anglein a quadrilateral when
the measurements of three angles of a quadrilateral are known, then subtract
the three angles from 360° to calculate the missing angle.
Solved examples to calculate the missing angle of a quadrilateral:
1. Three angles of a quadrilateral are 54°, 80° and 116°. Find the measure of the fourth angle.
Solution:
Let the measure of the fourth angle be x°.
We know that the sum of the angles of a quadrilateral is 360°.
Therefore, 54 + 80 + 116 + x = 360
⇒ 250 + x = 360
⇒ x = (360 - 250) = 110.
Hence, the measure of the fourth angle is 110°.
2. The three angles of the quadrilateral are 90°, 105°, 85°. Find the measure of the fourth angle of a quadrilateral.
Solution:
We know that sum of all the angles of a quadrilateral is 360°.
Let the unknown angle of the quadrilateral be x.
Then 90° + 105° + 85° + x = 360°
⇒ 280° + x = 360°
⇒ x = 360 - 280
⇒ x = 80°
Therefore, the measure of the fourth angle of the quadrilateral is 80°
3. The measures of two angles of a quadrilateral are 115°and 45°, and the other two angles are equal. Find the measure of each of the equal angles | 677.169 | 1 |
How Many Sides Does a Square Have?
A square is one of the most basic and recognizable shapes in geometry. It is a polygon with four equal sides and four equal angles. But have you ever wondered why a square has exactly four sides? In this article, we will explore the concept of a square, its properties, and the reasons behind its unique characteristics.
The Definition of a Square
Before we delve into the number of sides a square has, let's first establish what a square is. According to the Merriam-Webster dictionary, a square is defined as "a four-sided polygon that has four right angles and four congruent sides."
From this definition, we can gather two important properties of a square:
A square has four sides.
All four sides of a square are equal in length.
The Geometry of a Square
Now that we understand the basic definition of a square, let's explore its geometry in more detail. A square is a special type of rectangle, which is itself a special type of parallelogram. However, what sets a square apart from other rectangles and parallelograms is its unique combination of properties.
Here are some key characteristics of a square:
Equal sides: As mentioned earlier, all four sides of a square are equal in length. This property distinguishes a square from other quadrilaterals, such as rectangles or rhombuses, which may have different side lengths.
Right angles: A square has four right angles, meaning each angle measures exactly 90 degrees. This property is shared with rectangles, but not with other quadrilaterals like parallelograms or trapezoids.
Diagonals: The diagonals of a square are congruent and bisect each other at right angles. This means that the diagonals divide the square into four congruent right triangles.
Symmetry: A square possesses rotational symmetry of order 4, meaning it looks the same after a 90-degree rotation. This property makes squares popular in art, design, and architecture.
Why Does a Square Have Four Sides?
Now that we have explored the properties of a square, let's address the question at hand: why does a square have exactly four sides?
The answer lies in the definition of a polygon. A polygon is a closed figure with straight sides. The word "polygon" is derived from the Greek words "poly," meaning "many," and "gonia," meaning "angle." Therefore, a polygon must have multiple angles.
A square, being a polygon, must have at least three sides to form a closed figure. However, a triangle does not satisfy the definition of a square, as it does not have four equal sides or four right angles. Therefore, a square requires a minimum of four sides to meet the criteria of a polygon and possess its unique properties.
Real-World Examples of Squares
Squares are not just abstract mathematical concepts; they have practical applications in various fields. Let's explore some real-world examples of squares:
Tiles: Many flooring and wall tiles are square-shaped. The uniformity and symmetry of squares make them an ideal choice for creating visually appealing patterns.
Chessboard: A chessboard is a square grid consisting of 64 squares. The alternating colors of the squares create a visually pleasing pattern and aid in gameplay.
Windows: Some windows, particularly those with a modern design, are square-shaped. The simplicity and clean lines of squares make them a popular choice in architecture.
Computer screens: The majority of computer screens have a square or rectangular shape. This aspect ratio allows for efficient display of digital content.
Summary
In conclusion, a square is a four-sided polygon with four equal sides and four right angles. Its unique combination of properties, such as equal sides and right angles, distinguishes it from other quadrilaterals. A square has exactly four sides because it must meet the criteria of a polygon, which requires multiple angles. Squares have practical applications in various fields, including architecture, design, and mathematics.
Q&A
1. Can a square have more than four sides?
No, a square cannot have more than four sides. By definition, a square is a four-sided polygon.
2. What is the formula to calculate the perimeter of a square?
The formula to calculate the perimeter of a square is P = 4s, where P represents the perimeter and s represents the length of one side.
3. Are all squares rectangles?
Yes, all squares are rectangles, but not all rectangles are squares. A square is a special type of rectangle with equal sides.
4. Can a square have diagonals of different lengths?
No, the diagonals of a square are always congruent, meaning they have the same length.
5. What is the relationship between a square and a cube?
A square is a two-dimensional shape, while a cube is a three-dimensional shape. A cube is composed of six square faces. | 677.169 | 1 |
With the law of sine, you can find any unknown angle of a given triangle or the length of a particular side of a triangle or the length of a particular side of a triangle. This is a fundamental concept of trigonometry.
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We used Dev-C++ to compile the program, but you may use any other standard C compiler. This program make use of math header – math.h especially, two trig functions – sin () function and asin () function at lot. So if you are choosing a different compiler then use the correct math header file.
Before you try the example, learn following C programming concepts. You can skip it if already know it well. | 677.169 | 1 |
The difference between two adjacent angles = 24 degrees. Find the smaller of these angles.
We are given two adjacent angles, and we also know that their difference is 24 °. In order to find the degree measure of the smaller of them, we will compose and solve a linear equation.
Let's apply the property of adjacent corners to the solution. The sum of adjacent angles is 180 °.
Let's denote one of the angles as x °, then the second angle can be written as (x – 24) °.
We get a linear equation:
x + (x – 24) = 180;
x + x – 24 = 180;
2x = 180 + 24;
2x = 204;
x = 204: 2;
x = 102 ° is the largest of the adjacent angles and 102 – 24 = 78 ° is the smaller | 677.169 | 1 |
What is a polygon? | Area of Triangle in java
What is a polygon?
A polygon is any 2-dimensional shape formed with straight lines. Triangles, quadrilateral, pentagon, and hexagon are all examples of polygons.
A regular polygon is a polygon in which all sides of the polygon are of equal length.
We have a mathematical formula so that the area of a regular polygon can be calculated.
Area = (ln * ln * n / 4 * (tan (180 / n))
Where
ln = Length of the edge of the polygon
Number of sides in a polygon =
tan = touch function in degrees | 677.169 | 1 |
Hint:
In this question, we have to find the value of . we will solve this using vector theorem. For this we will use dot product and scalar product i.e., and the property that sum of square of direction cosine is 1 | 677.169 | 1 |
Finding the Direction Cosines
Okay I have been racking my brain with this one for over a week and still have no clue how to do this.I need to study this for a test I am having and can't seem to figure this out
Consider an arbitrary 3D vector: A=Axx+Ayy+Azz
a) Determine the direction cosines for this vector. These are cos[], cos[] and cos [], where is the angle between A and x , where is the angle between A and y, and is the angle of A and z.
b) Show that the direction cosines obey the relationship (cos[])2+(cos[])2+(cos [])2.
Solve the polygon by Law of Cosines as shown onATTACHMENT1.
Step 3.
Substitution of knowns into Law of Cosines should give you:
magnitude of C= 124 nt
Step 4.
For thedirection of force C, use Law of Sines to get the angle between A and C.
If you are drawing the triangle in the opposite order your directions will be reversed. But In that case the (h,k,l) direction will also altered. I didn't given the proof for thedirection cosine theorem. You can get it from the internet.
You can do this by using the Law of Cosines. In this case we know that two sides of the triangle are equal to the radius of 35/2 = 17.5 and the length of the chord is 26. You just plug into the law of cosines to find the angle.
Note: This particular problem can not handle complex numbers, so write your answer in terms of sines and cosines, rather than using e to a complex power. Please refer to the attachment for the solutions. Auxiliary equation help is given.
(b) The electron is moving in thedirection making equal angles with the three axis thus the velocity in vector form can be written as
(Writing directioncosines)
And thus the force is given by
Or
Or
Or
The magnitude of this
340904 Law of Cosines and Law of Sines In the triangle with sides a= 21 cm, b=45cm, and c = 60 cm, where the angle gamma is between the sides a and b, the angle beta between the sides a and c, and the angle alpha between sides b and c. | 677.169 | 1 |
Ewbank Kenny Geometry Calendar
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of Menelaus' theorem, the following theorems are given. 4.4 Theorem The line joining the midpoints of two sides of a triangle is parallel to the third side. Let M and N be the midpoints of sides AB and CA of triangle ABC. (See Fig. 4.4.) Let line MN meet side...
...QB-QB = 0 QB + QD = 0 Here is one more. Example 4 Prove by vectors that the line joining the mid-points of two sides of a triangle is parallel to the third side and half its length. Let D and E by the mid-points of AB and AC respectively. We have DE = M + ÄE Now...
...vectors represented by the other sides. (Punjab, 1986 S) 2. Prove that the line joining the mid-points of two sides of a triangle is parallel to the third side and half of it. (N. Bengal, 1989 ; Marathwada, 1990) 3. Prove that the lines joining the mid-points of...
...the slope of BC. Denote it by m. 4-0 _ 1 10 + 6~4 m We know that a segment that joins the midpoints of two sides of a triangle is parallel to the third side. So, DE is parallel to BC, and as a result they have the same slope. Therefore, the slope of DE is -... | 677.169 | 1 |
A point source S is placed midway between two converging mirrors having equal focal lengths f as shown in the figure. Find the values of d for which only
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A point source S is placed midway between two converging mirrors having equal focal lengths f as shown in the figure. Find the values of d for which only
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Both mirrors have equal focal lengths f. They will produce one image under two conditions. Case I: When the source is at a distance 2f from each mirror i.e. the source center of curvature of the mirrors, the image will be produced at the same point S. So, d=2f+2f=4f Case II: When the source S is at distance 'from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays after the reflection from the other mirror the object itself sine image is formed. Here, d=f+f=2f | 677.169 | 1 |
Help Calculating Angles For Woodworking Mathematics Stack Exchange
Miter Angles And Miter Saws Thisiscarpentry
Miter Angles And Miter Saws Thisiscarpentry
Dividing Angles For Woodworking Old School Compasses Method Youtube
How To Use A Speed Square And Bevel Gauge To Find Angles In Woodworking Youtube
Diy Inside Angles Without An Angle Finder Must Watch Youtube
A 60-degree angle divided by 2 results in two 30-degree angles and a 90-degree angle divided by 2 results in two 45-degree angles.
How to calculate angles carpentry.
In carpentry you usually do away with trigonometric functions to find angles.
A 135 degrees angle looks like this.
Instead you simply use two legs of a triangle to measure and mark the angle.
Finish carpenters are always bisecting corner angleswhich is easy to do with a protractor.
A 1 50 cm 50 cm.
On measuring our internal angle we find that this is 35 simply deduct this from the overall total of 180 and this will then give us our external angle in this case 145 Working out internal and external angles along a linear line.
If the units on the carpenters square is cm and the length of the sides of the square is 50 cm - the most exact angle can be made by multiplying these values with 50.
Leah demonstrates how to determine an angle without measuring when setting a porch post on a slanted surfaceThe level used in this video is an Empire Profe.
Adjust this tool to measure angles for carpentry by loosening the lock nut on the bevel square.
How to Calculate Cutting Angles.
Next take a compass and put the point in the bottom left hand corner where the angle starts or the two lines meet.
When designingbuilding and fitting staircases formulas are used to ensure the treads and risers are the right size and comply with local building codes.
The bevel square often referred to as an angle-finder is a short flat blade that swivels and locks on a handle.
A 135 degree angle is.
Carpenters often use formulas and math when carrying out carpentry jobs such as roofing to check a building is square and to calculate the length of rafters etc.
Place the bevel squares handle side against the base of a protractor and read the protractors degree scale to define the angle.
Please leave blank the values to be calculated.
Centers youll know two things about the right angle. | 677.169 | 1 |
MAT 102 GEOMETRY KCA Past Paper
UNIVERSITY EXAMINATIONS: 2011/2012 EXAMINATION FOR THE CERTIFICATE IN BRIDGING MATHEMATICS MAT 102 GEOMETRY DATE: APRIL 2012 TIME: 1½ HOURS INSTRUCTIONS: Answer Question One and Any other Two Questions
QUESTION TWO (15 MARKS)
a) Given that 0 0 0 90 ≤ ≤ θ , solve the equation 2 2 4cos 4cos 1 sin θ − += θ θ (5 Marks)
b) The length of the sides of an inscribed triangle are 10cm, 7cm and 5cm respectively. Calculate
i) The sizes of the angles of the triangle (8 Marks)
ii)The radius of the circle (2 Marks) QUESTION THREE (15 MARKS)
A point A lies on the latitude 500
N and longitude 100
W, B lies on latitude 500
N and 850
W and C lies on
350
S and 100
W. Calculate
i) The distance between A and C along a great circle (3 Marks)
ii) The radius of the parallel of latitude 500
N (3 Marks)
iii)The distance between A and B along a parallel of latitude (3 Marks)
iv)The distance of B from the equator (3 Marks)
v) The length of the chord AB (3 Marks) QUESTION FOUR (15 MARKS)
b) At a point A, Richard records the angle of elevation of the top (T) of the building as 220. He moves
further from the building by 28m to point B and notes that the angle of elevation of the top then is180
. Determine
i) The height of the building (5 Marks)
ii)How far he is from the bottom of the building when the angle of elevation of the top T is 100
(5 Marks) QUESTION FIVE (15 MARKS)
In a quadrilateral OACB, OA = a and OB = b. OA is parallel to BC. OA=3BC and M is a point on AB
such that AM:MB = 3:1.
i) Evaluate the vectors AB, BM, OM and MC in terms of a and b (12 Marks)
ii)Show that O, M and C are collinear. | 677.169 | 1 |
GK: MathNot everyone can answer any mathematical question brought to them, but the ones that can are always placed first to tutor other people and can only do this through practice. Brush up on your knowledge about general knowledge math with these quiz questions and score a perfect A.
Questions and Answers
1.
What is the shape where its angles add up to 720 degrees?
A.
Diamond
B.
Triangle
C.
Hexagon
D.
Rhombus
Correct Answer C. Hexagon
Explanation A hexagon is a shape with six sides and six angles. In a hexagon, the sum of all interior angles is always equal to 720 degrees. This is because each angle in a regular hexagon measures 120 degrees, and there are six angles in total. Therefore, a hexagon is the shape where its angles add up to 720 degrees.
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1
0
2.
Two angles, measuring 58 degrees and 32 degrees, would be considered what kind of angles?
A.
Vertical Angles
B.
Complementary Angles
C.
90 Degrees
D.
Supplementary Angles
Correct Answer B. Complementary Angles
Explanation Complementary angles are two angles that add up to 90 degrees. In this case, the two angles measuring 58 degrees and 32 degrees add up to 90 degrees, making them complementary angles.
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3.
What shape has 2 parallel sides and 2 perpendicular sides, angles adding up to 360 degrees?
A.
Cylinder
B.
Rectangle
C.
Trapezoid
D.
Polygon
Correct Answer C. Trapezoid
Explanation A trapezoid is a shape that has 2 parallel sides and 2 perpendicular sides, with angles adding up to 360 degrees. A cylinder is a three-dimensional shape, not a polygon, so it does not fit the description. A rectangle has 4 right angles, but all four sides are parallel, not just two. A polygon is a general term for any closed shape with straight sides, so it does not necessarily have 2 parallel and 2 perpendicular sides. Therefore, the correct answer is trapezoid.
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4.
What is the average weight loss for four men who dropped 55, 36, 15 and 74 pounds each?
A.
45
B.
18
C.
68
D.
80
Correct Answer A. 45
Explanation The average weight loss for four men can be calculated by adding up the weight losses of all four men (55 + 36 + 15 + 74 = 180) and then dividing it by the number of men (4). Therefore, the average weight loss is 180 divided by 4, which equals 45.
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5.
If g(x) = x/4, what is g(1/4)?
A.
1/256
B.
4
C.
4-Jan
D.
16-Jan
Correct Answer D. 16-Jan
Explanation The question asks for the value of g(1/4) when g(x) = x/4. To find this, we substitute x with 1/4 in the equation g(x) = x/4. Therefore, g(1/4) = (1/4)/4 = 1/16. However, none of the given options match this value, so the correct answer is not available.
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1
0
6.
What is the probability of rolling less than a 4 when tossing a single die?
A.
3-Feb
B.
4-Jan
C.
3-Jan
D.
2-Jan
Correct Answer D. 2-Jan
Explanation The probability of rolling less than a 4 when tossing a single die is 2 out of 6, or 1/3. This is because there are 3 possible outcomes (1, 2, or 3) out of the 6 total possible outcomes (1, 2, 3, 4, 5, or 6).
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7.
How many ways can a group of 7 choose a committee of 4?
A.
1
B.
840
C.
210
D.
35
Correct Answer D. 35
Explanation The question is asking for the number of ways a group of 7 people can choose a committee of 4. This can be calculated using the combination formula, which is nCr = n! / (r! * (n-r)!). In this case, n = 7 and r = 4. Plugging these values into the formula, we get 7! / (4! * (7-4)!) = 35. Therefore, there are 35 ways to choose a committee of 4 from a group of 7.
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0
1
8.
If m divided by 6 has 5 as a remainder, what is the remainder if 4m is divided by 6?
A.
5
B.
0
C.
2
D.
1
Correct Answer C. 2
Explanation When m is divided by 6, it leaves a remainder of 5. This means that m can be expressed as 6n + 5, where n is an integer. Now, if we multiply 4m by 4, we get 24n + 20. When this expression is divided by 6, the remainder is 2. Therefore, the remainder when 4m is divided by 6 is 2.
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9.
If 3x - x = 2x + x + 20, then x = ?
A.
0
B.
-5
C.
-20
D.
20
Correct Answer C. -20
Explanation The equation given is 3x - x = 2x + x + 20. By simplifying both sides of the equation, we get 2x = 20. Dividing both sides by 2, we find that x = 10. However, this contradicts the given answer of -20. Therefore, the given answer is incorrect and the correct answer cannot be determined from the given equation.
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10.
If a + b = 40 and a < 15, which of the following must be true?
A.
B > 25
B.
B < 25
C.
A > 0
D.
B < 40
Correct Answer A. B > 25
Explanation Since we know that a + b = 40 and a < 15, we can deduce that b must be greater than 25. This is because if a is less than 15, then the sum of a + b cannot be equal to 40 unless b is greater than 25. Therefore, the statement b > 25 must be true.
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11.
If 50,505 + x = 55,555, then 50,505 - 10x = ?
A.
0
B.
5
C.
-50.5
D.
5.05
Correct Answer B. 5
Explanation To find the value of x, we subtract 50,505 from both sides of the equation 50,505 + x = 55,555. This gives us x = 5,050.5. Now, to find the value of 50,505 - 10x, we substitute the value of x into the equation. Plugging in x = 5,050.5, we get 50,505 - 10(5,050.5) = 50,505 - 50,505 = 0. Therefore, the answer is 0.
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12.
Which fraction is equal to 2/3 + 4/5 + 1/2?
A.
87/60
B.
59/30
C.
49/30
D.
59/15
Correct Answer B. 59/30
Explanation To find the sum of the fractions 2/3, 4/5, and 1/2, we need to find a common denominator. The least common multiple of 3, 5, and 2 is 30. So, we can rewrite the fractions with the denominator of 30. 2/3 becomes 20/30, 4/5 becomes 24/30, and 1/2 becomes 15/30. Adding these fractions together gives us 20/30 + 24/30 + 15/30 = 59/30. Therefore, the correct answer is 59/30.
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13.
If your Canadian uncle tells you he got a ticket for driving 103 km/h, how fast was he going?
A.
64 MPH
B.
72 MPH
C.
79 MPH
D.
103 MPH
Correct Answer A. 64 MPH
Explanation Since the question states that the uncle got a ticket for driving 103 km/h, it implies that he was driving over the speed limit. However, the answer choices provided are in miles per hour (MPH). Therefore, to determine the speed in MPH, we convert 103 km/h to MPH. By using the conversion factor of 1 km = 0.621371 MPH, we can calculate that 103 km/h is approximately equal to 64 MPH.
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14.
If a two pound bag of grass seed covers 720 square feet, how many bags are needed to seed 1 acre?
A.
112
B.
47
C.
60.5
D.
89.3
Correct Answer C. 60.5
Explanation To find out how many bags are needed to seed 1 acre, we need to determine the number of square feet in 1 acre. Since 1 acre is equal to 43,560 square feet, we divide this number by the coverage of one bag (720 square feet). The result is approximately 60.5, which means that approximately 60.5 bags are needed to seed 1 acre of land.
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15.
If two lines both have a slope of 1/3 and they don't share any points, what are they called?
A.
Perpendicular
B.
Same Line
C.
Intersecting
D.
Parallel
Correct Answer D. Parallel
Explanation If two lines have the same slope and do not share any points, they are called parallel lines. Parallel lines never intersect and can be represented by the equation y = mx + b, where m is the slope. In this case, since both lines have a slope of 1/3 and do not intersect, they are parallel. | 677.169 | 1 |
Introduction to MongoDB $degreesToRadians Operator
In MongoDB, the $degreesToRadians operator is used to convert angle values from degrees to radians. It can be used in aggregation pipelines to convert angles to radians before performing other mathematical calculations.
Syntax
The syntax for the $degreesToRadians operator is as follows:
{$degreesToRadians:<angle>}
Here, <angle> is the angle value to be converted to radians, which can be a number or an expression that points to a field containing an angle value.
Use Cases
The $degreesToRadians operator is commonly used in aggregation pipelines for geospatial calculations. Geospatial calculations require radians as the unit of measurement, whereas degrees are commonly used. Therefore, it is necessary to convert degrees to radians before performing geospatial calculations.
Example
Suppose we have a collection called locations that contains documents, each of which includes a subdocument called location that contains an array called coordinates, which includes two values representing longitude and latitude in degrees. We want to convert these degree values to radians.
In this aggregation pipeline, we use the $project stage to keep the location subdocument and the coordinates array, and then use the $map operator to convert each value in the array to radians. Finally, we get a new field called coordinatesInRadians that contains the converted radian values. | 677.169 | 1 |
What does tangent to the line y 3 mean?
When a circle is tangent to a line, the line is perpendicular to the radius at the point of tangency. So since this circle is tangent to the y-axis at y=3, the y-coordinate of the center of the circle must also be 3.
How do you find the equation of the tangent to a circle?
A tangent to a circle at point P with coordinates is a straight line that touches the circle at P. The tangent is perpendicular to the radius which joins the centre of the circle to the point P. As the tangent is a straight line, the equation of the tangent will be of the form y = m x + c .
What is the tangent line to a curve?How do you find the center of a circle with two tangent lines?
Answers (1) Find the line bisecting the two tangent lines. This line will contain the center point of the circle. Form a right triangle with one of the tangent lines as a leg, the radius line as the other leg, and the bisecting line as the hypotenuse.
Can a tangent intersect a circle twice?
A line can only intersect a circle 0, 1, or 2 times: twice for chords and secants; once for tangents.
Can a tangent line intersect?
From geometry, you know that a line is tangent to a circle when the line intersects the circle at only one point (see Figure 11.13). Tangent lines to noncircular graphs, however, can intersect the graph at more than one point.
What does it mean when a ray is tangent to a circle?
A tangent to a circle is a straight line which touches the circle at only one point. This point is called the point of tangency. The tangent to a circle is perpendicular to the radius at the point of tangency. In the circle O , ↔PT is a tangent and ¯OP is the radius.
When can you say that a line intersecting the circle illustrates a second or a tangent?
Answer. Answer: Say can say that a line Intersecting the circle is a secant if the line intersects any two points of a circle while we can say that line Intersecting the circle is a tangent only if that line intersects of a circle at exactly one point.
What is difference between code and diameter?
A line that joins two points on the circumference of a circle is called a chord. A diameter divides a circle into two equal segments, each is called a semicircle. Diameter is the longest chord of a circle. The diameter of a circle is twice the radius. | 677.169 | 1 |
Eureka Math Kindergarten Module 6 Lesson 1 Exit Ticket Answer Key
Question 1.
Use your ruler.
First, draw a straight line from the dot.
Second, draw a different straight line from the dot.
Third, draw another straight line to make a triangle.
Answer:
Explanation:
Eureka Math Kindergarten Module 6 Lesson 1 Homework Answer Key
Question 1.
Follow the directions.
First, use your ruler to draw a line finishing the triangle.
Second, color the triangle green.
Third, use your ruler to draw a bigger triangle next to the green triangle.
Answer:
Explanation:
Question 2.
First, draw 2 lines to make a rectangle.
Second, circle all the corners in red.
Third, put an X on the longer sides.
Answer:
Explanation:
Question 3.
First, draw a line to complete the hexagon.
Second, color the hexagon blue.
Third, write the number of sides the hexagon has in the box below.
Answer:
Explanation:
Question 4.
On the back of your paper, draw:
A closed shape with 3 straight sides.
A closed shape with 4 straight sides.
A closed shape with 6 straight sides.
Answer:
Explanation:
Firstly draw a triangle with the help of a ruler
then draw a square with the help of a ruler
then draw a hexagon with the help of a ruler like in the picture below: | 677.169 | 1 |
Category: Geometry
Geometry is a branch of mathematics that deals with the study of shapes, sizes, positions, angles, and properties of space. It explores the relationships and properties of geometric figures such as points, lines, angles, surfaces, and solids. Geometry is a fundamental part of mathematics and has practical applications in various fields such as architecture, engineering, […]
In a right triangle with legs of length 5 units and 12 units, what is the length of the hypotenuse? If one leg of a right triangle is 8 centimeters and the hypotenuse is 10 centimeters, what is the length of the other leg? In a right-angled triangle with a hypotenuse of 13 units and […]
I was wondering if anyone can do 8 units for my ADV APP MATH B/Geometry 1B class. I'm in a program where i get 75 points worth of easy work and I would just give you my info and you would do the work through my Canvas account and work on kami. In my Canvas | 677.169 | 1 |
Pįgina vii ... hypotenuse and one leg of a right - angled triangle equal to 2045 and 1924 ; to find the remaining leg without squaring the given numbers . Answ . 693 . 2. If a side of a parallelogram be equal to one of the diagonals , the squares of ...
Pįgina ix ... hypotenuse is divided by the perpendicular to it from the right angle , and to compute also that perpendicular . Answ . The seg- ments are 25 and 144 , and the perpendicular 60 . 3. If the base of a triangle be 123 , and the other sides ...
Pįgina 26 ... that if these be equal , the lines are parallel . Ji ne A E G C D H Because the angle EGB 26 [ BOOK L THE ELEMENTS feet, respectively; to find the hypotenuse Answ 3145 feet Given the hypotenuse of a right-angled triangle equal.
Pįgina 40 ... hypotenuse , that is , the side subtending the right angle , is equal to the squares described upon the sides which contain the right angle . Let ABC be a right - angled triangle , having the right angle BAC ; the square described upon | 677.169 | 1 |
6. Plane A and plane B are two distinct planes that are both perpendicular to line ?. Which statement about planes A and B is true?
7. Triangle ABC is similar to triangle DEF. The lengths of the sides of triangle ABC are 5, 8, and 11. What is the length of the shortest side of triangle DEF if its perimeter is 60?
8. In the diagram below of right triangle ABC, altitude CD is drawn to hypotenuse AB. If AD = 3 and DB = 12, what is the length of altitude CD?
9. The diagram below shows the construction of an equilateral triangle. Which statement justifies this construction?
10. What is the slope of the line perpendicular to the line represented by the equation 2x + 4y = 12?
11. Triangle ABC is shown in the diagram below. If DE joins the midpoints of ADC and AEB which statement is not true?
12. The equations x2 + y2 = 25 and y = 5 are graphed on a set of axes. What is the solution of this system?
13. Square ABCD has vertices A(-2,-3), B(4,-1), C(2,5), and D(-4,3). What is the length of a side of the square?
14. The diagram below shows triangle ABD, with a ray ABC, BE ⊥ AD and ∠ EBD is congruent to ∠ CBD. If m ∠ ABE = 52, what is m ∠ D?
15. As shown in the diagram below, FD and CB intersect at point A and ET is perpendicular to both FD and CB at A. Which statement is not true?
16. Which set of numbers could not represent the lengths of the sides of a right triangle?
17. How many points are 5 units from a line and also equidistant from two points on the line?
18. The equation of a circle is (x - 2)2 + (y + 5)2 = 32. What are the coordinates of the center of this circle and the length of its radius?
19. The equation of a line is y = 2/3x + 5. What is an equation of the line that is perpendicular to the given line and that passes through the point (4,2)?
20. Consider the relationship between the two statements.
21. In the diagram of trapezoid ABCD below, AB || DC, AD = BC, m ∠ A = 4x + 20, and m ∠ C = 3x - 15.
22. In circle R shown below, diameter DE is perpendicular to chord ST at point L. Which statement is not always true?
23. Which equation represents circle A shown in the diagram below?
24. Which equation represents a line that is parallel to the line whose equation is 3x - 2y = 7?
25. In the diagram below of circle O, PAC and PBD are secants. If mCD = 70 and mAB = 20, what is the degree measure of ?P? | 677.169 | 1 |
What is Coriolis component of acceleration in polar coordinate system?
Coriolis acceleration is the acceleration due to the rotation of the earth, experienced by particles (water parcels, for example) moving along the earth's surface. ... Coriolis acceleration is generated by the eastward rotation of the earth around the N-S axis.
How do you find polar coordinates?
What is Z in polar coordinates?
The representation of a complex number as a sum of a real and imaginary number, z = x + iy, is called its Cartesian representation. θ = arg(z) = tan -1(y / x). ... The values x and y are called the Cartesian coordinates of z, while r and θ are its polar coordinates. Note that r is real and r 3 0.
What does R equal in polar coordinates?
In polar coordinates, a point in the plane is determined by its distance r from the origin and the angle theta (in radians) between the line from the origin to the point and the x-axis (see the figure below).
What is polar and Cartesian coordinates?
In Cartesian coordinates there is exactly one set of coordinates for any given point. With polar coordinates this isn't true. In polar coordinates there is literally an infinite number of coordinates for a given point. For instance, the following four points are all coordinates for the same point.
How do polar coordinates work?
The polar coordinates of a point describe its position in terms of a distance from a fixed point (the origin) and an angle measured from a fixed direction which, interestingly, is not "north'' (or up on a page) but "east'' (to the right). That is in the direction Ox on Cartesian axes.
Why do we use polar coordinates?
Polar coordinates are used often in navigation as the destination or direction of travel can be given as an angle and distance from the object being considered. For instance, aircraft use a slightly modified version of the polar coordinates for navigation.
What is a polar equation?
Polar Equations A polar equation is any equation that describes a relation between r r r and θ \theta θ, where r r r represents the distance from the pole (origin) to a point on a curve, and θ \theta θ represents the counterclockwise angle made by a point on a curve, the pole, and the positive x x x-axis.
Are Cartesian and rectangular coordinates the same?
The Cartesian coordinates (also called rectangular coordinates) of a point are a pair of numbers (in two-dimensions) or a triplet of numbers (in three-dimensions) that specified signed distances from the coordinate axis.
How do you convert to rectangular coordinates?
To convert from polar coordinates to rectangular coordinates, use the formulas x=rcosθ and y=rsinθ.
How do you read Cartesian coordinates?
The position of any point on the Cartesian plane is described by using two numbers: (x, y). The first number, x, is the horizontal position of the point from the origin. It is called the x-coordinate. The second number, y, is the vertical position of the point from the origin.
How do you explain coordinates?
Coordinates are two numbers (Cartesian coordinates), or sometimes a letter and a number, that locate a specific point on a grid, known as a coordinate plane. A coordinate plane has four quadrants and two axes: the x axis (horizontal) and y axis (vertical).
What do you mean by coordinates?
Coordinates are distances or angles, represented by numbers, that uniquely identify points on surfaces of two dimensions (2D) or in space of three dimensions ( 3D ). ... Cartesian coordinates , also called rectangular coordinates, have two or three straight-line axes that define the positions of points in 2D or 3D.
What do we use coordinates for?
Coordinates systems are often used to specify the position of a point, but they may also be used to specify the position of more complex figures such as lines, planes, circles or spheres. For example, Plücker coordinates are used to determine the position of a line in space.
What is effective coordination?
An effective Coordination activities and structures can bring a sense of order to the resulting chaos. Coordination in general may be defined as intentional actions to harmonize individual responses to maximize impact and achieve synergy. a situation where the overall effect is greater than the sum of the parts.
What are the type of coordination?
The two primary types of coordination are internal coordination or establishing a relationship between all the employees, departments, etc. and external coordination or establishing a relationship between the employees and the outsiders.
What is Type 2 coordination chart?
Type II - Under short circuit conditions, the contactor or starter shall cause no danger to persons or installation and shall be suitable for further use. The risk of contact welding is recognized, in which case the manufacturer shall indicate the measures to be taken in regards to equipment maintenance.
What is coordination example?
The definition of coordination is being able to move and use your body effectively and multiple people or things working well together. An example of coordination is when a gymnast walks on a tightrope without falling. An example of coordination is when two people work together to plan or coordinate a party. noun.
What are the three types of coordination?
There are three basic coordinating mechanisms: mutual adjustment, direct supervision, and standardization (of which there are three types: of work processes, of work outputs, and of worker skills).
What are good coordination skills?
These Coordinating skills are always in high demand:Organize staff-wide meetings, create meeting agendas, and assign action items after meetings.Create and manage project timelines, deadlines, and budgets.Work cross-functionally with other departments to ensure organizational objectives are met.Ещё
What are some coordination exercises?
5 Coordination Exercises to Include in Your ProgrammingBall or Balloon Toss. Catch and bump a balloon back and forth using your hands, head, and other body parts. ... Jump Rope. This classic coordination exercise works to synchronize your hand-foot-eye movements. ... Balance Exercises. ... Target Exercises. ... Juggling and Dribbling.
How do you coordinate different departments?
9 tips to encourage collaboration across departmentsMake sure everyone is on the same page. ... Encourage consistent open communication. ... Practice transparency—from the top. ... Enable empathy and understanding. ... Lead by example. ... Encourage open feedback. ... Create a sense of community and collaborative culture. ... Share technology and information.Ещё•12 авг. 2020 г.
How do you join two departments together?
As project managers, we have the ability to encourage collaboration between departments from the ground level by structuring our team's interactions with other departments.Provide Context. ... Cultivate Empathy. ... Develop a Common Language. ... Get Involved In Other Departments' Processes. ... Facilitate Consistent Communications.Ещё•4 дек. 2020 г.
What are the key factors you take into consideration when building an action plan?
What is the effect of lack of coordination in an organization?
A lack of coordination in an organization can decrease productivity, complicate processes and delay the completion of tasks. In order to coordinate the efforts of an entire organization, the organization requires a systematic integration of a process that creates accountability within the organization.
What does loss of coordination mean?
Ataxia describes a lack of muscle control or coordination of voluntary movements, such as walking or picking up objects. A sign of an underlying condition, ataxia can affect various movements and create difficulties with speech, eye movement and swallowing.
What are coordination problems?
A situation in which the interests of agents coincide, and the aim is to try to reach an outcome in which those interests are satisfied. Informally, this is a situation in which each person has an interest in doing something that chimes in with what the others do.
Why is coordination effective and efficient?
Coordination activates each function of management and makes them effective and purposeful. It helps in achieving harmony among individual efforts for attaining organisational goals. It is present in all the activities of an organisation such as production, sales, finance etc. | 677.169 | 1 |
Lesson
Lesson 5
5.1: Notice and Wonder: Midpoints
Here's a triangle \(ABC\) with midpoints \(L, M\), and \(N\).
Description: <p>Triangle A B C with midpoints M, L and N drawn forming a triangle. Segments A N and C N are marked congruent with two ticks. Segments B L and C L are marked congruent with one tick. Segments A M and B M are marked congruent with three ticks.</p>
What do you notice? What do you wonder?
5.2: Dilation or Violation?
Here's a triangle \(ABC\). Points \(M\) and \(N\) are the midpoints of 2 sides.
Convince yourself triangle \(ABC\) is a dilation of triangle \(AMN\). What is the center of the dilation? What is the scale factor?
Convince your partner that triangle \(ABC\) is a dilation of triangle \(AMN\), with the center and scale factor you found.
With your partner, check the definition of dilation on your reference chart and make sure both of you could convince a skeptic that \(ABC\) definitely fits the definition of dilation.
Convince your partner that segment \(BC\) is twice as long as segment \(MN\).
Prove that \(BC=2MN\). Convince a skeptic.
5.3: A Little Bit Farther Now
Here's a triangle \(ABC\). \(M\) is \(\frac23\) of the way from \(A\) to \(B\). \(N\) is \(\frac23\) of the way from \(A\) to \(C\).
What can you say about segment \(MN\), compared to segment \(BC\)? Provide a reason for each of your conjectures.
Dilate triangle \(DEF\) using a scale factor of -1 and center \(F\).
How does \(DF\) compare to \(D'F'\)?
Are \(E\), \(F\), and \(E'\) collinear? Explain or show your reasoning.
Summary
Let's examine a segment whose endpoints are the midpoints of 2 sides of the triangle. If \(D\) is the midpoint of segment \(BC\) and \(E\) is the midpoint of segment \(BA\), then what can we say about \(ED\) and triangle \(ABC\)?
Segment \(ED\) is parallel to the third side of the triangle and half the length of the third side of the triangle. For example, if \(AC=10\), then \(ED=5\). This happens because the entire triangle \(EBD\) is a dilation of triangle \(ABC\) with a scale factor of \(\frac12\).
\(\overline{BD}\cong\overline{DC},\overline{BE}\cong\overline{EA}\)
Caption:
\(\overline{BD}\cong\overline{DC},\overline{BE}\cong\overline{EA}\)
In triangle \(ABC\), segment \(FG\) divides segments \(AB\) and \(CB\) proportionally. In other words, \(\frac{BG}{GA}\)=\(\frac{BF}{FC}\). Again, there is a dilation that takes triangle \(ABC\) to triangle \(GBF\), so \(FG\) is parallel to \(AC\) and we can calculate its length using the same scale factor.
\(\overleftrightarrow{FG} \parallel \overleftrightarrow{AC}\)
Caption:
\(\overleftrightarrow{FG} \parallel \overleftrightarrow{AC}\)
Glossary Entries
dilation
A dilation with center \(P\) and positive scale factor \(k\) takes a point \(A\) along the ray \(PA\) to another point whose distance is \(k\) times farther away from \(P\) than \(A\) is.
Triangle \(A'B'C'\) is the result of applying a dilation with center \(P\) and scale factor 3 to triangle \(ABC\).
scale factor
The factor by which every length in an original figure is increased or decreased when you make a scaled copy. For example, if you draw a copy of a figure in which every length is magnified by 2, then you have a scaled copy with a scale factor of 2 | 677.169 | 1 |
Types Of Triangles Worksheet Grade 5
Students classify triangles as equilateral 3 equal sides isosceles 2 equal sides scalene all sides have different lengths or as a right triangle one angle of 90 degrees. Incorporated here is an array of topics like finding the area of a triangle with dimensions in integers decimals and fractions finding the area involving unit conversions finding the area of the three types of triangles and more.
Find here an unlimited supply worksheets for classifying triangles by their sides angles or both one of the focus areas of 5th grade geometry.
Types of triangles worksheet grade 5. Types of triangles grade 6 geometry worksheet identify if the triangle is scalene isosceles or equilateral. In these pdf worksheets for 4th grade and 5th grade kids learn to distinguish between various triangles based on the length of the sides and tell whether the triangle provided with measures is an equilateral scalene or isosceles triangle. Grade 6 geometry worksheet types of triangles.
The worksheet are available in both pdf and html formats. Types of triangles grade 6 geometry worksheet identify if the triangle is scalene isosceles or equilateral. Worksheets math grade 5 geometry classifying triangles.
Home worksheets classify triangles worksheets for classifying triangles by sides angles or both. Classify the triangles into acute obtuse and right angled triangle with following angles. These worksheets are printable pdf files.
4 types of triangles this math worksheet gives your child practice identifying equilateral isosceles scalene and right triangles. Types of triangles grade 6 geometry worksheet identify if the triangle is scalene isosceles or equilateral. This extensive collection of worksheets on triangles for grades 3 through high school is incredibly useful in imparting a clear understanding of a variety of topics like classifying triangles similar triangles congruence of triangles median and centroid of a triangle inequality theorem pythagorean inequalities area perimeter and angles in a triangle and much more.
Classifying triangles based on side measures. Explore this assortment of the area of triangles worksheets to elevate the practice of students in grade 5 through high school. 5cm 3cm 6cm if two angles of a triangle measures 50 and 60 | 677.169 | 1 |
Midpoint
Midpoint Formula - The midpoint formula calculates the midpoint between two given points in a coordinate plane.
Midpoint of a Line Segment - The midpoint of a line segment in a Cartesian plane is calculated by taking the average of the x-coordinates of the endpoints for the x-coordinate of the midpoint, and similarly for the y-coordinate. This calculator helps find the midpoint. | 677.169 | 1 |
Are you struggling to grasp the complexities of geometry? Fear not! Welcome to Geometry Spot: All You Need To Know – your ultimate guide to navigating the world of shapes, angles, and formulas. Whether you're a student seeking clarity or a curious mind delving into the realm of mathematics, this comprehensive resource promises to demystify geometry and ignite your passion for its beauty.
Dive into the World of Geometry Spot:
What Is Geometry Spot?
Geometry Spot isn't just another educational website; it's a gateway to mathematical enlightenment. Here, we unravel the mysteries of geometry, offering intuitive explanations and interactive explorations. Say goodbye to confusion and hello to clarity as you embark on a journey through shapes and figures.
Exploring Interactive Features:
At Geometry Spot, learning is anything but dull. Dive into our interactive explorations, where you can manipulate geometric shapes with ease. Rotate, resize, and move shapes to unlock their secrets, all while honing your spatial reasoning skills. Prepare to witness mathematics come to life before your eyes!
Understanding Fundamental Concepts:
Embracing the Beauty of Shapes:
Shapes are the building blocks of geometry, and Geometry Spot celebrates their elegance. From the simplicity of circles to the complexity of polygons, each shape has its own unique properties waiting to be discovered.
Geometry Spot: All You Need To KnowYou're sitting in math class staring at shapes on the board – circles, triangles, squares. Your teacher starts going on about angles and formulas and your eyes glaze over. Geometry can seem so abstract and confusing, but it doesn't have to be! This geometry spot has got you covered with easy explanations of geometry concepts like the properties of shapes, perimeter and area formulas, and the Pythagorean theorem. We'll break down those complicated theorems into simple steps you can understand. You'll be an expert on lines, rays, angles, and all things geometry by the time we're done. So sharpen your pencils and get ready to become a geometry whiz! This geometry spot will make you love math's beautiful shapes.
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Perimeter and area formulas can be daunting at first glance, but fear not! Geometry Spot breaks down these calculations into simple steps, empowering you to conquer rectangles, triangles, and more. Discover the joy of measuring shapes with precision and accuracy.
Unveiling the Power of the Pythagorean Theorem:
The Pythagorean theorem is a cornerstone of geometry, linking the lengths of a right triangle's sides. At Geometry Spot, we unravel this theorem's mysteries, guiding you through its proof and practical applications. Prepare to marvel at the elegance of mathematical logic!
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Conclusion:
Embark on a journey of mathematical discovery with Geometry Spot: All You Need To Know. From exploring shapes to mastering formulas, our user-friendly platform offers endless opportunities for learning and growth. Say goodbye to math anxiety and hello to a newfound love for geometry! | 677.169 | 1 |
Hexagon Form
Hexagon Form - Web a regular hexagon has: Interior angles of 120° exterior angles of 60° area = (1.5√3) × s 2, or approximately 2.5980762 × s 2 (where s=side length) radius equals side length; Area = 3√3 2 × side2in an irregular hexagon, the sides are of unequal length, and each. Web a hexagon is a 6 sided polygon which can also be described as a closed 6 sided shape where all the sides are straight and equal, as well as the angles. Given the length of the side = 6 inches.
Web in a hexagon, the sum of all 6 interior angles is always 720º. Also, the shape of a hexagon is given in the figure below: Area of hexagon = (3√3 s 2 )/2 = (3 × √3 × 6 2 )/2 = 54√3 = 93.5 in 2. Interior angles of 120° exterior angles of 60° area = (1.5√3) × s 2, or approximately 2.5980762 × s 2 (where s=side length) radius equals side length; Several special types of hexagons are illustrated above. Area (a) = 3 √ 3/2 x (s)2, here s = side. The sum of all the internal angles of a hexagon is 720°.
Hexagon Shape Printable Customize and Print
Web a hexagonal pyramid is a 3d shaped pyramid that has the base shaped like a hexagon along with the sides or faces in the shape of isosceles triangles that form the hexagonal pyramid at.
What is a Hexagon? 6 Sided Shape How many sides?
The hexagon on the left is a regular hexagon and is likely what most people think of when they hear the word hexagon. The formula to find the diagonals of hexagons is: Web the word.
6 Printable Hexagon Template
Since a regular hexagon is comprised of six equilateral triangles, the formula for finding the area of a hexagon is derived from the formula of finding the area of an equilateral triangle. The prefix hexa.
Hexagon Formula, Properties, Examples, Definition
Area = 3√3 2 × side2in an irregular hexagon, the sides are of unequal length, and each. Sum of interior angles of 720° 9 diagonals; The radius is the side length. Web in greek, "hex".
Hexagon Definition, Shape, Properties, Formulas
= 3√3/2 x 9.9 x 9.9. Web the formula is given below: What is a regular hexagon shape? A hexagonal pyramid has a base with 6 sides along with 6 isosceles triangular lateral faces. The.
Hexagon Definition, Formula & Examples Cuemath
Find the area of a regular hexagon whose each side measures 9.9 cm. Interior angles of 120° exterior angles of 60° area = (1.5√3) × s 2, or approximately 2.5980762 × s 2 (where s=side.
Hexagon Picture Images of Shapes
Write down the formula for finding the area of a hexagon if you know the side length. Empowering an autonomous, sustainable future. Area of hexagon with apothem. Given the length of the side = 6.
Printable Hexagon Template
Hex meaning six and gonia meaning corners. Web the formula is given below: The sum of all the internal angles of a hexagon is 720°. It is also made of 6 regular triangles! The area.
Hexagon Form Also, the shape of a hexagon is given in the figure below: A regular hexagon can be divided into six equilateral triangles. In a regular hexagon, all sides are the same length, and each internal angle is 120 degrees. A regular hexagon can be inscribed into a circle or circumscribed by a circle. Finding the area of a regular hexagon when only the perimeter is known. | 677.169 | 1 |
2. A line segment joining two points on a circle is called: A. arc B. tangent C. sector
D. chord
3. Sand is pouring to form a conical pile such that its altitude is always twice its radius. If the volume of a conical pile is increasing at the rate of 25 pi cu.ft/min, how fast is the radius is increasing when the radius is 5 feet? A. 0.5 ft/min B. 0.5 pi ft/min C. 5 ft/min D. 5 pi ft/min SOLUTION: h = 2r, r = 5ft Vcone =
6. An observer wishes to determine the height of a tower. He takes sights at the top of the tower from A to B, which are 50 ft. apart, 92.54 ft SOLUTION: β = 180 − 40 = 140° α = 180 − 30 − 140 = 10° 50
x
= sin 30 ; x = 143.969621 sin θ h = 143.969621 sin(40) = 92.54 ft 7. A tangent to a conic is a line A. which is parallel to the normal B. which touches the conic at only one point C. which passed inside the conic D. all of the above 8. Find the area of the triangle which the line 2x – 3y + 6 = 0 forms with the coordinate axes. A. 3 B. 4 C. 5 D. 2 SOLUTION: 2x − 3(0) + 6 = 0 x=
17. A wall 8 feet high is 3.375 feet from a house. Find the shortest ladder that will reach from the ground to the house when leaning over the wall. A. 16.526 ft B. 15.625 ft C. 14.625 ft D. 17.525 ft SOLUTION: 2
25. A conic section whose eccentricity is less than one is known as: A. circle B. parabola C. hyperbola
D. ellipse
26. The plate number of a vehicle consists of 5-alphanumeric sequence is arranged such that the first 2 characters are alphabet and the remaining 3 are digits. How many arrangements are possible if the first character is a vowel and repetitions are not allowed? A. 90 B. 900 C. 9,000 D. 90,000 SOLUTION: Vowel = a , e , i , o , u = 5 ; =(5)(25)(10)(9)(8) = 90,000 27. The axis of the hyperbola, which is parallel to its directrices, is known as: A. conjugate axis B. transverse axis C. major axis D. minor axis 28. The minute hand of a clock is 8 units long. What is the distance traveled by the tip of the minute hand in 75 minutes. A. 10pi B. 20pi C. 25pi D. 40pi SOLUTION: 1 min = 6° 6°
30. The probability of a defect of a collection of bolts is 5%. If a man picks 2 bolts, what is the probability that does not pick 2 defective bolts? A. 0.950 B. 0.9975 C. 0.0025 D. 0.9025 SOLUTION: 1 − (0.05)(0.05) = 0.9975 1
34. The cable of a suspension bridge hangs in the form if a parabola when the load is uniformly distributed horizontally. The distance between two towers is 150m, the points of the cable on the towers are 22 m above the roadway, and the lowest point on the cables is 7 m above the roadway. Find the vertical distance to the cable from a point in the roadway15 m from the foot of a tower.
44. If the equation is unchanged by the substitution of –x for x, its curve is symmetric with respect to the A. y-axis C. origin B. x-axis D. line 45 degrees with the axis 45. Find the number of sides of a regular polygon if each interior angle measures 108 degrees. A. 7 B. 8 C.5 D. 6 SOLUTION: (n−2)(180) n
= 108
n= 5 46. The integer part of common logarithm is called the________. A. radicand B. root C. characteristic
47. The constant "e" is named in honor of: A. Euler B. Eigen
D. mantissa
C. Euclid
D. Einstein
48. A man rows upstream and back in 12 hours. If the rate of the current is 1.5 kph and that of the man in still water is 4 kph, what was time spent downstream? A. 1.75 hrs B. 2.75 hrs. C. 3.75 hrs D. 4.75 hrs SOLUTION: S
T = Tup + Tdown C = 1.5kph, v = 4kph
S
S
T = 2.5 + 5.5 = 20.625 km
Tdown =? Tdown =
S = vt S
S
Tdown = V+C = 5.5
20.625 5.5
= 3.75 hrs
S
Tup = V−C = 2.5
49. The probability that A can solve a given problem is 4/5, that B can solve it is 2/3, and that C can solve it is 3/7. If all three try, compute the probability that the problem will be solved. A. 101/105 B. 102/105 C. 103/105 D. 104/105 SOLUTION: 4
57. If 9 ounces of cereal will feed 2 adults or 3 children, then 90 ounces of cereal, eaten at the same rate, will feed 8 adults and how many children? A. 8 B. 12 C.15 D. 18 SOLUTION: rate of children and adult 9oz 2
= 4.5 oz/adult
9oz 3
= 3oz/children
formulate an equation: (8)(4.5) + (x)(3) = 90 x = 18 children
58. Mary is twice as old as Helen. If 8 is subtracted from Helen's age and 4 is added to Mary's age, Mary will then be four times as old as Helen. How old is Helen now? A. 24 B. 36 C. 18 D. 16
59. A point on the curve where the second the derivative of a function is equal to zero is called. A. maxima B. minima C. point of inflection D. point of intersection 60. Find the area of the triangle whose sides are 25, 39, and 40. A. 46 B. 684 C. 486 SOLUTION:
θ = tan−1 (√3) = 60° 68. A sequence of numbers where the succeeding term is greater than the preceding term is called: A. dissonant resonance C. Isometric series B. convergent series D.divergent series
72. A statement of the truth of which is admitted without proof is called: A. an axiom B. a postulate C. a theorem D. a corollary 73. A rectangular trough is 8 feet long, 2 feet across the top and 4 feet deep. If water flows in at a rate of 2 ft3/min, how fast is the surface rising when the water is 1 ft deep? A. ¼ ft/min B. ½ ft.min C. 1/8 ft/min D. 1/6 ft/min SOLUTION: V = LWH dv dt
81. In debate on two issues among 32 people, 16 agreed with the first issue, 10 agreed with the second issue and of these 7 agreed with both. What is the probability of selecting a person at random who did not agree with either issue? A. 1/32 B. 13/32 C. 3/8 D. 3/10 SOLUTION: 32 people
91. Determine the probability of throwing a total of 8 in a single throw with two dice, each of whose faces is numbered from 1 to 6. A. 1/3 B. 1/18 C. 5/36 D. 2/9 92. Find the distance between the point (3, 2, -1) and the plane 7x – 6y + 6z + 8 = 0. A. 1 B. 2 C. 3 D. 4 SOLUTION: d=
7(3)−6(2)+6(−1)+8
= 1
1√7²+6²+6²
93. How many parallelograms are formed by a set of 4 parallel lines intersecting another set of 7 parallel lines? A. 123 B. 124 C. 125 D. 126 SOLUTION: 𝐦(𝐦−𝟏)𝐧(𝐧−𝟏) 𝟒 [𝟕(𝟕−𝟏)(𝟒)(𝟒−𝟏)] 𝟒
= 𝟏𝟐𝟔
94. The graphical representation of the cumulative frequency distribution in a set of statistical data is called: A. Ogive B. Histogram C. Frequency polyhedron D. mass diagram 95. Find the area bounded by the curve defined by the equation x2 = 8y and its latus rectum. A. 11/3 SOLUTION:
25. The GCF of two numbers is 34, and their LCM is 4284. If one of the number is 204, the other number is A.714 B. 716 C. 2124 D. 3125 Solution: Other Number =
(34)(4284) 204
= 714
26. Jonas, star player of Adamson University has free throw shooting of 83%. The game is tied at 87-87. He is fouled and given 2 free throws. What is the probability that the game will go overtime? A. 0.3111
B. 0.6889
C. 0.0289
D. 0.9711
27. Find the work done in moving an object along a vector a = 31 + 4j if the force applied is b = 21 + j. A. 8
40. Two stones are 1 mile apart and are of the same level as the foot of a hill. The angles of depression of the two stones viewed from the top of the hill are 5 degrees and 15 degrees respectively. Find the height of the hill. A. 109.1 m
(8+7i)(8+7i) = 15 + 112i 43. How far is the directrix of the parabola (x-4)^2 = -8(y-2) from the x-axis? A. 2 Solution:
B. 3
1
y = − 8 (𝑥 − 4)2 + 2 1
Where: a = − 8 , b = 1, c = 0 y=k–p 𝑦=
4𝑎𝑐−𝑏 2 −1 4𝑎
y =4
C. 4
D. 1
44. A weight W is attached to a rope 21 ft long which passes through a pulley at P, 12 ft above the ground. The other end of the rope is attached to a truck at a point A, 3 ft above the ground. If the truck moves off at the rate of 10ft/sec, how fast is the weight rising when it is 7 ft above the ground? A. 9.56 ft/sec C. 8.27 ft/sec
B. 7.82 ft/sec D. 6.25 ft/sec
45. The first farm of GP is 160 and the common ratio is 3/2. How many consecutive terms must be taken to give a sum of 2110? A. 5
B. 6
C. 7
D. 8
Solution: 2𝑛
2110 =
160( 1− ) 3 1−3 2
n=5
46. Steve earned a 96% on his first math test, a 74% his second test, and 85% on 3 tests average. What is his third test? A. 82%
P(2) = (2) (5) (4)(0.15)2(0.85)3 = 0.138178 P (0 or 1 or 2) = 0.9734 50. Find the average rate of the area of a square with respect to its side x as x changes from 4 to 7. A. 9 B. 3 C. 11 D. 18 51. The equations for two lines are 3y – 2x = 6 and 3x + ky = -7. For what value of k will the two lines be parallel? A. -9/2
54. A fair coin is tossed three times. Find the probability that there will appear three heads. A. 1/4
B. 1/2
C. 1/8
D. 1/6
Solution: You have a fair coin: this means it has a 50% chance of landing heads up and a 50% chance of landing tails up. pH=pT=1/2 pHxpTxpH=1/2×1/2×1/2 = 1/8 = C. 1
1
𝟏
P3H = C(3,3) (2)3 (2)3-3 = 𝟖 55. A spherical balloon inflated with r = 3(cube root of t) as t is greater than zero and t is less than equal or equal to 10. Find the rate of change of volume in cubic cm at t = 8. A. 37.70 Solution:
56. Joe and his dad are bricklayers. Joe can lay bricks for a well in 5 days. With his father's help, he can build it in 2 days. How long would it take his father to build it alone? A. 3-1/4 days C. 2-1/3 days Solution: 1
71. The probability that a man, age 60, will survive to age 70 is 0.80 the probability that a woman of the same age will live up to age 70 is 0.90. What is the probability that only one of the survives? A. 0.72 B. 0.26 C. 0.28 D. 0.0
81. Timothy leaves home for Legaspi City 400 miles away. After 2 hours, he has to reduce his speed by 20 mph due to rain. If he takes 1 hour for lunch and gas and reaches Legaspi City 9 hours after left home, what was his initial speed? A. 63 mph B. 62 mph C. 65mph D. 64 mph 82. How many arrangements of the letters in the word "VOLTAGE" begin with a vowel and end with a consonant? A. 1490 B.1440 C.1460 D.1450 Solution: 3! (4!) (10) = 1440 83. An airplane flying with the wind, took 2 hours to travel 1000 km and 2.5 hours in flying back. What was the wind velocity in kph? A. 50
B. 60
Solution: 100 2
–x=
C. 70
1000 2.5
D. 40
+x
X = 50 mph 84. A woman is paid $ 20 for each day she works and the forfeits $ 5 for each day she is idle. At the end of 25 days she nets $ 450. How many days did she work? A. 21
87. The positive value of k which make 4x^2 – 4kx + 4k + 5 a perfect square trinomial is A. 6 B. 5 C. 4 D. 3 88. A tree is broken over by a windstorm. The tree was 90 feet high and the top of the tree is 25 feet from the foot of the tree. What is the height of the standing part of the tree? A. 48.47 ft B. 41.53 ft C. 45.69 ft D. 44.31 ft 89. The Rotary Club and the Jaycee Club had a joint party. 120 members of the Rotary Club and 100 members of the Jaycees Club also attended but 30 of those attended are members of both clubs. How many persons attended the party? A. 190
97. A particle moves along a line with acceleration 2 + 6t at time t. When t = 0, its velocity equals 3 and it is at position s = 2. When t =1, it is at position s = A. 2 B. 5 C. 6 D. 7 Solution: @t = 0 A = 2 +6(0) A=2
@t = 1 A = 2 + 6(1) A=8
at = 10 S = 10 - 3 = 7
98. The edge of a cube has length 10 in., with a possible error of 1 %. The possible error, in cubic inches, in the volume of cube is A. 3 B. 1 C. 10 D. 30 Solution: v = s3 dv/ds = 3s2 dv/v = (3s2ds)/s3 =3 99. What is the rate of change of the area if an equilateral triangle with respect to its side s when s = 2? A. 0.43 B. 0.50 C.10 D. 1.73 Solution: A=
5 1.500 C. 1.200 D. 0.222 Let X - number of children who recognize their mother's voice X has Binomial distribution (n=20, p= 0.90) E(X)=m= np= 20* 0.90=18 P(x = 20) = P(x ≤ 20) – P(x ≤ 19) = = 1 – 0.878 = 0.122 6. Find the differential equation of the family of lines passing through the origin. A. xdx – ydy = 0 C. xdx – ydy = 0 B. xdy – ydx = 0
D. ydx – xdy = 0
Solution Let y = mx be the family of lines through origin. Therefore, dy dx = m Eliminating m, x dy – ydx = 0. 7. A chord passing through the focus of the parabola y2 = 8x has one end at the point (8, 8). Where is the other end of the chord? A. (1/2, 2) B. (-1/2, -2) C. (-1/2, 2) D. (1/2, -2) 8. Find the radius of the circle inscribed in the triangle determined by the line 2
y= x+4, y= -x -4, and y = 7x + 2. A. 2.29
B. 0.24
C. 1.57
D. 0.35
9. What would happen to the volume of a sphere if the radius is tripled? A. Multiplied by 3 C. Multiplied by 27 B. Multiplied by 9 D. Multiplied by 6 Solution: V1/V2 = (r1/ r2)3 = (r1/ 3r1)3 Therefore: V2= 27V1 10. Six non- parallel lines are drawn in a plan. What is the maximum number of point of intersection of these lines? A. 20 B. 12 C. 8 D. 15
Solution: 𝑁(𝑁−1) 2
=
N=6 6(6−1) 2
= 15
11. In a triangle ABC where AC=4 and angle ACB=90 degrees, an altitude t is drawn from C to the hypotenuse. If t = 1, what is the area of the triangle ABC? A. 1.82 B. 1.78 C. 2.07 D. 2.28 Solution: Using sine law: (4/sin45) = (x/sin90) X=AB=4.2 Side CB= sq.rt of (4.2^2-4^2) CB= 1.289 Area=(1/2)(b)(h)sin theta = (1/2)(1.289)(4)sin90 = 2.07 12. In a 15 multiple choice test questions, with five possible choices if which only on is correct, what is the standard deviation of getting a correct answer? A.1.55 B. 1.65 C. 1.42 D. 1.72 Solution: square root of [15×(1/5)×(4/5)] = 1.55 13. What is the area bounded by the curve y = tan 2 x and the lines y = 0 and x = pi/2? A. 0 B. infinity C. 1 D. Ɵ
28. The plane rectangular coordinate system is divided into four parts which are known as: A. octants B. quadrants C. axis D. coordinates 29. A student already finished 70% of his homework in 42 minutes. How many minutes does she still have to work? A. 18 B. 15 C. 20 D. 24 Solution: Equation; 0.70 x total time(t) = 42min Total time(t) = 60min 60 – 42 = 18min 30. In how many ways can 5 people be lined up to get on a bus, if a certain 2 persons refuse to follow each other? A. 36 B. 48 C. 96 D. 72 Solution: Using calculator 3!(3)(4)= 72 31. Water is being pumped into a conical tank at the rate of 12 cu.ft/min. The height of the tank is 10 ft and its radius is 5ft. How fast is the water level rising when the water height is 6ft? A. 2/3 pi ft/min
34. Find the area bounded by one arch of the companions to the cycloid x = a theta, y = a (1cos theta) and the y-axis. A. 2pi a^2
B. 4pi a^2
C. pi a^2
D. 3pi a^2
35. A rectangular plate 6m by 8m is submerged vertically in a water. Find the force on one face if the shorter side is uppermost and lies in the surface of the liquid. A. 941.76 kN
B. 1,583.52 kN
C. 3,767.04 kN
D. 470.88 kN
36. Michael is four times as old as his son Carlos. If Michael was 18 years old when Carlos was born, how old is Michael now? A. 36 yrs.
B. 20 yrs.
C. 24 yrs.
D. 32 yrs.
Solution: Given:
Then
x + 4x = x + x + 18
X – Carlo's age
x – Carlo's age was born
5x = 2x + 18
4x – Michael's age
x + 18 – Michael's age
x=6
Substitute value of x=6 to x + 18: x + 18 = 24 yrs. 37. In polar coordinate system the distance from a point to the pole is known as: A. polar angle
C. X-coordinates
B. radius vector
D. Y-coordinates
38. A certain man sold his ballot at Php 1.13 per piece. If there 100 balots sold all in all, how much is his total collection? A. Php 113.00
B. Php 115.00
C. Php 112.00
D. 116.00
Solution: X = 1.13(100) = Php 113.00
39. A certain population of bacteria grows such that its rate of change is always proportional to the amount present. It doubles in 2 years. If in 3 years there are 20,000 of bacteria present, how much is present initially? A. 9,071
B. 10.071
C. 7,071
D. 8,071
Solution: 1
Q=2𝑄𝑜
Q = 𝑄𝑜 22𝑡
Q = 𝑄𝑜 𝑒 𝑟𝑡
20000 = 𝑄𝑜 (2)3/2
2𝑄𝑜 = 𝑄𝑜 𝑒 2𝑟
𝑄𝑜 = 20000 / (2)3/2
2 = (𝑒 2𝑟 )1/2
𝑸𝒐 = 7,071
𝑒 𝑟 = 21/2
40. In throwing a pair of dice, what is the probability of getting of 5? A. 1/36
45. Two vertical poles are 10 m apart. The poles are 5 m and 8 m, respectively. They are to be stayed by guy wires fastened to a single stake on the ground and attached to the tops of the poles. Where should the stake be placed to use the least amount of wire? A. 6.15 m from 5 m pole C. 6.51 m from 5 m pole
54. What is the area of the ellipse whose eccentricity is 0.60 and whose major axis has a length of 6? A. 40.21 B. 41.20 C. 42.10 D. 40.12 Solution: 2𝑎 = 6 𝑎=3 𝑐 𝑒=𝑎 𝑐 = .6 ∗ 3 = 1.8 𝑏 = √𝑎2 − 𝑐 2 𝑏 = √32 − 1.82 = 2.4 𝐴 = 𝜋𝑎𝑏 𝐴 = 𝜋(3)(2.4) 𝐴 = 22.61 55. Tickets to the school play sold at $4 each for adults and $1.50 each for children. If there were four times as many adult's tickets sold as children's tickets, and the total were $3500. How many children's tickets were sold? A. 160 B. 180 C. 200 D. 240 Solution:
57. The Rotary Club and the Jaycees Club had a joint party. 120 members of the Rotary Club attended and 100 members of the Jaycees Club also attended but 30 of those who attended are members of both parts. How many persons attended the party? A. 190 B. 220 C. 250 D. 150 Solution: 120 + 100 = 220 220 − 30 = 𝟏𝟗𝟎 58. Find the value of k for which the graph of y = x^3 + kx^2 + 4 will have an inflection point at x = -1. A. 3 B. 4 C. 2 D. 1 Solution: 𝑦 ′ = 3𝑥 2 + 𝑘𝑥+ 0 𝑦 ′′ = 6𝑥 + 2𝑘 2𝑘 = −6𝑥 𝑘 = −3𝑥 𝑘 = −3(−1) 𝒌=𝟑
59. Solve for x if log4x = 5. A. 2048 B. 256 Solution: 45 = 𝑥
C. 625
D. 1024
𝒙 = 𝟏𝟎𝟐𝟒 6092.54 ft Solution: ℎ 𝑡𝑎𝑛 𝜃 = 𝑥 ℎ
ℎ
𝑡𝑎𝑛 30 = (50+𝑥)
𝑡𝑎𝑛 40 = 𝑥
𝑡𝑎𝑛 30(50 + 𝑥) = 𝑡𝑎𝑛 40𝑥 50+𝑥 𝑥
𝑡𝑎𝑛40
= 𝑡𝑎𝑛30
𝑥 = 110.29 ℎ = 𝑡𝑎𝑛 40𝑥 ℎ = 𝑡𝑎𝑛 40(110.29) 𝒉 = 𝟗𝟐. 𝟓𝟒 𝒇𝒕.
62. If four babies are born per minute, how many babies are born in one hour? A. 230 B. 250 C. 240 D. 260 𝑀=
75. In an ellipse, a chord which contains a focus and is in a line perpendicular to the major axis is a: A. latus rectum C. focal width B. minor axis D. conjugate axis
76. Determine the rate of a woman rowing in still water and the rate of the river current, if it takes her 2 hours to row 9 miles with the current and 6 hours to return against the current. 1 mph B. 2 mph C. 3 mph D. 4 mph d1 = d 2
3. A reflecting telescope has a parabolic mirror for which the distance from the vertex to the focus is 30 ft. If the distance across the top of the mirror is 64 in., how deep is the mirror of the center? A. 32/45 in. B. 30/43 in. C. 32/47 in. D. 35/46 in.
Age of ben =2X =2(8.25) = 16.5 11. What percentage of the volume of a cone is the maximum volume right circular cylinder that can be inscribed in it? A. 24% B. 34% C. 44% D. 54% 12. A balloon rising vertically, 150 m from an observer. At exactly 1 min, the angle of elevation is 29 deg 28 min. How fast is the balloon using at that instant? A. 104m/min B. 102m/min C. 106m/min D. 108m/min 13. A conic section whose eccentricity is less than one (1) is known as: A. a parabola B. an ellipse C. a circle D. a hyperbola 14. A tangent to a conic is a line A. which is parallel to the normal B. which touches the conic at only one point C. which passes inside the conic D. all of the above 15. A die and a coin are tossed. What is the probability that a three and a head will appear? A. 1/4 B. 1/2 C. 2/3 D.1/12 5 5 16. Find the integral of 12sin xcos xdx if lower limit = 0 and upper limit = pi/2. A. 0.8 B.0.6 C.0.2 D.0.4 17. 12 oz of chocolate is added to 10 oz of flavoring is equivalent to A.1 lb and 8 oz B. 1 lb and 6 oz C.1 lb and 4 oz D.1 lb and 10 oz 18. The Ford company increased its assets price from 22 to 29 pesos. What is the percentage of increase? A.24.14% B.31.82% C.41.24% D.28.31% 19. Find the area bounded by outside the first curve and inside the second curve, r = 5, r = 10sinθ A. 47.83 B.34.68 C.73.68 D.54.25 20. In two intersecting lines, the angles opposite to each other are termed as: A. opposite angles C. horizontal angles B. vertical angles D. inscribed angles Soln.B = vertical angles 21. The area in the second quadrant of the circle x^2 + y^2 = 36 is revolved about the line y + 10 = 0. What is the volume generated? A. 2932 c.u. B. 2392 c.u. C. 2229 c.u. D. 2292 c.u.
22. A cardboard 20 in x 20 in is to be formed into a box by cutting four equal squares and folding the edges. Find the volume of the largest box. A.592 cu.in. B.529 cu.in. C.696 cu.in. D.689 cu.in.
27. Water is running out of a conical funnel at the rate of 1 cubic inch per sec. If the radius of the base of the funnel is 4 in. and the altitude is 8 in., find the rate at which the water level is dropping when it is 2 in. from the top. ` A. -1/pi in./sec B. -2/pi in./sec C. -1/9pi in./secD.-2/9pi in./sec Soln. 1
V = 3 𝜋𝑟 2 ℎ 𝑅 𝑟 1 = = 𝐻 ℎ 2 𝜋 ℎ 2
𝜋
3
12
V= ( ) ℎ= 2
;𝑟 =
ℎ 2
ℎ3 𝑑𝑉 3𝜋 2 𝑑ℎ = ℎ 𝑑𝑡 12 𝑑𝑡 −1 =
3𝜋 𝑑ℎ (2)2 12 𝑑𝑡
𝑑ℎ 𝟏 = − 𝝅/𝒔𝒆𝒄 𝑑𝑡 𝟗
28. How many inches is 4 feet? A. 36 B. 48
C. 12
D. 56
4ft x 12inch / 1ft = 48inch 29. A rectangular trough is 8 ft. long, 2 ft. across the top, and 4 ft. deep. If water flows in at a rate of 2 cu. ft./min., how fast is the surface rising when the water is 1 ft. deep? A. 1/5 ft./min B. 1/8 ft./min C. 1/6 ft./min D. 1/16 ft./min
Soln. V = (8)(2)(1)h 𝑑𝑉 𝑑ℎ = 16 𝑑𝑡 𝑑𝑡 𝑑ℎ 𝟏 = 𝒇𝒕/𝒎𝒊𝒏 𝑑𝑡 𝟖
30. Five tables and eight chairs cost $115; three tables and five chairs cost $70. Determine the total cost of each table.
33. For what value of k will the line kx +5y = 2k have a y-intercept 4? A. 8 B. 7 C. 9 D.10 34. If a bug moves a distance of 3pi cm along a circular arc and if this arc subtends a central angle of 45 degrees, what is the radius of the circle? A. 8 B. 12 C. 14 D. 16 35. Two vertices of a rectangle are on the positive x-axis. The other two vertices are on the lines y = 4x and y = -5x + 6. What is the maximum possible area of the rectangle? A.2/5 B.5/2 C.5/4 D. 4/5 36. Find the length of the arc of 6xy = x^4 + 3 from x = 1 to x = 2. A.12/17 B.17/12 C.10/17 D.17/10 37. A certain radioactive substance has half-life of 3 years. If 10 grams are present initially, how much of the substance remain after 9 years? A.2.50g B.5.20g C. 1.25g D.10.20g 38. A cubical box is tobuilt so that it holds 125 cu. cm. How precisely should the edge be made so that the volume will be correct to within 3 cu. cm.? A.0.02 B.0.03 C.0.01 D.0.04 39. Find the eccentricity of the ellipse when the length of its latus rectum is 2/3 of the length of its major axis. A.0.62 B. 0.64 C.0.58 D.0.56 40. Find k so that A = and B = are perpendicular. A. 2/3 B.3/2 C.5/3 D.3/5 41. Find the moment of inertia of the area bounded by the curve x^2 = 8y, the line x = 4 and the x-axis on the first quadrant with respect to y-axis. A.25.6 B. 21.8 C.31.6 D.36.4 42. Find the force on one face of a right triangle of sides 4m and altitude of 3m. The altitude is submerged vertically with the 4m side in the surface.
A.62.64 kN
B.58.86 kN
C.66.27 kN
D.53.22 kN
43. In how many ways can 6 people be seated in a row of 9 seats? A. 30,240 B. 30,420 C.60,840 D. 60,480 SOLUTIONS: 9P6 = 60,480 44. The arc of a sector is 9 units and its radius is 3 units. What is the area of the sector? A.12.5 B.13.5 C.14.5 D.15.5 SOLUTIONS: 1
A = 2 𝑟𝐶 1
A = 2 (3)(9) A = 13.5 45. The sides of a triangle are 195, 157, and 210, respectively. What is the area of the triangle? A.73,250 B.10,250 C.14,586 D.11,260 SOLUTIONS: S=
195+157+210 2
= 281
A = √281 (281 − 195(281 − 157)(281 − 210) A = 14586.21 46. A box contains 9 red balls and 6 blue balls. If two balls are drawn in succession, what is the probability that one of them is red and the other is blue? A.18/35 B.18/37 C.16/35 D.16/37 47. A car goes 14 kph faster than a truck and requires 2 hours and 20 minutes less time to travel 300 km. Find the rate of the car. A.40 kph B.50 kph C.60 kph D.70 kph 48. Find the slope of the line defined by y – x = 5. A.1 B.1/4 C.-1/2 SOLUTIONS: y = mx+b y–x=5
D.5
y=x+5 by inspection, the slope is equal to 1 49. The probability of John's winning whenever he plays a certain game is 1/3. If he plays 4 times, find the probability that he wins just twice. A.0.2963 B.0.2936 C.0.2693 D.0.2639 SOLUTIONS: nCrpq n=4 , r=2 , therefore :
p = 1/3 q = 2/3
1 2 2 2
4C2 x (3) (3) = 0.2963 50. A man row upstream and back in 12 hours. If the rate of the current is 1.5 kph and that of the man in still water is 4 kph, what was the time spent downstream? A.1.75 hr B.2.75 hr C.3.75 hr D. 4.75 hr SOLUTIONS: d=d (V + c)(t) = (V – c)(t) (4 + 1.5)(x) = (4 – 1.5)(12 - x) x = 3.75 hrs 51. If cot A = -24/7 and A is in the 2nd quadrant, find sin 2A. A.336/625 B.-336/625 C.363/625 SOLUTION: Cot A = 1 tan 𝐴
=
D. -363/625
−24
7 −24 7 7
tan A = −24 7
A = tan−1 (−24) A = -16.260 sin 2A = sin (2x = 16.250) =
−336 625
52. The volume of a square pyramid is 384 cu. cm. Its altitude is 8 cm. How long is an edge of the base? A.11 B.12 C.13 D.14
x=5 y=0 61. What is the height of the parabolic arch which has span of 48 ft. and having a height of 20 ft. at a distance of 16 ft. from the center of the span? A. 30 ft. B. 40 ft. C. 36 ft. D.34ft. SOLUTION:
73. Boyles's law states that when a gas is compressed at constant temperature, the product of its pressure and volume remains constant. If the pressure gas is 80 lb/sq.in. when the volume is 40 cu.in., find the rate of change of pressure with respect to volume when the volume is 20 cu.in. A. -8 B. -10 C. -6 D.-9 SOLUTION: 74. Find the average rate of change of the area of a square with respect to its side x as x changes from 4 to 7. A. 8 B. 11 C. 6 D. 21 SOLUTION: A= x^2 Limits 7-4=3 lim A'= 2x = 2(3) = 6 75. How many cubic feet is equivalent to 100 gallons of water? A. 74.80 B. 1.337 C. 13.37 SOLUTION: 100L = 1m 3
D. 133.7
1m = 3.28 ft 1L = 0.2642 gal 1𝐿
1𝐿3
100 gal = 0.2642 𝐿 1000𝐿 𝐿 (
3.18 𝐿𝐿 3 1𝐿
) = 13.37
76. A merchant purchased two lots of shoes. One lot he purchased for $32 per pair and the second lot he purchased for $40 per pair. There were 50 pairs in
the first lot. How many pairs in the second lot if he sold them all at $60 per pair and made a gain of $2800 on the entire transaction? A. 50 B. 40 C. 70 D. 60 SOLUTION: PB=50(32)=1600 2800 = 1400 + PR2
82. In an ellipse, a chord which contains a focus and is in a line perpendicular to the major axis is a: A.latus rectum B. minor axis C. focal width D. major axis 83. With 17 consonant and 5 vowels, how many words of four letters can be four letters can be formed having 2 different vowels in the middle and 1 consonant (repeated or different) at each end? A.5780 B. 5785 C. 5790 D. 5795 2 84. Evaluate tan (j0.78). A.0.653 B.-0.653 C.0.426 D. -0.426 85. A particle moves along a line with velocity v = 3t^2 – 6t. The total distance traveled from t = 0 to t = 3 equals A.8 B. 4 C. 2 D. 16 86. An observer at sea is 30 ft. above the surface of the water. How much of the ocean can he sea? A.124.60 sq. mi. C. 154.90 sq. mi. B.142.80 sq. mi. D. 132.70 sq. mi. 87. There are three consecutive integers. The sum of the smallest and the largest is 36. Find the largest number. A.17 B. 18 C.19 D. 20 88. If y = sqrt. of (3 – 2x), find y. A.1/sqrt. of (3 – 2x) C. 2/sqrt. of (3 – 2x) B. -1/sqrt. of (3 – 2x) D. -2/sqrt. of (3 – 2x) 89. The logarithm of MN is 6 and the logarithm of N/M is 2, find the value of logarithm of N. A.3 B. 4 C. 5 D.6 90. A woman is paid $20 for each day she works and forfeits $5 for each day she is idle. At the end of 25 days she nets $450. How many days did she work? A.21 days B. 22 days C. 23 days D.24 days 91. Francis runs 600 yards in one minute. What is his rate in feet per second? A.25 B. 30 C.35 D.40 92. For a complex number z = 3 + j4 the modulus is: A.3 B. 4 C. 5 D. 6 93. Which of the following is an exact DE? A. (x^2 + 1)dx – xydy = 0 C. 2xydx + (2 + x^2)dy = 0 B. xdy + (3x – 2y)dy = 0 D. x^2 ydy – ydx = 0 94. There are 8 different colors, 3 of which are red, blue and green. In how many ways can 5 colors be selected out of the 8 colors if red and blue are always included but green is excluded? A.12 B.11 C. 10 D.9 95. Five cards are drawn from a pack of 52 well – shuffled cards. Find the probability that 3 are 10's and 2 are queens.
7. An air balloon flying vertically upward at constant speed is situated 150 m horizontally from an observer. After one minute, it is found that the angle of elevation from the observer is 28 deg 59 min. What will be then the angle of elevation after 3 minutes from its initial position?
12. A cylindrical container open at the top with minimum surface area at a given volume. What is the relationship of its radius to height? A. radius = height B. radius = 2height C. radius = height/2 D. radius = 3height
13. A water tank is shaped in such a way that the volume of water in the tank is V = 2y3/2cu. in. when its depth is y inches. If water flows out through a hole at the bottom of the tank at the rate of 3(sqrt. Of y) cu. in/min. At what rate does the water level in the tank fall? A. 11 in/min
B. 1 in/min
C. 0.11 in/min
D. 1/11 in/min
SOLUTION: 14. A family's electricity bill averages $80 a month for seven months of the year and $20 a month for the rest of the year. If the family's bill were averaged over the entire year, what would the monthly bill be? A. $45
B. $50
C. $55
D. $60
SOLUTION:
12months = 560 + 100 = 660 × 1/12 = $55
15. When a baby born he weighs 8 lbs and 12 oz. After two weeks during his check-up he gains 6 oz. What is his weight now in lbs and oz? A. 8 lbs and 10 oz B. 9 lbs and 4 oz 8 lbs and 12 oz. SOLUTION:
C. 9lbs and 2 oz D. 10 lbs and 4 oz
12 𝑜𝑧 × 0
.0625𝑙𝑏𝑠
= 0.75 𝑙𝑏𝑠 𝑜𝑧
0.0625𝑙𝑏𝑠 8 𝑜𝑧 × 𝑜𝑧
𝑡𝑜𝑡𝑎𝑙 = 9.25 𝑙𝑏𝑠 𝑜𝑟 9𝑙𝑏𝑠 𝑎𝑛𝑑 4 𝑜𝑧
= 0.5 𝑙𝑏𝑠
8 + 0.75 + 0.5 = 9.25 𝑙𝑏𝑠 16. A given function f(t) can be represented by a Fourier series if it A. is periodic B. is singled valued C. is periodic, single valued and has a finite number of maxima and minima in any one period D.has a finite number of maxima and minima in any one period
24. An audience of 450 persons is seated in rows having the same number of persons in each row. If 3 more persons seat in each row, it would require 5 rows less to seat the audience. How many rows? A. 27
B. 32
C. 24
SOLUTION: r - rows; n - number of persons 450 = rn = (r-5)(n+3)
D. 30
rn = rn - 5n + 3r - 15 n = (3r-15)/5 450 = r*(3r-15)/5 750 = r² - 5r r²-5r-750=0 (r-30)(r+25)=0 then r=30 25. The volume of a cube becomes three times when its edge is increased by 1 inch. What is the edge of a cube? A. 2.62
B. 2.26
C. 3.26
D. 3.62
SOLUTION: a3 = V; (a+1)3 = 3V ; (a+1)3=3a3; a = 2.26
26. What is the angle of the sun above the horizon, when the building 150 ft high cast a shadow of 405 ft? A. 21.74
B. 68.26 deg
C. 20.32 deg
D. 69.68 deg
SOLUTION: Arc tan(105/405)=20.32 27. Water ir running out of a conical tunnel at the rate of 1 cu. in/sec. If the radius of the base of the tunnel is 4 in and the altitude is 8 in, find the rate at which the water level is dropping when it is 2 in from the top. A. -1/9pi in/sec B. -1/2pi in/sec
28. A statistic department is contacting alumni by telephone asking for donations to help fund a new computer laboratory. Past history shows that 80% of the alumni contacted in this manner will make a contribution of at least P50.00. A random sample of 20 alumni is selected. What is the probability that between 14 to 18 alumni will make a contribution of at least P50.00? A. 0.421 B. 0.589
C. 0.844
D. 0.301
29. Jun rows has banca a river at 4 km/hr. What is the width of the river if he goes at a point 1/3 km. A. 5.33 km
B. 2.25 km
C. 34.25 km
D. 2.44
30. Find the volume generated by revolving about the x-axis, the area bounded by = cosh x from x = 0 to x = 1. A. 5.34
36. What is a so that the points (-2, -1, -3), (-1, 0, -1) and (a, b, 3) are in straight line? A. 2 B. 4 C. 3 D. 1 37 Find the volume generated when the area bounded by y = 2x – x and y = (x – 1)2 is revolved about the x-axis A. 2.34 B. 3.34 C. 4.43 D. 1.34 38. Find the centroid of a semi-ellipse given the area of semi-ellipse as A = ab and volume of 4
the ellipsed as V = 3 𝜋ab2 A. 2b/3𝜋 B. b/2𝜋
C. 4b/3𝝅
39. How many 5 poker hands are there in a standard deck of cards? A. 2,595,960 B. 2,959,960 C. 2,429,956
43. Using power series expansion about 0, find cosx by differentiating from sinx A. 1- (x^2/2!)+(x^4/4!)-(x^6/6!)+ C. 1-(x^3/3!)+(x^5/5!)-(x^7/7!)+
B.x-(x^2/2!)+(x^4/4!)-(x^5/5!)+ D.x-(x^3/3!)+(x^5/5!)-(x^7/7!)+
44. Find the area bounded by y = √4 A. 7.8
B 6.7
𝑥in the first quadrant and the lines x = and x= 3 C. 5.5
D. 6.5
C. 24 x 10
D. 2.4 x 105
45. Express 2,400,000 in scientific notation A. 2.4 x 10
B. 2.4 x 106
SOLUTION: 2.40000x106
46.An interior designer has to design two offices, each office containing 1 table, 1 chair, 1 mirror, 2 cabinets. A supplier gives him options between 4 tables, 5 chairs, 5 mirrors and 10 cabinets. In how many ways can he design the offices assuming there is no repetition? A. 14100
B. 2400
C. 21600
D. 1740
47. What is the equation of a circle that passes through the vertex and the points of latus rectum of y 2 = x
51. The value of all the quarters and dimes in a parking meter is $18. There are twice as many quarters as dimes. What is the total number of dimes in the parking meter? A. 40 B. 20 C. 60 D. 80 x+y=18
2x+y=19
x=18-x
2x+18+x=18
3x=18 X=6
52. A ball is dropped from height of 12 m and it rebounds ½ of the distance it falls. If it continues to fall and rebound in this way, how far will it travel before coming to rest? B. 30 m C. 48 m D. 60 m
A. 36 m
53. At t = o, a particle starts at rest and moves along a line in such a way that at time t its acceleration is 24t2 feet per second per second. Through how many feet does the particle move during the first 2 seconds? A. 32 B. 48 C. 64 D. 96 SOLUTION: S = wot+at = 0+24(2) = 48 ft.
54. If a trip takes 4 hours at an average speed of 55 miles per hour, which of the following is closest to the time the same trip would take at an average speed of 65 miles per hour? A. 3.0 hours B. 3.4 hours C. 3.8 hours D. 4.1 hours SOLUTION: V1t1 = V2t2; t2 = 55 65
(4)
= 𝟑. 𝟒 𝒉𝒓𝒔
55. A laboratory has a 75-gram sample of radioactive materials. The half-life of the material. The half life on the material is 10 days. What is the mass of the laboratory's sample remaining after 30 days? A. 9,375 grams B. 11.25 grams C. 12.5 grams D. 22.5 grams SOLUTION: 𝑑𝑜
59. Which of the following is equivalent to the expression below? (x2 – 3x + 1) – (4x – 2)
A. x2 – 7x – 1
B. x2 – 7x + 3
C. -3x2 – 7x + 3 D. x2 + 12x+ 2
SOLUTION : (x2-2x+1)-(4x-3)=0; x2-7x+3=0
60. For what value of k will x + have a relative maximum at x = -2? 𝑥
A. -4
B. -2
C. 2
D. 4
SOLUTION: x-k/x=0
; x=-2
-2-K/-2=0; k=4
61. When the area in sq. units of an expanding circle is increasing twice as fast as its radius in linear units, the radius is
A. 1/4 𝜋𝝅
C. 1 1/4
B. 0
D. 1
62. If the function f is defined by f(x)= f(0) = x5 – 1, then f-1, the inverse function of f, is defined by f-1(x) =
A.
B.
C.
D.
SOLUTION: f (0) = x5 – 1 =
f (x) = f-1(x) =
63. A school has 5 divisions in a class IX having 60, 50, 55, 62, and 58 students. Mean marks obtained in a History test were 56, 64, 72, 63 and 50 by each division respectively. What is overall average of the marks per student?
64. The number n of ways that an organization consisting of twenty-six members can elect a president, treasury, and secretary (assuming no reason is elected to more than one position) is A. 15600 B. 15400 C. 15200 D. 15000 SOLUTION: 26!/(26-3)! = 15600
65. Find the equation of the line that passes through (3, -8) and is parallel to 2x + 3y = 2
67. In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls?
A. 11760 B. 17640
C. 48620
D. none of these
SOLUTION: 10!/(10-5)! + 8!/(8-4)! = 31920
68. The area of a right triangle is 50. One of its angles is 45°. Find the hypothenuse of the triangle A. 10 B. C. 10 D. 10 SOLUTION:
A=50 A=1/2 bh = 1/2 (h/sinǾ)(h)
1
Ǿ=45 sinǾ=h/b b=h/sinǾ 1
h=
69. Each side of the square pyramid is 10inches. The slant height, H, of this pyramid measures 12 in. What is the area in square inches, of the base of the pyramid? A. 100 B. 144 C. 120 D. 240 SOLUTION:
Ab= S2 =102 =100 sq. inches tan25°+tan 50°
70. Find the exact value of
1−tan 25° tan 50°
A. 1.732
B. 3.732
C. 2.732
SOLUTION:
= 𝟑. 𝟕𝟑𝟐
71. Which term of the arithmetic sequence 2, 5, 8, … is equal to 227? A. 74
Solution: 3 12 83. What is the probability of a three with a single die exactly 4 times out of 5 trials? 𝑃=
A. 25/776
B. 125/3888
C. 625/3888
D. 1/7776
84. A man is on a wharf 4 m above the water surface. He pulls in a rope to which is attached a coat at the rate of 2 m/sec. How fast is the angle between the rope and the water surface changing when there are 20 m of rope out? A. 0.804 rad/sec B. 0.0408 rad/sec C. 0.0402 rad/sec D. 0.0204 rad/sec
85. Find the area of the largest rectangle that can be inscribed in the ellipse 25x^2 + 16x^2 = 400 A. 30
91. The force required to stretch a spring is proportional to the elongation. If 24 N stretches a spring 3 mm, find the force required to stretch a spring 2 mm. A. 16
B. 18
C. 14
D.12
SOLUTION: F= (24x 2mm)/ 3mm = 16N
92. A is 3 times as old as B. Three years ago, A is four times as old as B. Find the sum of their ages. A. 30 Solution: 4( X-3)- (X-3)=3X-X 4X- 12- X+3= 2X 3X-9= 2X X= 9 B=X = 9 A=3X= 27 B+A= 9+27 = 36
B. 36
C. 26
D. 28
93. The area of a rhombus is 264 sq. cm. If one of the diagonals is 24 cm long, find the length of the other diagonal. A. 22
B. 20
C. 26
D. 28
SOLUTION: A= 1/2(D1D2) 264= 1/2(24xD2) D2= 22 94. In a triangle ABC, angle A= 60 degree and angle B =45 degree. What is the ratio of side BC to side AC? A. 1:22
B. 1:36
C. 1:48
D. 1:19
95. Solve the equation cos^2 A= 1 – cos^A. A. 45o, 315o
B. 45o,225o
C. 45o,135o
96. Find the distance from the point (6, -2) to the line 3x + 4y + 10 = 0. A.4
14. From past experience it is known 90 percent of one year old children can distinguish their mothers voice of a similar sounding female. A random sample of one years old are given this voice recognize test. Find the probability that atleast 3 children did not recognize their mothers voice. SOLUTION
20. A statue 3 m high is standing on a base of 4m high. If an observers eye is 1.5 m above the ground how far should he stand from the base in order that the angle subtended by the statue is a maximum? SOLUTION X=√𝐻1𝐻2 =√(3)(4) X=3.71m ANSWER: C. 3.71 21. What is the number in the series below? 3, 16, 6, 12, 12, 6, SOLUTION 3, 16, (3x2), (16-2^2), (3x2^2), (16-2^3), (3x2^3) =(3x2^2) =24 ANSWER: D. 24 22. A man who is on diet losses 24 lb in 3 months 16 lb in the next 3 months and so on for a long time. What is the maximum total weight loss? A. 72 B. 64 C. 54 D. 81
23. What is the slope of the linear equation 3y-x=9? SOLUTION 3y-x=9 3y=x+9 y=1/3x+(9)(1/3) m=1/3 ANSWER: A. 1/3 24. Each of the following figures has exactly two pairs of parallel sides except a A. parallelogram B.rhombus C. trapezoid D. square 25. A points A and B are 100 m apart and are of the same elevation as the foot of the building. The angles of elevation of the top of the building from points A and B are 21 degrees and 32 respectively. How far is A from the building? SOLUTION
ℎ
Tan32=
Tan21=
𝑥
ℎ
100+𝑥
𝑥𝑡𝑎𝑛32
Tan21=
100+𝑥
X=159.276 100+x= 100+159.276 =259.28m
ANSWER: A. 259.28 26. What is the area in sq.m.of the zone of a spherical segment having a volume of 1470.265 cu.m if the diameter of the sphere is 30m. A. 655.487 B. 565.487 C. 756.847 D. 465.748 SOLUTION A=2 πrh V=
31. The sum of the base and altitude of an isosceles triangle is 36cm. Find the altitude of the ttriangle if its area is to be a maximum. SOLUTION: x + y = 36 x = 36 - y 1 A= 2 bh 1
A = 2 ( 36 − 𝑦 )𝑦 1
A= 2 ( 36 -𝑦 2 ) 𝑦2
A = 18 − 2 0 = 18 - y y = 18 ANSWER: C 18cm
32. An insurance policy pays 80 percent of the first P20,000 of a certain patients medical expenses, 60 percent of the next P40,000 and 40 percent of the P40,000 after that. If the patients total medical bill is P92,000 how much will the policy pay? ANSWER: C. 52,800 33. A scientist found 12mg of radioactive isotope is a soil sample. After 2 hours, only 8.2 mg of the isotope remained. Determine the half life of the isotope? SOLUTION: 𝑙𝑛𝑥1 𝑡 = 𝑡1 𝑙𝑛𝑥 2
SOLUTION √52 + (−12)2 A= π𝑟 2 =π(3)2 = 𝟐𝟖. 𝟐𝟕 ANSWER: B. 28.27 42. If a flat circular plate of radius r = 2 m is submerged horizontally in water so that the top surface is at a depth of 3m, then the force on the top surface of the plate is SOLUTION F= WhA =w = 9810N F=(9810)(3)(𝜋(2)2 ) F=369828.29N =369,829.15N ANSWER: A. 369,829.15N 43. A hemispherical tank with a diameter of 8 ft is full of water find the work done in ft-lb in pumping all the liquid out of the top of the tank. B. 12,546 𝑑2 𝑦
ANSWER:D. pi/8 57. Which of the following is a vector? A. kinetic energy B. electric field intensity C. entropy D. work 58. In how many ways can 6 people be lined up to get on a bus if certain 3 persons refuse to follow each other? SOLUTION:. 6P3 =120 ways ANSWER:D. 480 59. The bases of a frustum of a pyramid are 18cm by 18cm and 10cm by 10cm. Its lateral area is 448 sq. cm. what is the altitude of the frustum? ANSWER:B. 6.93cm 60. A store advertises a 20 percent off sale. If an article marked for sale at $24.48, what is the regular price? SOLUTION:20 % discount $24.48 discounted price $24.48 = x – 20% (x) X = $30.60 ANSWER:C. $30.60
ANSWER: B. 12 62. Dave is 46 yrs old. Twice as old as rave. How old is rave? SOLUTION: D=46 yrs R=2X 2x=46 X = 23yrs old ANSWER: C. 23 yrs 63. The angles of elevation of the top of a tower at two points 30 m and 80 m from the foot of the tower, on a horizontal line are complementary. What is the height of the tower? SOLUTION: A+B = 90 A= 90-B 𝐻
tanѲ=80
𝐻
B =tan−1 30 equation no. 2 𝐻
𝐻
Tan 90- (tan−1(30)) = 80 H= 49m
𝐻
tan(90-B)=80 equation no. 1 ANSWER: C. 49m 64.A large tank filled with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. What is the concentration of the solution in the tank at t = 5 min?
ANSWER: C. 0.0795 lb/ gal 65. The intensity I of light at a depth of x meters below the surface of a lake satisfies the differential dldx = (-1/4)I. At what depth will the intensity be 1 percent of thtat at the surface? ANSWER: B. 2.29m 66. What is the discriminant of the equation4𝑥 2 = 8𝑥 − 5? ANSWER: B. -16 67. Find the percentage error in the area of a square of side s caused by increasing the side by 1 percent. ANSWER: B. 2 percent 68. What is the height of a right circular cone having a slant height of 3.162 m and base diameter of 2 m? SOLUTION: H=√(3.162)2 − (1)2 H= 3m ANSWER: C. 3 69. In how many orders can 7 different pictures be hung in a row so that 1 specified picture is at the center? SOLUTION: 6i = 720 ways ANSWER: D. 720 70. What is the x-intercept of the line passing through (1,4) and (4,1)? ANSWER: B. 5 71. One ball is drawn at random from a box containing 3 red balls, 2 white balls, and 4 blue balls. Determine the probability that is not red. SOLUTION # 𝑜𝑓𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑃= # 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 6 𝑃= 9 𝟐 𝑷= 𝟑 ANSWER: B. 2/3 72. An airplane flying with the wind took 2 hours to travel 1000 km and 2.5 hours flying back. What was the wind velocity in kph?
77. A rubber ball is made to all from height of 50 ft and is observed to rebound 2/3 of the distance it falls. How far will the ball travel before coming to rest if the ball continues to fall in this manner? SOLUTION a1= 50 x 2/3 = 33.33 𝑎1
81. Find the inclination of the line passing through (5,3) and (10,7) SOLUTION: p1(-5,3) p2(10,7) Tan theta = (y2-y1) / (x2-x1) = (7-3) / 10-(-5) = 14.92 degrees ANSWER: B. 14.93֯ 82. An ellipse has an eccentricity of 1/3. Find the distance between the two directrix if the distance between the foci us 4.
87. A man bought 5 tickets in a lottery for aprize of P 2,000.00. If there are total 400 tickets, what is his mathematical expectation? SOLUTION: . 5/400 = x/2000 ; x = 25 ANSWER: A. P25.00 88. In what quadrants will Ѳ be terminated if cos Ѳis negative?
SOLUTION: Quadrant II, the x direction is negative, and both cosine and tangent become negative Quadrant III, sine and cosine are negative Therefore 2,3 ANSWER: B. 2,3 89. For what value of the constant k is the lie x + y = k normal to the curve 𝑦 = 𝑥 2 SOLUTION: So the slope of the normal is -1, which means that the slope of the tangent is 1. dy/dx = 2x Find out where the slope is 1: 2x = 1 --> x = 1/2 So we have the coordinates (1/2, 1/4). So eqn of normal is: y - 1/4 = -1*(x - 1/2) y =-x + 1/2 + 1/4 k = 1/2 + 1/4 = 3/4
94. If in the fourier series of a periodic function, the coefficient aჿ = 0 and aⁿ = 0, then it must be having ____________ symmetry. A. odd B. odd quarter wave C. even D. either A or B 95. Tickets number 1 to 20 are mixed up then and then a ticket is drawn has a number which is a multiple of 3 or 5? SOLUTION Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E)/n(S) = 9/20. ANSWER: D. 9/20 96. A car travels 90 kph. What is its speed in meter per second? SOLUTION: 90 km/hr x 1000 meter/1 km 1 hr/3600 sec. = 25 ANSWER: C. 25 97. The line y = 3x = b passes through the point (2,4) Find b. SOLUTION:
10. For the formula R= E/C, find the maximum error if C= 20 with possible error 0.1 and E= 120 with a possible error of 0.05 A. 0.0325
B. 0.0275
Solution: dR = 1/C dE – E/C² dC
C. 0.0235
D. 0.0572
dR = 1/20(0.05) – 120/20² (-0.01) = 0.0325 11. The probability that a married man watches a certain television show is 0.4 and the probability that a married woman watches the show is 0.5. The probability that a man watches the show, given that his wife does is 0.7. Find the probability that a wife watches the show given that her husband does. A. 0.875
B. 0.745
C. 0.635
D. 0.925
Solution: Let :
M – the event that the man watch the show W - the event that the woman watch the show
19. Three circle of radai 3, 4, and 5 inches respectively, are tangent to each other extremely. Find the largest angle of a triangle found by joining the center of the circles. A. 72.6 degrees B. 75.1 degrees C. 73.4 degrees D. 73.5 degrees
20. A reflecting telescopes has a parabolic mirror for witch the distance from the vertex to the focus is 30 ft. If the distance across the top of the mirror is 64 in, how deep is the mirror of the center? A. 32/45 in
B. 30/43 in
C. 32/47 in
D. 35/46 in
Solution: Given: x = 64/2 = 32 , p = 30x12 = 360 at origin at the center X2 = 4py y = x2/4p = 322/4(360) = 32/45 in 21 C. 110.29 ft B. 143.97 ft D. 95.24 ft
22. The average of six scores is 83. If the highest score is removed, the average of the remaining scores is 81.2. Find the highest score. A. 91 C. 93 B. 92 D. 94 Solution: Given: 𝐴𝑣𝑒 𝑜𝑓 𝑠𝑖𝑥 𝑠𝑐𝑜𝑟𝑒𝑠 = 83
V=x3 Dv=3x2dx Dv=3(10)2(0.01) Dv=3 24. The area in the second quadrant of the circle x^2+y^2=36 is revolved about the line y+10=0. What is the volume generated? A. 2932 c.u C. 2229 c.u B. 2392 c.u D. 2292 c.u Solution:
27 0.500 C. 1.200 D. 0.222 Solution: .9022=0.122
28. If Jose is is 10% taller than Pedro and Pedro is 10% taller than Mario, then Jose taller than Mario by _______%. A. 18 B.20 C.21 D.23 Solution: Jose Pedro Mario
31. A pole which lean 11 degrees from the vertical toward the sun cast a shadow 12m long when the angle of the elevation of the sun is 40 degrees. Find the length of the pole. A. 15.26 m
B. 14.26 m
C. 13.26 m
D. 12.26 m
Solution: X= 180 - 40 – 90 – 11=39 Z
=
12
Sin40
= 12.26
sin39
32. A tree stands vertically on a sloping hillside. At a distance of 16 m down the hill, the tree subtends an angle of 34 degrees. If the inclination of the hill is 20 degrees. Find the height of the tree. A. 12.5 m
B. 13.4 m
C. 14.3 m
=
=14.3m
D. 15.2 m
Solution: 16 Sin56
h sin14
33. Robin flies to San Francisco from Santa Barbara in 3 hours. He flies back in 2 hours. If the wind are blowing from the north at the velocity of 40mph going but changed in 20mph from north returning. What was the air speed of the plane.
A. 140mph
B. 150mph
C. 160mph
D. 170mph
Solution: (x-3) (40) = (x+2)(20) 40x-120 = 20x+40 x= 40+ 120 x= 160mph 34. What would happen in the volume of a sphere if the radius is tripled? A. Multiplied by 3
B. Multiplied by 9
C. Multiplied by 27
D. Multiplied by 6
Solution: V=4/3 πr^3 V(3) = 4/5π(3)^3 = 4/3π = 27
therefore: multiplied by 27
35 The distance between the center of the 3 circles which are mutually tangent each other are 10,12, and 14 units. Find the area of the largest circle. A. 72pi
B. 64pi
C. 23pi
Solution: A= πr^2 A= π(8)^2
=64π
D. 16pi
36. What is the vector which is orthogonal both to 9i + 9j and 9i + 9k? A. 81i+ 81j – 81k
B. 81i - 81j – 81k
C. 81i - 81j + 81k
D. 81 i+ 81j + 81k
Solution: (9i + 9j) (9i + 9k) = 81(i + j) (i + k) = 81 (i –j –k) = 81 -81i -81k 37. Good costs in merchants P72 at what price should he mark them so that he may sell the at discount of P10 form marked price and still make a profit of 20% on the selling price? A. P150
B. P200
C. P100
Capital
P72
Worked price
X
D. P250
Solution:
Selling price
0.20X
Profit
0.20(0.90)
Profit = Income + capital = 0.20 (0.90) = 0.90X-72
X= 100
38. A ranch has cattle and horses in a ratio of 9.5. If there are 80 more heads of a cattle than horses. How many animals are on the ranch? Solution: (9/5) = (x+80/x) 9x = 5(x+80) 9x-5x = 400 X = 100 + 80 Y = 100 Total= 180+100 = 280 A. 140
B. 150
C. 238
D. 280
39. A group of students plan to pay equal amount in hiring a vehicle for an excursion trip at a cost of P6000. However, by adding two more students to the original group, th cost of each student will be reduced by P150. Find the number of students in the original group. A. 10
B. 9
C. 8
D. 7
Solution: 6000= n (2a1 + (n-1) ( 600 ) 2
n= 8 n
40. The volume of the sphere is 36pi cu.m. The surface area of the sphere in sq. m is. A. 36pi
B. 24pi
C. 18pi
D. 12pi
Solution: V = 4/3 πr^3 36π = 4/3 πr^3 r=3 A = 4π(3)^2 A = 36π 41. The logarithm of MN is 6 and the logarithm of N/M is 2. Find the value of logarithm of N. A. 3
55. In Jones family, each daughter has as many brothers as sisters and each son has three times as many sisters as brothers. How many daughters and sons are there in the Jones family? A. 3, 2 B. 4, 2 C. 5, 2 D. 6, 3 Solution: 𝐺 = 𝑛𝑜. 𝑜𝑓 𝑠𝑖𝑠𝑡𝑒𝑟𝑠 𝐺−1=𝐵 3(𝐵 − 1) = 𝐺
𝑦 = 0.01333𝑥 2 + 3 𝑦 = 0.01333(15) + 3 𝑦 = 5.99 ≈ 𝟔𝒎 63. A target with a black circular center and a white ring of uniform width is to be made. If the radius of the center is to be 3 cm, how wide should the ring be so that the area of the ring is the same as the area of the center? A. 1.232 𝑐𝑚 B. 1.263 𝑐𝑚 C. 1,252 𝑐𝑚 D. 1.243 𝑐𝑚
Solution: 70. Two buildings with flat roofs are 60 m apart. From the roof of the shorter building 40 m in height, the angle of elevation to the edge of the roof of the taller building is 40°. How high is the taller building? A. 60 m B. 70 m C. 80 m D. 90 m tan 40 =
x
𝑥 60
40 = 𝜃
𝑥 = (tan 40)(60) 𝑥 = 50 𝐻𝑡𝑎𝑙𝑙 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 = 40 + 50 = 𝟗𝟎 𝒎
40m
60m
71. Three ships are situated as follows A is 225 mi due north of C, and B is 375 mi due to east of C. What is the bearing of B from A? A. N 56° E
B. S 56° E
C. N 59° E
D. S 59° E
Solution: 𝐭𝐚𝐧 θ =
225 375
225
= tan−1 375 = 30. 96
θ = 90° − 30. 96° = 𝟓𝟗. 𝟎𝟒 ∴ Bearing of B from A is 𝐒 𝟓𝟗° 𝐄
72. The longest diagonal of a cube is 6 cm. The total area of the cube is A. 32√2 sq. m
B. 72 sq. m
C. 24√2 sq. m
D. 36 sq. m
Solution: 𝐴𝑆 = 6 𝑎2 𝑑 = √3 𝑎 𝑎=
𝑑 √3
6 √3
=
= 2√3
𝐴𝑆 = 6 (2√3)2 = 𝟕𝟐 𝒎𝟐 73. A support wire is anchored 12 m up from the base of a flagpole and the wire makes a 15° angle with the ground. How long is the wire? A. 12 m
B. 92 m
C. 46 m
D. 24 m
Solution: Tan 15° =
12 𝑎𝑑𝑗
𝑎𝑑𝑗 = 44. 78 𝑚 𝑐 = √44. 782 + 122 = 𝟒𝟔. 𝟑𝟓 𝒎 ∴ 𝑤𝑖𝑟𝑒 𝑖𝑠 𝟒𝟔 𝒎 𝑙𝑜𝑛𝑔
74. A motorboat weighs 32000 lb and its motor provides a thrust of 5000 lb. Assume that the water resistance is 𝑑𝑣 100 pounds for each foot per second of the speed v of the boat. Then 1000 𝑑𝑡 = 5000 – 100 v. If the boats starts from the rest, what is the maximum velocity that it can attain? A. 20 ft/s
77. A snack machine accepts only quarters. Candy bars cost 25₵ packages of peanuts cost 75₵ and cans of cola cost 50₵. How many quarters are needed to buy two candy bars, one package of peanuts and one can of cola? A. 8
B. 7
C. 6
D. 5
Solution:
78. A ball is dropped from a height of 18 m. On each rebound it rises 2/3 of the height from which it last fell. What is the total distance it travels in coming to rest? A. 80 m
B. 90 m
C. 72 m
D. 86 m
Solution:
79. Find the work done in moving an object along the vector a=3i + 4j if the force applied is b= 2i + j A. 11.2
B. 10
C.12.6
D. 9
Solution: 𝑊 = 𝐹 𝑥 𝑣 = ( 3𝑖 + 4𝑗 )( 2𝑖 + 𝑗) = 𝟏𝟎
80. By stringing together 9 differently colored beads. How many different bracelets can be made? A. 362, 880
∴ 𝒚′ = 𝒎 = 𝒔𝒍𝒐𝒑𝒆 = 𝟑 84. Points A and B are 100 m apart and are of the same elevation as the foot of the building. The angles of elevation of the top of the building from points A and B are 21 degrees and 32 degrees respectively. How far is A from the building? A. 259.28 m B. 265.42 m C. 271.62 m D. 277.92 m
91. A toll road averages 300,000 cars a day when the toll is $2.00 per car. A study has shown that for each 10-cent increase in the toll, 10,000 fewer cars will use the road each day. What toll will maximize the revenue? A. $2.25 B. $2.75 C. $3.00 D. $2.50 Solution: Let: n = cars
93. Find the eccentricity of a hyperbola whose transverse and conjugate axes are equal in length. A. √𝟐 Solution:
B. √3
C. 2 √2
D. 2 √3
(x2/a2) – (y2/b2) = 1
e=
√𝑎2 +𝑏2 𝑎 a=b
e=
√2𝑎2
e=a
𝑎 √2
𝑎 e = √𝟐
94. For what values of x is |x-3| = 1? A. 4 B. 2
C. 2, 4
D. -2, -4
Solution: By inspection and substituting all the given in the equation: |x-3| = 1 |x-3| = 1 |2-3| = 1 |4-3| = 1 |1|= 1 |1|= 1 95. Susan's age in 20 years will be the same as Thelma's age now. Ten years from now, Thelma's age will be twice Susan's. What is the present age of Susan? A. 45 B. 40 C. 50 D. 30 Solution:
1. Joseph gave ¼ of his candies to Joy and Joy gave 1/5 of what she got to Tim. If Tim received 2 candies, how many candies did Joseph have originally? A. 30 B. 20 C. 50 D. 40 2. What conic section is described by the equation 4x2-y2+8x+4y=15? A. parabola B. hyperbola C. circle D. ellipse 3. Find the maximum area of a rectangle which can be inscribed in an ellipse having the equation x2 + 4y2 = 4 A. 4 B. 3 C. 2 D. 5
15. What is the length of the shortest line segment in the first quadrant drawn tangent to the ellipse b2x2 + a2y2 = a2b2 and meeting to the coordinates axes? A. a/b B. a + b C. ab D. b/a 16. Find the radius of the circle inscribed in the triangle determined by the lines y=x+4, y= x-4 and y = 7x-2. 5
25. Melissa is 4 times as old as Jun. Pat is 5 years older than Melissa. If Jim is y, how old is Pat? A. 4y + 5 B. y + 5 C. 5y + 4 D. 4 + 5y Sol'n: Melissa – 4y Jim – y Pat – 4y + 5 Therefore, Pat = 4y + 5 26. A conic section whose eccentricity, is less than one is known as: A. a parabola B. an ellipse C. a circle D. a hyperbola 27. Two lines passing through the point (2,3) make an angle of 45 degrees with each other. If the pipe of one of the lines is 2, find the slope of the other. A. -2 B. -1 C. -3 D. 0 Sol'n: (2,3) 𝜃 = 45 m1= 2 Tan 𝜃 = m2 –m1 / 1+ m2 m1 Tan 45 = m2 –2 / 1+ m2 (2) m2 = -3 28. From the top of a building the angle of depression of the foot of a pole is 48 deg 10 min. From the foot of a building the angle of elevation of the top of a pole is 18 deg 50min. Both building and pole are on a level ground. If the height of a pole is 4m, how high is the building? A. 13.10m B. 12.10m C. 10.90m D. 11.60m Sol'n: Tan 𝜃 = y / x x= y / Tan 𝜃 = 4 / tan 18°50' x= 12.13
38. An equilateral triangle with side "a" is revolved about its altitude. Find the volume of the solid generated. A. 0.32a3
B. 0.23a3
C. 0.41a3
D. 0.14a3
39. If the area bounded by the parabolas y=x2-C2 and y=C2-x2 is 576 square units, find the value of C. A. 5
B. 6
C. 7
D. 8
40. Solve y"-5y'+4y = sin 3x. 1
A. y= 25 (3 cos 3𝑥 − sin 3𝑥) + 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 4𝑥 1
B. y= 25 (3 sin 3𝑥 − cos 3𝑥) + 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 4𝑥 𝟏
C. y= 𝟓𝟎 (𝟑 𝐜𝐨𝐬 𝟑𝒙 − 𝐬𝐢𝐧 𝟑𝒙) + 𝑪𝟏 𝒆𝒙 + 𝑪𝟐 𝒆𝟒𝒙 1
D. y= 50 (3 sin 3𝑥 − cos 3𝑥) + 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 4𝑥 41. A car is travelling at a rate of 36 m/s towards a statue of height 6m. What is the rate of change of a distance of the car towards the top of the statue when it is 8m from the statue? A. 32.4 m/s
ℒ(t cos kt) = (𝐬𝟐+𝐤𝟐)𝟐 44. Carmela and Marian got summer jobs at the ice cream shop and were supposed to work 15 hours per week each for 8 weeks. During that time Marian was ill for one week and Carmela took her shifts. How many hours did Carmela work during the 8 weeks? A. 120
B. 135
Solution: Total hours in 8 weeks 15 ℎ𝑜𝑢𝑟𝑠 𝑤𝑒𝑒𝑘
𝑥 8 𝑤𝑒𝑒𝑘𝑠 = 120 ℎ𝑜𝑢𝑟𝑠
C. 150
D. 185
Total hours Carmela works when Marian was ill for 1 week 120 hours + 15 hours = 135 hours 45. Manuelita had 75 stuffed animals. Her grandmother gave 15 of them to her. What percentage of the stuffed animals did her grandmother give her? A. 20%
B. 15%
C. 25%
D. 10%
Solution: 75 15
=5
100% / 5 = 20 % 46. Find the coordinates of an object that has been displaced from the point (-4,9) by the vector 4i-5i. A. (0,4)
54. The Rotary Club and the Jaycee Club had a joint party, 120 members of the Rotary Club and 100 members of the Jaycees Club also attended but 30 of those attended are members of both clubs. How many persons attended the party? A. 220
B. 190
C. 150
D. 250
55. Two numbers have a harmonic mean of 9 and a geometric mean of 6. Determine the arithmetic mean. A. ¼
B. 4
C. 1/9
D. 9
Solution: HM = 9 GM = 6 GM2 = (HM)(AM) AM =
𝐺𝑀2 𝐻𝑀
=
62 9
=4
56. Find the force on one force of a right triangle of sides 4m and altitude of 3m. The altitude is submerged vertically with the 4m side in the surface. A. 58.86 kN
B. 62.64 kN
Solution: W(0) = 4 m W(3) = 0 0−4 3−0
=
−4 3 4
W (h) = 4 - 3 ℎ F = ∫ 𝛾𝐻2𝑂 ℎ 𝑤(ℎ) 𝑑ℎ 3
F = ∫0 (9810)(ℎ) (4 −
4 3
ℎ ) 𝑑ℎ
C. 53.22 kN
D. 66.67 kN
3
F = (9810) ∫0 (4ℎ −
4 3
𝑥 2 ) 𝑑ℎ
= 58.86 kN
57. An airplane flying with the wind, took 2 hours to travel 1000 km and 2.5hours in flying back. What was the wind velocity in kph? A. 40
k = −𝟐 62. What is the allowable error in measuring the edge of the cube that is intended to hold 8 cu. M. of the error of the computed volume is not to exceed 0.03 cu. m? A. 0.002 Solution:
B. 0.003
C. 0.0025
D. 0.001
3
3
Edge = √𝑣 = √8
=2
dv = 3E2dE 𝑑𝑣
0.03 (3)(2)2
dE = 3𝐸2 = dE = 0.0025
63. A man can do a job in 8 days. After the man has worked for 3 days, his son joins him together they complete the job in 3 more days. How long will it take the son to do job alone? A. 12 days
B. 10 days
C. 13 days
D. 11 days
Solution: Let x = For son Man = 1/8 1
Son = 1/x 1
1
3 (8) + 3 (𝑥 + 8) = 1 3
8x (8 +
3 𝑥
+
3 8
= 1) 8𝑥
3x + 24 + 3x = 8x 6X + 24 + 8X X = 12 days 64. The probability that a randomly chosen safes prospects will make a purchase is 0.18. If a salesman calls on 5 prospects, what is the probability that the salesmen will make exactly 3 sales? A. 0.0392
B. 0.0239
Solution: ( 5 C3 ) ( 0.18 )3 (1 – 0.18 )2
C. 0.0329
D. 0.0293
X = 0.0392 5
65. If 𝑠𝑒𝑐 2 𝐴 = 2 , 𝑡ℎ𝑒𝑛 1 − 𝑠𝑖𝑛2 𝐴 = A. 0.20
B. 0.30
C. 0.40
D. 0.50
Solution: 5
𝑠𝑒𝑐 2 𝐴 = 2
------- 1
1 − 𝑠𝑖𝑛2 𝐴 = 𝑐𝑜𝑠 2 𝐴 𝑠𝑒𝑐 2 𝐴 + 𝑐𝑜𝑠 2 𝐴 cos 𝐴 =
1 𝑠𝑒𝑐𝐴
--------- 2
=1
= 𝑐𝑜𝑠 2 𝐴 =
1 − 𝑠𝑖𝑛2 𝐴 =
1 𝑠𝑒𝑐 2 𝐴
=
1 5 2
1 𝑠𝑒𝑐 2 𝐴
= 𝟎. 𝟒𝟎
66. What is the angle between the diagonal of a cube and one of its edges? A. 44.74°
B. 54.74°
C. 64.74°
D. 74.74°
Solution: A (1, 1, 1) B (0, 0, 1) Cos𝜃 =
(𝑎)(𝑏) 𝑙𝑎𝑙𝑙𝑏𝑙
= cos −1(
(1,1,1)(0,0,1) √3
)
𝜽 = 𝟓𝟒. 𝟕 °
67. The line 3x-4y=5 is perpendicular to the line A. 3x-4y=1 B. 4x-3y=1 Solution : 3x-4y=5 4y= 3x-5
72. A person had a rectangular-shaped garden with sides of lengths 16 feet and 9 feet. The garden was changed into square design with the same area as the original rectangularshaped garden. How many feet in length are each sides of the new square-shape garden. A. 7
74. Martin , a motel housekeeper, has finished cleaning about 40% of the 32 rooms he's been assigned. About how many more rooms does he have left to clean? A. 29
B. 25
Solution: Room left to clean = 60% (30) = 19 Room
C. 21
D. 19
75. A horse tied to a post with twenty-foot rope. What is the longest path that the horse can walk? A. 20 feet
B. 40 feet
C. 62.83 feet
D.125.66feet
76. Doming wants to know the height of a telephone pole. He measures his shadow, which is 3 feet long , and the pole's shadow, whcih 10 feet long . Domingo's height is 6 feet. How tall is the pole ? A. 40 ft
B. 30 ft
C. 20 ft
D. 10 ft
77. A weight of 60 pounds rests on the end of an 8-foot lever and is 3 feet from the fulcrum. What weight must be placed on the other and of the lever to balance the 60 pound weight. A. 36pounds
B. 32pounds
C. 40pounds
D. 46pounds
Solution : 5x =60(3) =180 X= 36 78. A number is 1 more than twice another. Their squares differ by 176. What is the larger number? A. 9
B. 7
C. 15
D. 16
79. The sides of a right triangle is in arithmetic progression whose common difference is 6cm. Its area is A. 216sq.cm
B. 270sq.cm
C. 360sq.cm
D. 144sq.cm
Solution : A.P. (x)
(x+12) ^2 = x^2 + (x+6)^2
(x+6)
X^2+24x+144 = x^2 + x^2 +12x +36
(x+12) ---- hypo
X^2 – 12x – 108 = 0
C^2 = A^2 + B^2
X^2 – 18x + 6x – 108 = 0 X(x-18) + 6(x-18)=0 (x-18)(x+6)=0
X=-6 or 18 Area = ½ (18)(24) = 216
80. A tank has 100 liters of brine with 40 N dissolved salt. Pure water enters the tank at the rate of 2 liters per minute abd the resulting mixture leaves the tank at the same rate. When will the concentration in the tank be 0.20 N/L A. 24.6min | 677.169 | 1 |
23.000 --> 00:00:27.000
So, last week we learned how to
do triple integrals in
00:00:27.000 --> 00:00:31.000
rectangular and cylindrical
coordinates.
00:00:31.000 --> 00:00:41.000
And, now we have to learn about
spherical coordinates,
00:00:41.000 --> 00:00:49.000
which you will see are a lot of
fun.
00:00:49.000 --> 00:00:53.000
So, what's the idea of
spherical coordinates?
00:00:53.000 --> 00:00:58.000
Well, you're going to represent
a point in space using the
00:00:58.000 --> 00:01:02.000
distance to the origin and two
angles.
00:01:02.000 --> 00:01:06.000
So, in a way,
you can think of these as a
00:01:06.000 --> 00:01:09.000
space analog of polar
coordinates because you just use
00:01:09.000 --> 00:01:12.000
distance to the origin,
and then you have to use angles
00:01:12.000 --> 00:01:15.000
to determine in which direction
you're going.
00:01:15.000 --> 00:01:22.000
So, somehow they are more polar
than cylindrical coordinates.
00:01:22.000 --> 00:01:25.000
So, how do we do that?
So, let's say that you have a
00:01:25.000 --> 00:01:29.000
point in space at coordinates x,
y, z.
00:01:29.000 --> 00:01:32.000
Then, instead of using x,
y, z, you will use,
00:01:32.000 --> 00:01:37.000
well, one thing you'll use is
the distance from the origin.
00:01:37.000 --> 00:01:41.000
OK, and that is denoted by the
Greek letter which looks like a
00:01:41.000 --> 00:01:43.000
curly p, but actually it's the
Greek R.
00:01:43.000 --> 00:01:57.000
So -- That's the distance from
the origin.
00:01:57.000 --> 00:02:02.000
And so, that can take values
anywhere between zero and
00:02:02.000 --> 00:02:06.000
infinity.
Then, we have to use two other
00:02:06.000 --> 00:02:09.000
angles.
And, so for that,
00:02:09.000 --> 00:02:16.000
let me actually draw the
vertical half plane that
00:02:16.000 --> 00:02:22.000
contains our point starting from
the z axis.
00:02:22.000 --> 00:02:25.000
OK, so then we have two new
angles.
00:02:25.000 --> 00:02:27.000
Well, one of them is not really
new.
00:02:27.000 --> 00:02:30.000
One is new.
That's phi is the angle
00:02:30.000 --> 00:02:34.000
downwards from the z axis.
And the other one,
00:02:34.000 --> 00:02:39.000
theta, is the angle
counterclockwise from the x
00:02:39.000 --> 00:02:51.000
axis.
OK, so phi, let me do it better.
00:02:51.000 --> 00:02:54.000
So, there's two ways to draw
the letter phi,
00:02:54.000 --> 00:02:56.000
by the way.
And, I recommend this one
00:02:56.000 --> 00:02:58.000
because it doesn't look like a
rho.
00:02:58.000 --> 00:03:07.000
So, that's easier.
That's the angle that you have
00:03:07.000 --> 00:03:13.000
to go down from the positive z
axis.
00:03:13.000 --> 00:03:17.000
And,
so that angle varies from zero
00:03:17.000 --> 00:03:23.000
when you're on the z axis,
increase to pi over two when
00:03:23.000 --> 00:03:28.000
you are on the xy plane all the
way to pi or 180� when you are
00:03:28.000 --> 00:03:38.000
on the negative z axis.
It doesn't go beyond that.
00:03:38.000 --> 00:03:44.000
OK, so -- Phi is always between
zero and pi.
00:03:44.000 --> 00:03:46.000
And, finally,
the last one,
00:03:46.000 --> 00:03:51.000
theta, is just going to be the
same as before.
00:03:51.000 --> 00:03:55.000
So, it's the angle after you
project to the xy plane.
00:03:55.000 --> 00:04:00.000
That's the angle
counterclockwise from the x
00:04:00.000 --> 00:04:02.000
axis.
OK, so that's a little bit
00:04:02.000 --> 00:04:05.000
overwhelming not just because of
the new letters,
00:04:05.000 --> 00:04:07.000
but also because there is a lot
of angles in there.
00:04:07.000 --> 00:04:11.000
So, let me just try to,
you know, suggest two things
00:04:11.000 --> 00:04:13.000
that might help you a little
bit.
00:04:13.000 --> 00:04:17.000
So, one is, these are called
spherical coordinates because if
00:04:17.000 --> 00:04:21.000
you fix the value of rho,
then you are moving on a sphere
00:04:21.000 --> 00:04:27.000
centered at the origin.
OK, so let's look at what
00:04:27.000 --> 00:04:34.000
happens on a sphere centered at
the origin, so,
00:04:34.000 --> 00:04:41.000
with equation rho equals a.
Well, then phi measures how far
00:04:41.000 --> 00:04:44.000
south you are going,
measures the distance from the
00:04:44.000 --> 00:04:46.000
North Pole.
So,
00:04:46.000 --> 00:04:48.000
if you've learned about
latitude and longitude in
00:04:48.000 --> 00:04:51.000
geography,
well, phi and theta you can
00:04:51.000 --> 00:04:54.000
think of as latitude and
longitude except with slightly
00:04:54.000 --> 00:04:59.000
different conventions.
OK, so, phi is more or less the
00:04:59.000 --> 00:05:05.000
same thing as latitude in the
sense that it measures how far
00:05:05.000 --> 00:05:09.000
north or south you are.
The only difference is in
00:05:09.000 --> 00:05:12.000
geography,
latitude is zero on the equator
00:05:12.000 --> 00:05:16.000
and becomes something north,
something south,
00:05:16.000 --> 00:05:18.000
depending on how far you go
from the equator.
00:05:18.000 --> 00:05:21.000
Here, you measure a latitude
starting from the North Pole
00:05:21.000 --> 00:05:24.000
which is zero,
increasing all the way to the
00:05:24.000 --> 00:05:29.000
South Pole, which is at pi.
And, theta or you can think of
00:05:29.000 --> 00:05:33.000
as longitude,
which measures how far you are
00:05:33.000 --> 00:05:36.000
east or west.
So, the Greenwich Meridian
00:05:36.000 --> 00:05:39.000
would be here,
now, the one on the x axis.
00:05:39.000 --> 00:05:46.000
That's the one you use as the
origin for longitude,
00:05:46.000 --> 00:05:48.000
OK?
Now, if you don't like
00:05:48.000 --> 00:05:51.000
geography, here's another way to
think about it.
00:05:51.000 --> 00:05:56.000
So -- Let's start again from
cylindrical coordinates,
00:05:56.000 --> 00:06:02.000
which hopefully you're kind of
comfortable with now.
00:06:02.000 --> 00:06:06.000
OK, so you know about
cylindrical coordinates where we
00:06:06.000 --> 00:06:09.000
have the z coordinates stay z,
and the xy plane we do R and
00:06:09.000 --> 00:06:13.000
theta polar coordinates.
And now, let's think about what
00:06:13.000 --> 00:06:18.000
happens when you look at just
one of these vertical planes
00:06:18.000 --> 00:06:21.000
containing the z axis.
So, you have the z axis,
00:06:21.000 --> 00:06:25.000
and then you have the direction
away from the z axis,
00:06:25.000 --> 00:06:29.000
which I will call r,
just because that's what r
00:06:29.000 --> 00:06:32.000
measures.
Of course, r goes all around
00:06:32.000 --> 00:06:35.000
the z axis, but I'm just doing a
slice through one of these
00:06:35.000 --> 00:06:39.000
vertical half planes,
fixing the value of theta.
00:06:39.000 --> 00:06:43.000
Then, r of course is a polar
coordinate seen from the point
00:06:43.000 --> 00:06:46.000
of view of the xy plane.
But here, it looks more like
00:06:46.000 --> 00:06:48.000
you have rectangular coordinates
again.
00:06:48.000 --> 00:06:51.000
So the idea of spherical
coordinate is you're going to
00:06:51.000 --> 00:06:54.000
polar coordinates again in the
rz plane.
00:06:54.000 --> 00:07:01.000
OK, so if I have a point here,
then rho will be the distance
00:07:01.000 --> 00:07:06.000
from the origin.
And phi will be the angle,
00:07:06.000 --> 00:07:10.000
except it's measured from the
positive z axis,
00:07:10.000 --> 00:07:16.000
not from the horizontal axis.
But, the idea in here,
00:07:16.000 --> 00:07:18.000
see,
let me put that between quotes
00:07:18.000 --> 00:07:21.000
because I'm not sure how correct
that is,
00:07:21.000 --> 00:07:28.000
but in a way,
you can think of this as polar
00:07:28.000 --> 00:07:34.000
coordinates in the rz plane.
So, in particular,
00:07:34.000 --> 00:07:38.000
that's the key to understanding
how to switch between spherical
00:07:38.000 --> 00:07:41.000
coordinates and cylindrical
coordinates,
00:07:41.000 --> 00:07:44.000
and then all the way to x,
y, z if you want,
00:07:44.000 --> 00:07:48.000
right,
because this picture here tells
00:07:48.000 --> 00:07:53.000
us how to express z and r in
terms of rho and phi.
00:07:53.000 --> 00:08:03.000
So, let's see how that works.
If I project here or here,
00:08:03.000 --> 00:08:12.000
so, this line is z.
But, it's also rho times cosine
00:08:12.000 --> 00:08:19.000
phi.
So, I get z equals rho cos phi.
00:08:19.000 --> 00:08:21.000
And, if I look at r,
it's the same thing,
00:08:21.000 --> 00:08:31.000
but on the other side.
So, r will be rho sine phi.
00:08:31.000 --> 00:08:34.000
OK, so you can use this to
switch back and forth between
00:08:34.000 --> 00:08:37.000
spherical and cylindrical.
And of course,
00:08:37.000 --> 00:08:43.000
if you remember what x and y
were in terms of r and theta,
00:08:43.000 --> 00:08:49.000
you can also keep doing this to
figure out, oops.
00:08:49.000 --> 00:08:57.000
So, x is r cos theta.
That becomes rho sine phi cos
00:08:57.000 --> 00:09:01.000
theta.
Y is r sine theta.
00:09:01.000 --> 00:09:06.000
So, that becomes rho sine phi
sine theta.
00:09:06.000 --> 00:09:15.000
And z is rho cos phi.
But, basically you don't really
00:09:15.000 --> 00:09:19.000
need to remember these formulas
as long as you remember how to
00:09:19.000 --> 00:09:22.000
express r in terms of rho sine
phi,
00:09:22.000 --> 00:09:29.000
and x equals r cos theta.
So, now, of course,
00:09:29.000 --> 00:09:31.000
we're going to use spherical
coordinates in situations where
00:09:31.000 --> 00:09:33.000
we have a lot of symmetry,
and in particular,
00:09:33.000 --> 00:09:35.000
where the z axis plays a
special role.
00:09:35.000 --> 00:09:38.000
Actually, that's the same with
cylindrical coordinates.
00:09:38.000 --> 00:09:40.000
Cylindrical and secure
coordinates are set up so that
00:09:40.000 --> 00:09:44.000
the z axis plays a special role.
So, that means whenever you
00:09:44.000 --> 00:09:47.000
have a geometric problem,
and you are not told how to
00:09:47.000 --> 00:09:51.000
choose your coordinates,
it's probably wiser to try to
00:09:51.000 --> 00:09:57.000
center things on the z axis.
That's where these coordinates
00:09:57.000 --> 00:10:01.000
are the best adapted.
And,
00:10:01.000 --> 00:10:03.000
in case you ever need to switch
backwards,
00:10:03.000 --> 00:10:07.000
I just want to point out,
so, rho is the square root of r
00:10:07.000 --> 00:10:11.000
squared plus z squared,
which means it's the square
00:10:11.000 --> 00:10:15.000
root of x squared plus y squared
plus z squared.
00:10:15.000 --> 00:10:20.000
OK, so that's basically all the
formulas about spherical
00:10:20.000 --> 00:10:28.000
coordinates.
OK, any questions about that?
00:10:28.000 --> 00:10:31.000
OK, let's see,
who had seen spherical
00:10:31.000 --> 00:10:34.000
coordinates before just to see?
OK, that's not very many.
00:10:34.000 --> 00:10:36.000
So, I'm sure for,
one of you saw it twice.
00:10:36.000 --> 00:10:42.000
That's great.
Sorry, oops,
00:10:42.000 --> 00:10:49.000
OK, so let's just look quickly
at equations of some of the
00:10:49.000 --> 00:10:54.000
things.
So, as I've said,
00:10:54.000 --> 00:11:02.000
if I set rho equals a,
that will be just a sphere of
00:11:02.000 --> 00:11:11.000
radius a centered at the origin.
More interesting things:
00:11:11.000 --> 00:11:14.000
let's say I give you phi equals
pi over four.
00:11:14.000 --> 00:11:18.000
What do you think that looks
like?
00:11:18.000 --> 00:11:29.000
Actually, let's take a quick
poll on things.
00:11:29.000 --> 00:11:31.000
OK, yeah, everyone seems to be
saying it's a cone,
00:11:31.000 --> 00:11:33.000
and that's indeed the correct
answer.
00:11:33.000 --> 00:11:41.000
So, how do we see that?
Well, remember,
00:11:41.000 --> 00:11:44.000
phi is the angle downward from
the z axis.
00:11:44.000 --> 00:11:49.000
So, let's say that I'm going to
look first at what happens if
00:11:49.000 --> 00:11:53.000
I'm in the right half of a plane
of a blackboard,
00:11:53.000 --> 00:11:56.000
so, in the yz plane.
Then, phi is the angle downward
00:11:56.000 --> 00:11:58.000
from here.
So, if I want to get pi over
00:11:58.000 --> 00:12:01.000
four, that's 45�.
That means I'm going to go
00:12:01.000 --> 00:12:03.000
diagonally like this.
Of course, if I'm in the left
00:12:03.000 --> 00:12:06.000
half of a plane of a blackboard,
it's going to be the same.
00:12:06.000 --> 00:12:10.000
I also take pi over four.
And, I get the other half.
00:12:10.000 --> 00:12:13.000
And, because the equation does
not involve theta,
00:12:13.000 --> 00:12:17.000
it's all the same if I rotate
my vertical plane around the z
00:12:17.000 --> 00:12:21.000
axis.
So, I get the same picture in
00:12:21.000 --> 00:12:27.000
any of these vertical half
planes, actually.
00:12:27.000 --> 00:12:32.000
OK, now, so this is phi equals
pi over four.
00:12:32.000 --> 00:12:35.000
And, just in case,
to point out to you what's
00:12:35.000 --> 00:12:39.000
going on, when phi equals pi
over four, cosine and sine are
00:12:39.000 --> 00:12:42.000
equal to each other.
They are both one over root two.
00:12:42.000 --> 00:12:46.000
So, you can find,
again, the equation of this
00:12:46.000 --> 00:12:51.000
thing in cylindrical
coordinates, which I'll remind
00:12:51.000 --> 00:12:54.000
you was z equals r.
OK, in general,
00:12:54.000 --> 00:12:58.000
phi equals some given number,
or z equals some number times
00:12:58.000 --> 00:13:01.000
r.
That will be a cone centered on
00:13:01.000 --> 00:13:04.000
the z axis.
OK, a special case:
00:13:04.000 --> 00:13:07.000
what if I say phi equals pi
over two?
00:13:07.000 --> 00:13:09.000
Yeah, it's just going to be the
xy plane.
00:13:09.000 --> 00:13:13.000
OK, that's the flattest of all
cones.
00:13:13.000 --> 00:13:20.000
OK, so phi equals pi over two
is going to be just the xy
00:13:20.000 --> 00:13:22.000
plane.
And, in general,
00:13:22.000 --> 00:13:24.000
if phi is less than pi over
two, then you are in the upper
00:13:24.000 --> 00:13:28.000
half space.
If phi is more than pi over
00:13:28.000 --> 00:13:32.000
two, you'll be in the lower half
space.
00:13:32.000 --> 00:13:36.000
OK, so that's pretty much all
we need to know at this point.
00:13:36.000 --> 00:13:45.000
So, what's next?
Well, remember we were trying
00:13:45.000 --> 00:13:52.000
to do triple integrals.
So now we're going to triple
00:13:52.000 --> 00:13:59.000
integrals in spherical
coordinates.
00:13:59.000 --> 00:14:01.000
And, for that,
we first need to understand
00:14:01.000 --> 00:14:06.000
what the volume element is.
What will be dV?
00:14:06.000 --> 00:14:12.000
OK, so dV will be something,
d rho, d phi,
00:14:12.000 --> 00:14:18.000
d theta, or in any order that
you want.
00:14:18.000 --> 00:14:23.000
But, this one is usually the
most convenient.
00:14:23.000 --> 00:14:27.000
So, to find out what it is,
well, we should look at how we
00:14:27.000 --> 00:14:29.000
are going to be slicing things
now.
00:14:29.000 --> 00:14:32.000
OK, so if you integrate d rho,
d phi, d theta,
00:14:32.000 --> 00:14:37.000
it means that you are actually
slicing your solid into little
00:14:37.000 --> 00:14:40.000
pieces that live,
somehow,
00:14:40.000 --> 00:14:45.000
if you set an interval of rows,
OK,
00:14:45.000 --> 00:14:48.000
sorry, maybe I should,
so, if you first integrate over
00:14:48.000 --> 00:14:51.000
rho,
it means that you will actually
00:14:51.000 --> 00:14:57.000
choose first the direction from
the origin even by phi and
00:14:57.000 --> 00:15:00.000
theta.
And, in that direction,
00:15:00.000 --> 00:15:04.000
you will try to figure out,
how far does your region
00:15:04.000 --> 00:15:07.000
extend?
And, of course,
00:15:07.000 --> 00:15:11.000
how far that goes might depend
on phi and theta.
00:15:11.000 --> 00:15:16.000
Then, you will vary phi.
So, you have to know,
00:15:16.000 --> 00:15:21.000
for a given value of theta,
how far down does your solid
00:15:21.000 --> 00:15:22.000
extend?
And, finally,
00:15:22.000 --> 00:15:25.000
the value of theta will
correspond to,
00:15:25.000 --> 00:15:28.000
in which directions around the
z axis do we go?
00:15:28.000 --> 00:15:31.000
So, we're going to see that in
examples.
00:15:31.000 --> 00:15:34.000
But before we can do that,
we need to get the volume
00:15:34.000 --> 00:15:36.000
element.
So, what I would like to
00:15:36.000 --> 00:15:40.000
suggest is that we need to
figure out,
00:15:40.000 --> 00:15:46.000
what is the volume of a small
piece of solid which corresponds
00:15:46.000 --> 00:15:49.000
to a certain change,
delta rho,
00:15:49.000 --> 00:15:52.000
delta phi,
and delta theta?
00:15:52.000 --> 00:15:56.000
So, delta rho means that you
have two concentric spheres,
00:15:56.000 --> 00:16:01.000
and you are looking at a very
thin shell in between them.
00:16:01.000 --> 00:16:05.000
And then, you would be looking
at a piece of that spherical
00:16:05.000 --> 00:16:08.000
shell corresponding to small
values of phi and theta.
00:16:08.000 --> 00:16:14.000
So, because I am stretching the
limits of my ability to draw on
00:16:14.000 --> 00:16:18.000
the board, here's a picture.
I'm going to try to reproduce
00:16:18.000 --> 00:16:21.000
on the board,
but so let's start by looking
00:16:21.000 --> 00:16:24.000
just at what happens on the
sphere of radius a,
00:16:24.000 --> 00:16:28.000
and let's try to figure out the
surface area elements on the
00:16:28.000 --> 00:16:30.000
sphere in terms of phi and
theta.
00:16:30.000 --> 00:16:39.000
And then, we'll add the rho
direction.
00:16:39.000 --> 00:16:49.000
OK, so -- So,
let me say, let's start by
00:16:49.000 --> 00:17:02.000
understanding surface area on a
sphere of radius a.
00:17:02.000 --> 00:17:12.000
So, that means we'll be looking
at a little piece of the sphere
00:17:12.000 --> 00:17:21.000
corresponding to angles delta
phi and in that direction here
00:17:21.000 --> 00:17:26.000
delta theta.
OK, so when you draw a map of
00:17:26.000 --> 00:17:29.000
the world on a globe,
that's exactly what the grid
00:17:29.000 --> 00:17:33.000
lines form for you.
So, what's the area of this guy?
00:17:33.000 --> 00:17:35.000
Well, of course,
all the sides are curvy.
00:17:35.000 --> 00:17:37.000
They are all on the sphere.
None of them are straight.
00:17:37.000 --> 00:17:41.000
But still, if it's small enough
and it looks like a rectangle,
00:17:41.000 --> 00:17:46.000
so let's just try to figure
out, what are the sides of your
00:17:46.000 --> 00:17:49.000
rectangle?
OK, so, let's see,
00:17:49.000 --> 00:17:55.000
well, I think I need to draw a
bigger picture of this guy.
00:17:55.000 --> 00:17:59.000
OK, so this guy,
so that's a piece of what's
00:17:59.000 --> 00:18:05.000
called a parallel in geography.
That's a circle that goes
00:18:05.000 --> 00:18:07.000
east-west.
So now,
00:18:07.000 --> 00:18:10.000
this parallel as a circle of
radius,
00:18:10.000 --> 00:18:14.000
well, the radius is less than a
because if your vertical is to
00:18:14.000 --> 00:18:17.000
the North Pole,
it will be actually much
00:18:17.000 --> 00:18:19.000
smaller.
So, that's why when you say
00:18:19.000 --> 00:18:22.000
you're going around the world it
depends on whether you do it at
00:18:22.000 --> 00:18:28.000
the equator or the North Pole.
It's much easier at the North
00:18:28.000 --> 00:18:33.000
Pole.
So, anyway, this is a piece of
00:18:33.000 --> 00:18:40.000
a circle of radius,
well, the radius is what I
00:18:40.000 --> 00:18:49.000
would call r because that's the
distance from the z axis.
00:18:49.000 --> 00:18:51.000
OK, that's actually pretty hard
to see now.
00:18:51.000 --> 00:18:58.000
So if you can see it better on
this one, then so this guy here,
00:18:58.000 --> 00:19:03.000
this length is r.
And, r is just rho,
00:19:03.000 --> 00:19:07.000
well, what was a times sine
phi.
00:19:07.000 --> 00:19:09.000
Remember, we have this angle
phi in here.
00:19:09.000 --> 00:19:14.000
I should use some color.
It's getting very cluttered.
00:19:14.000 --> 00:19:19.000
So, we have this phi,
and so r is going to be rho
00:19:19.000 --> 00:19:21.000
sine phi.
That rho is a.
00:19:21.000 --> 00:19:29.000
So, let me just put a sine phi.
OK, and the corresponding angle
00:19:29.000 --> 00:19:32.000
is going to be measured by
theta.
00:19:32.000 --> 00:19:48.000
So, the length of this is going
to be a sine phi delta theta.
00:19:48.000 --> 00:19:54.000
That's for this side.
Now, what about that side,
00:19:54.000 --> 00:19:56.000
the north-south side?
Well, if you're moving
00:19:56.000 --> 00:19:58.000
north-south, it's not like
east-west.
00:19:58.000 --> 00:20:01.000
You always have to go all the
way from the North Pole to the
00:20:01.000 --> 00:20:04.000
South Pole.
So, that's actually a great
00:20:04.000 --> 00:20:08.000
circle meridian of length,
well, I mean,
00:20:08.000 --> 00:20:13.000
well, the radius is the radius
of the sphere.
00:20:13.000 --> 00:20:22.000
Total length is 2pi a.
So, this is a piece of a circle
00:20:22.000 --> 00:20:27.000
of radius a.
And so, now,
00:20:27.000 --> 00:20:34.000
the length of this one is going
to be a delta phi.
00:20:34.000 --> 00:20:41.000
OK, so, just to recap,
this is a sine phi delta theta.
00:20:41.000 --> 00:20:46.000
And, this guy here is a delta
phi.
00:20:46.000 --> 00:20:59.000
So, you can't read it because
it's -- And so,
00:20:59.000 --> 00:21:02.000
that tells us if I take a small
piece of the sphere,
00:21:02.000 --> 00:21:06.000
then its surface area,
delta s,
00:21:06.000 --> 00:21:15.000
is going to be approximately a
sine phi delta theta times a
00:21:15.000 --> 00:21:22.000
delta phi,
which I'm going to rewrite as a
00:21:22.000 --> 00:21:27.000
squared sine phi delta phi delta
theta.
00:21:27.000 --> 00:21:31.000
So, what that means is,
say that I want to integrate
00:21:31.000 --> 00:21:34.000
something just on the surface of
a sphere.
00:21:34.000 --> 00:21:37.000
Well, I would use phi and theta
as my coordinates.
00:21:37.000 --> 00:21:46.000
And then, to know how big a
piece of a sphere is,
00:21:46.000 --> 00:21:55.000
I would just take a squared
sine phi d phi d theta.
00:21:55.000 --> 00:21:59.000
OK, so that's the surface
element in a sphere.
00:21:59.000 --> 00:22:03.000
And now, what about going back
into the third dimension,
00:22:03.000 --> 00:22:05.000
so, adding some depth to these
things?
00:22:05.000 --> 00:22:10.000
Well, I'm not going to try to
draw a picture because you've
00:22:10.000 --> 00:22:17.000
seen that's slightly tricky.
Well, let me try anyway just
00:22:17.000 --> 00:22:24.000
you can have fun with my
completely unreadable diagrams.
00:22:24.000 --> 00:22:28.000
So anyway, if you look at,
now, something that's a bit
00:22:28.000 --> 00:22:33.000
like that piece of sphere,
but with some thickness to it.
00:22:33.000 --> 00:22:38.000
The thickness will be delta
rho, and so the volume will be
00:22:38.000 --> 00:22:44.000
roughly the area of the thing on
the sphere times the thickness.
00:22:44.000 --> 00:22:48.000
So, I claim that we will get
basically the volume element
00:22:48.000 --> 00:22:51.000
just by multiplying things by d
rho.
00:22:51.000 --> 00:23:10.000
So, let's see that.
So now, if I have a sphere of
00:23:10.000 --> 00:23:19.000
radius rho, and another one
that's slightly bigger of radius
00:23:19.000 --> 00:23:27.000
rho plus delta rho,
and then I have a little box in
00:23:27.000 --> 00:23:29.000
here.
Then,
00:23:29.000 --> 00:23:34.000
I know that the volume of this
thing will be essentially,
00:23:34.000 --> 00:23:38.000
well, its thickness,
the thickness is going to be
00:23:38.000 --> 00:23:42.000
delta rho times the area of its
base,
00:23:42.000 --> 00:23:44.000
although it doesn't really
matter,
00:23:44.000 --> 00:23:48.000
which is what we've called
delta s.
00:23:48.000 --> 00:23:55.000
OK, so we will get,
sorry, a becomes rho now.
00:23:55.000 --> 00:23:57.000
Square sine phi,
delta rho,
00:23:57.000 --> 00:24:00.000
delta phi,
delta theta,
00:24:00.000 --> 00:24:04.000
and so out of that we get the
volume element and spherical
00:24:04.000 --> 00:24:08.000
coordinates,
which is rho squared sine phi d
00:24:08.000 --> 00:24:09.000
rho,
d phi,
00:24:09.000 --> 00:24:14.000
d theta.
And, that's a formula that you
00:24:14.000 --> 00:24:17.000
should remember.
OK, so whenever we integrate a
00:24:17.000 --> 00:24:20.000
function,
and we decide to switch to
00:24:20.000 --> 00:24:25.000
spherical coordinates,
then dx dy dz or r dr d theta
00:24:25.000 --> 00:24:33.000
dz will become rho squared sine
phi d rho d phi d theta.
00:24:33.000 --> 00:24:40.000
OK, any questions on that?
No?
00:24:40.000 --> 00:24:58.000
OK, so let's -- Let's see how
that works.
00:24:58.000 --> 00:25:04.000
So, as an example,
remember at the end of the last
00:25:04.000 --> 00:25:11.000
lecture, I tried to set up an
example where we were looking at
00:25:11.000 --> 00:25:16.000
a sphere sliced by a slanted
plane.
00:25:16.000 --> 00:25:20.000
And now, we're going to try to
find the volume of that
00:25:20.000 --> 00:25:23.000
spherical cap again,
but using spherical coordinates
00:25:23.000 --> 00:25:26.000
instead.
So, I'm going to just be
00:25:26.000 --> 00:25:29.000
smarter than last time.
So, last time,
00:25:29.000 --> 00:25:33.000
we had set up these things with
a slanted plane that was cutting
00:25:33.000 --> 00:25:35.000
things diagonally.
And,
00:25:35.000 --> 00:25:37.000
if I just want to find the
volume of this cap,
00:25:37.000 --> 00:25:41.000
then maybe it makes more sense
to rotate things so that my
00:25:41.000 --> 00:25:45.000
plane is actually horizontal,
and things are going to be
00:25:45.000 --> 00:25:49.000
centered on the z axis.
So, in case you see that it's
00:25:49.000 --> 00:25:52.000
the same, then that's great.
If not, then it doesn't really
00:25:52.000 --> 00:25:55.000
matter.
You can just think of this as a
00:25:55.000 --> 00:26:01.000
new example.
So, I'm going to try to find
00:26:01.000 --> 00:26:10.000
the volume of a portion of the
unit sphere -- -- that lies
00:26:10.000 --> 00:26:20.000
above the horizontal plane,
z equals one over root two.
00:26:20.000 --> 00:26:22.000
OK, one over root two was the
distance from the origin to our
00:26:22.000 --> 00:26:24.000
slanted plane.
So, after you rotate,
00:26:24.000 --> 00:26:28.000
that say you get this value.
Anyway, it's not very important.
00:26:28.000 --> 00:26:31.000
You can just treat that as a
good example if you want.
00:26:31.000 --> 00:26:36.000
OK, so we can compute this in
actually pretty much any
00:26:36.000 --> 00:26:39.000
coordinate system.
And also, of course,
00:26:39.000 --> 00:26:42.000
we can set up not only the
volume, but we can try to find
00:26:42.000 --> 00:26:44.000
the moment of inertia about the
central axis,
00:26:44.000 --> 00:26:47.000
or all sorts of things.
But, we are just doing the
00:26:47.000 --> 00:26:49.000
volume for simplicity.
So, actually,
00:26:49.000 --> 00:26:52.000
this would go pretty well in
cylindrical coordinates.
00:26:52.000 --> 00:26:55.000
But let's do it in spherical
coordinates because that's the
00:26:55.000 --> 00:26:57.000
topic of today.
A good exercise:
00:26:57.000 --> 00:27:01.000
do it in cylindrical and see if
you get the same thing.
00:27:01.000 --> 00:27:08.000
So, how do we do that?
Well, we have to figure out how
00:27:08.000 --> 00:27:14.000
to set up our triple integral in
spherical coordinates.
00:27:14.000 --> 00:27:18.000
So, remember we'll be
integrating one dV.
00:27:18.000 --> 00:27:28.000
So, dV will become rho squared
sign phi d rho d phi d theta.
00:27:28.000 --> 00:27:32.000
And, now as we start,
we're already facing some
00:27:32.000 --> 00:27:35.000
serious problem.
We want to set up the bounds
00:27:35.000 --> 00:27:37.000
for rho for a given,
phi and theta.
00:27:37.000 --> 00:27:39.000
So, that means we choose
latitude/longitude.
00:27:39.000 --> 00:27:42.000
We choose which direction we
want to aim for,
00:27:42.000 --> 00:27:45.000
you know, which point of the
sphere we want to aim at.
00:27:45.000 --> 00:27:50.000
And, we are going to shoot a
ray from the origin towards this
00:27:50.000 --> 00:27:55.000
point, and we want to know what
portion of the ray is in our
00:27:55.000 --> 00:28:03.000
solid.
So -- We are going to choose a
00:28:03.000 --> 00:28:11.000
value of phi and theta.
And, we are going to try to
00:28:11.000 --> 00:28:16.000
figure out what part of our ray
is inside this side.
00:28:16.000 --> 00:28:20.000
So, what should be clear is at
which point we leave the solid,
00:28:20.000 --> 00:28:23.000
right?
What's the value of rho here?
00:28:23.000 --> 00:28:25.000
It's just one.
The sphere is rho equals one.
00:28:25.000 --> 00:28:29.000
That's pretty good.
The question is,
00:28:29.000 --> 00:28:33.000
where do we enter the region?
So, we enter the region when we
00:28:33.000 --> 00:28:38.000
go through this plane.
And, the plane is z equals one
00:28:38.000 --> 00:28:41.000
over root two.
So, what does that tell us
00:28:41.000 --> 00:28:44.000
about rho?
Well, it tells us,
00:28:44.000 --> 00:28:50.000
so remember,
z is rho cosine phi.
00:28:50.000 --> 00:28:55.000
So, the plane is z equals one
over root two.
00:28:55.000 --> 00:29:00.000
That means rho cosine phi is
one over root two.
00:29:00.000 --> 00:29:05.000
That means rho equals one over
root two cosine phi or,
00:29:05.000 --> 00:29:11.000
as some of you know it,
one over root two times second
00:29:11.000 --> 00:29:17.000
phi.
OK, so if we want to set up the
00:29:17.000 --> 00:29:27.000
bounds, then we'll start with
one over root two second phi all
00:29:27.000 --> 00:29:32.000
the way to one.
Now, what's next?
00:29:32.000 --> 00:29:35.000
Well, so we've done,
I think that's basically the
00:29:35.000 --> 00:29:38.000
hardest part of the job.
Next, we have to figure out,
00:29:38.000 --> 00:29:41.000
what's the range for phi?
So, the range for phi,
00:29:41.000 --> 00:29:44.000
well, we have to figure out how
far to the north and to the
00:29:44.000 --> 00:29:48.000
south our region goes.
Well, the lower bound for phi
00:29:48.000 --> 00:29:51.000
is pretty easy,
right, because we go all the
00:29:51.000 --> 00:29:56.000
way to the North Pole direction.
So, phi starts at zero.
00:29:56.000 --> 00:29:59.000
The question is,
where does it stop?
00:29:59.000 --> 00:30:02.000
To find out where it stops,
we have to figure out,
00:30:02.000 --> 00:30:06.000
what is the value of phi when
we hit the edge of the region?
00:30:06.000 --> 00:30:10.000
OK, so maybe you see it.
Maybe you don't.
00:30:10.000 --> 00:30:15.000
One way to do it geometrically
is to just, it's always great to
00:30:15.000 --> 00:30:19.000
draw a slice of your region.
So, if you slice these things
00:30:19.000 --> 00:30:22.000
by a vertical plane,
or actually even better,
00:30:22.000 --> 00:30:25.000
a vertical half plane,
something to delete one half of
00:30:25.000 --> 00:30:28.000
the picture.
So, I'm going to draw these r
00:30:28.000 --> 00:30:33.000
and z directions as before.
So, my sphere is here.
00:30:33.000 --> 00:30:38.000
My plane is here at one over
root two.
00:30:38.000 --> 00:30:43.000
And, my solid is here.
So now, the question is what is
00:30:43.000 --> 00:30:49.000
the value of phi when I'm going
to stop hitting the region?
00:30:49.000 --> 00:30:54.000
And, if you try to figure out
first what is this direction
00:30:54.000 --> 00:30:57.000
here, that's also one over root
two.
00:30:57.000 --> 00:31:03.000
And so, this is actually 45�,
also known as pi over four.
00:31:03.000 --> 00:31:09.000
The other way to think about it
is at this point,
00:31:09.000 --> 00:31:16.000
well, rho is equal to one
because you are on the sphere.
00:31:16.000 --> 00:31:22.000
But, you are also on the plane.
So, rho cos phi is one over
00:31:22.000 --> 00:31:26.000
root two.
So, if you plug rho equals one
00:31:26.000 --> 00:31:31.000
into here, you get cos phi
equals one over root two which
00:31:31.000 --> 00:31:34.000
gives you phi equals pi over
four.
00:31:34.000 --> 00:31:37.000
That's the other way to do it.
You can do it either by
00:31:37.000 --> 00:31:39.000
calculation or by looking at the
picture.
00:31:39.000 --> 00:31:43.000
OK, so either way,
we've decided that phi goes
00:31:43.000 --> 00:31:48.000
from zero to pi over four.
So, this is pi over four.
00:31:48.000 --> 00:31:54.000
Finally, what about theta?
Well, because we go all around
00:31:54.000 --> 00:32:00.000
the z axis we are going to go
just zero to 2pi.
00:32:00.000 --> 00:32:06.000
OK, any questions about these
bounds?
00:32:06.000 --> 00:32:10.000
OK, so note how the equation of
this horizontal plane in
00:32:10.000 --> 00:32:13.000
spherical coordinates has become
a little bit weird.
00:32:13.000 --> 00:32:16.000
But,
if you remember how we do
00:32:16.000 --> 00:32:19.000
things,
say that you have a line in
00:32:19.000 --> 00:32:21.000
polar coordinates,
and that line does not pass
00:32:21.000 --> 00:32:23.000
through the origin,
then you also end up with
00:32:23.000 --> 00:32:26.000
something like that.
You get something like r equals
00:32:26.000 --> 00:32:31.000
a second theta or a cos second
theta for horizontal or vertical
00:32:31.000 --> 00:32:33.000
lines.
And so, it's not surprising you
00:32:33.000 --> 00:32:38.000
should get this.
That's in line with the idea
00:32:38.000 --> 00:32:44.000
that we are just doing again,
polar coordinates in the rz
00:32:44.000 --> 00:32:46.000
directions.
So of course,
00:32:46.000 --> 00:32:48.000
in general, things can be very
messy.
00:32:48.000 --> 00:32:51.000
But, generally speaking,
the kinds of regions that we
00:32:51.000 --> 00:32:55.000
will be setting up things for
are no more complicated or no
00:32:55.000 --> 00:32:59.000
less complicated than what we
would do in the plane in polar
00:32:59.000 --> 00:33:00.000
coordinates.
OK, so there's,
00:33:00.000 --> 00:33:03.000
you know, a small list of
things that you should know how
00:33:03.000 --> 00:33:07.000
to set up.
But, you won't have some
00:33:07.000 --> 00:33:18.000
really, really strange thing.
Yes?
00:33:18.000 --> 00:33:20.000
D rho?
Oh, you mean the bounds for rho?
00:33:20.000 --> 00:33:23.000
Yes.
So, in the inner integral,
00:33:23.000 --> 00:33:26.000
we are going to fix values of
phi and theta.
00:33:26.000 --> 00:33:29.000
So, that means we fix in
advance the direction in which
00:33:29.000 --> 00:33:31.000
we are going to shoot a ray from
the origin.
00:33:31.000 --> 00:33:35.000
So now, as we shoot this ray,
we are going to hit our region
00:33:35.000 --> 00:33:37.000
somewhere.
And, we are going to exit,
00:33:37.000 --> 00:33:40.000
again, somewhere else.
OK, so first of all we have to
00:33:40.000 --> 00:33:43.000
figure out where we enter,
where we leave.
00:33:43.000 --> 00:33:46.000
Well, we enter when the ray
hits the flat face,
00:33:46.000 --> 00:33:50.000
when we hit the plane.
And, we would leave when we hit
00:33:50.000 --> 00:33:52.000
the sphere.
So, the lower bound will be
00:33:52.000 --> 00:33:56.000
given by the plane.
The upper bound will be given
00:33:56.000 --> 00:33:58.000
by the sphere.
So now, you have to get
00:33:58.000 --> 00:34:01.000
spherical coordinate equations
for both the plane and the
00:34:01.000 --> 00:34:02.000
sphere.
For the sphere, that's easy.
00:34:02.000 --> 00:34:05.000
That's rho equals one.
For the plane,
00:34:05.000 --> 00:34:08.000
you start with z equals one
over root two.
00:34:08.000 --> 00:34:11.000
And, you switch it into
spherical coordinates.
00:34:11.000 --> 00:34:14.000
And then, you solve for rho.
And, that's how you get these
00:34:14.000 --> 00:34:19.000
bounds.
Is that OK?
00:34:19.000 --> 00:34:26.000
All right, so that's the setup
part.
00:34:26.000 --> 00:34:29.000
And, of course,
the evaluation part goes as
00:34:29.000 --> 00:34:30.000
usual.
00:34:42.000 --> 00:34:46.000
And, since I'm running short of
time, I'm not going to actually
00:34:46.000 --> 00:34:52.000
do the evaluation.
I'm going to let you figure out
00:34:52.000 --> 00:34:58.000
how it goes.
Let me just say in case you
00:34:58.000 --> 00:35:07.000
want to check your answers,
so, at the end you get 2pi over
00:35:07.000 --> 00:35:13.000
three minus 5pi over six root
two.
00:35:13.000 --> 00:35:17.000
Yes, it looks quite complicated.
That's basically because you
00:35:17.000 --> 00:35:20.000
get one over,
well, you get a second square
00:35:20.000 --> 00:35:23.000
when you integrate C.
When you integrate rho squared,
00:35:23.000 --> 00:35:24.000
you will get rho cubed over
three.
00:35:24.000 --> 00:35:27.000
But that rho cubed will give
you a second cube for the lower
00:35:27.000 --> 00:35:29.000
bound.
And, when you integrate sine
00:35:29.000 --> 00:35:31.000
phi second cubed phi,
you do a substitution.
00:35:31.000 --> 00:35:37.000
You see that integrates to one
over second squared with a
00:35:37.000 --> 00:35:42.000
factor in front.
So, in the second square,
00:35:42.000 --> 00:35:49.000
when you plug in,
no, that's not quite all of it.
00:35:49.000 --> 00:35:51.000
Yeah, well, the second square
is one thing,
00:35:51.000 --> 00:35:53.000
and also the other bound you
get sine phi which integrates to
00:35:53.000 --> 00:35:56.000
cosine phi.
So, anyways,
00:35:56.000 --> 00:36:04.000
you get lots of things.
OK, enough about it.
00:36:04.000 --> 00:36:07.000
So, next, I have to tell you
about applications.
00:36:07.000 --> 00:36:13.000
And, of course,
well, there's the same
00:36:13.000 --> 00:36:14.000
applications that we've seen
that last time,
00:36:14.000 --> 00:36:16.000
finding volumes,
finding masses,
00:36:16.000 --> 00:36:19.000
finding average values of
functions.
00:36:19.000 --> 00:36:22.000
In particular,
now, we could say to find the
00:36:22.000 --> 00:36:26.000
average distance of a point in
this solid to the origin.
00:36:26.000 --> 00:36:28.000
Well,
spherical coordinates become
00:36:28.000 --> 00:36:32.000
appealing because the function
you are averaging is just rho
00:36:32.000 --> 00:36:35.000
while in other coordinate
systems it's a more complicated
00:36:35.000 --> 00:36:37.000
function.
So, if you are asked to find
00:36:37.000 --> 00:36:41.000
the average distance from the
origin, spherical coordinates
00:36:41.000 --> 00:36:43.000
can be interesting.
Also,
00:36:43.000 --> 00:36:47.000
well, there's moments of
inertia,
00:36:47.000 --> 00:36:50.000
preferably the one about the z
axis because if you have to
00:36:50.000 --> 00:36:52.000
integrate something that
involves x or y,
00:36:52.000 --> 00:36:55.000
then your integrand will
contain that awful rho sine phi
00:36:55.000 --> 00:36:57.000
sine theta or rho sine phi
cosine theta,
00:36:57.000 --> 00:37:00.000
and then it won't be much fun
to evaluate.
00:37:00.000 --> 00:37:05.000
So, that anyway,
there's the usual ones.
00:37:05.000 --> 00:37:08.000
And then there's a new one.
So, in physics,
00:37:08.000 --> 00:37:16.000
you've probably seen things
about gravitational attraction.
00:37:16.000 --> 00:37:19.000
If not, well,
it's what causes apples to fall
00:37:19.000 --> 00:37:22.000
and other things like that as
well.
00:37:22.000 --> 00:37:26.000
So, anyway, physics tells you
that if you have two masses,
00:37:26.000 --> 00:37:30.000
then they attract each other
with a force that's directed
00:37:30.000 --> 00:37:33.000
towards each other.
And in intensity,
00:37:33.000 --> 00:37:37.000
it's proportional to the two
masses, and inversely
00:37:37.000 --> 00:37:41.000
proportional to the square of
the distance between them.
00:37:41.000 --> 00:37:45.000
So,
if you have a given solid with
00:37:45.000 --> 00:37:50.000
a certain mass distribution,
and you want to know how it
00:37:50.000 --> 00:37:53.000
attracts something else that you
will put nearby,
00:37:53.000 --> 00:37:58.000
then you actually have to,
the first approximation will be
00:37:58.000 --> 00:37:59.000
to say,
well, let's just put a point
00:37:59.000 --> 00:38:02.000
mass at its center of mass.
But, if you're solid is
00:38:02.000 --> 00:38:04.000
actually not homogenous,
or has a weird shape,
00:38:04.000 --> 00:38:07.000
then that's not actually the
exact answer.
00:38:07.000 --> 00:38:09.000
So, in general,
you would have to just take
00:38:09.000 --> 00:38:12.000
every single piece of your
object and figure out how it
00:38:12.000 --> 00:38:14.000
attracts you,
and then compute the sum of
00:38:14.000 --> 00:38:15.000
these.
So, for example,
00:38:15.000 --> 00:38:18.000
if you want to understand why
anything that you drop in this
00:38:18.000 --> 00:38:21.000
room will fall down,
you have to understand that
00:38:21.000 --> 00:38:24.000
Boston is actually attracting it
towards Boston.
00:38:24.000 --> 00:38:26.000
And, Somerville's attracting it
towards Somerville,
00:38:26.000 --> 00:38:29.000
and lots of things like that.
And, China, which is much
00:38:29.000 --> 00:38:33.000
further on the other side is
going to attract towards China.
00:38:33.000 --> 00:38:35.000
But, there's a lot of stuff on
the other side of the Earth.
00:38:35.000 --> 00:38:37.000
And so, overall,
it's supposed to end up just
00:38:37.000 --> 00:38:41.000
going down.
OK, so now, how to find this
00:38:41.000 --> 00:38:47.000
out, well, you have to just
integrate over the entire Earth.
00:38:47.000 --> 00:38:52.000
OK, so let's try to see how
that goes.
00:38:52.000 --> 00:38:56.000
So, the setup that's going to
be easiest for us to do
00:38:56.000 --> 00:39:01.000
computations is going to be that
we are going to be the test mass
00:39:01.000 --> 00:39:04.000
that's going to be falling.
And, we are going to put
00:39:04.000 --> 00:39:07.000
ourselves at the origin.
And, the solid that's going to
00:39:07.000 --> 00:39:10.000
attract us is going to be
wherever we want in space.
00:39:10.000 --> 00:39:13.000
You'll see, putting yourself at
the origin is going to be
00:39:13.000 --> 00:39:15.000
better.
Well, you have to put something
00:39:15.000 --> 00:39:17.000
at the origin.
And, the one that will stay a
00:39:17.000 --> 00:39:21.000
point mass, I mean,
in my case not really a point,
00:39:21.000 --> 00:39:24.000
but anyway, let's say that I'm
a point.
00:39:24.000 --> 00:39:27.000
And then, I have a solid
attracting me.
00:39:27.000 --> 00:39:32.000
Well,
so then if I take a small piece
00:39:32.000 --> 00:39:37.000
of it with the mass delta M,
then that portion of the solid
00:39:37.000 --> 00:39:42.000
exerts a force on me,
which is going to be directed
00:39:42.000 --> 00:39:47.000
towards it,
and we'll have intensity.
00:39:47.000 --> 00:39:59.000
So, the gravitational force --
-- exerted by the mass delta M
00:39:59.000 --> 00:40:09.000
at the point of x,
y, z in space on a mass at the
00:40:09.000 --> 00:40:13.000
origin.
Well, we know how to express
00:40:13.000 --> 00:40:16.000
that.
Physics tells us that the
00:40:16.000 --> 00:40:21.000
magnitude of this force is going
to be, well, G is just a
00:40:21.000 --> 00:40:23.000
constant.
It's the gravitational
00:40:23.000 --> 00:40:27.000
constant, and its value depends
on which unit system you use.
00:40:27.000 --> 00:40:33.000
Usually it's pretty small,
times the mass delta M,
00:40:33.000 --> 00:40:39.000
times the test mass little m,
divided by the square of the
00:40:39.000 --> 00:40:43.000
distance.
And, the distance from U to
00:40:43.000 --> 00:40:48.000
that thing is conveniently
called rho since we've been
00:40:48.000 --> 00:40:51.000
introducing spherical
coordinates.
00:40:51.000 --> 00:40:54.000
So, that's the size,
that's the magnitude of the
00:40:54.000 --> 00:40:56.000
force.
We also need to know the
00:40:56.000 --> 00:41:01.000
direction of the force.
And, the direction is going to
00:41:01.000 --> 00:41:07.000
be towards that point.
So, the direction of the force
00:41:07.000 --> 00:41:11.000
is going to be that of x,
y, z.
00:41:11.000 --> 00:41:13.000
But if I want a unit vector,
then I should scale this down
00:41:13.000 --> 00:41:22.000
to length one.
So, let me divide this by rho
00:41:22.000 --> 00:41:32.000
to get a unit vector.
So, that means that the force
00:41:32.000 --> 00:41:40.000
I'm getting from this guy is
actually going to be G delta M m
00:41:40.000 --> 00:41:44.000
over rho cubed times x,
y, z.
00:41:44.000 --> 00:41:50.000
I'm just multiplying the
magnitude by the unit vector in
00:41:50.000 --> 00:41:54.000
the correct direction.
OK, so now if I have not just
00:41:54.000 --> 00:41:56.000
that little p is delta M,
but an entire solid,
00:41:56.000 --> 00:41:59.000
then I have to sum all these
guys together.
00:41:59.000 --> 00:42:04.000
And, I will get the vector that
gives me the total force
00:42:04.000 --> 00:42:06.000
exerted, OK?
So, of course,
00:42:06.000 --> 00:42:09.000
there's actually three
different calculations in one
00:42:09.000 --> 00:42:12.000
because you have to sum the x
components to get the x
00:42:12.000 --> 00:42:16.000
components of a total force.
Same with the y,
00:42:16.000 --> 00:42:28.000
and same with the z.
So, let me first write down the
00:42:28.000 --> 00:42:36.000
actual formula.
So, if you integrate over the
00:42:36.000 --> 00:42:39.000
entire solid,
oh, and I have to remind you,
00:42:39.000 --> 00:42:42.000
well, what's the mass,
delta M of a small piece of
00:42:42.000 --> 00:42:45.000
volume delta V?
Well, it's the density times
00:42:45.000 --> 00:42:48.000
the volume.
So, the mass is going to be,
00:42:48.000 --> 00:42:54.000
sorry, density is delta.
There is a lot of Greek letters
00:42:54.000 --> 00:43:04.000
there, times the volume element.
So, you will get that the force
00:43:04.000 --> 00:43:12.000
is the triple integral over your
solid of G m x,
00:43:12.000 --> 00:43:18.000
y, z over rho cubed,
delta dV.
00:43:18.000 --> 00:43:21.000
Now, two observations about
that.
00:43:21.000 --> 00:43:23.000
So, the first one,
well, of course,
00:43:23.000 --> 00:43:29.000
these are just constants.
So, they can go out.
00:43:29.000 --> 00:43:31.000
The second observation,
so here, we are integrating a
00:43:31.000 --> 00:43:33.000
vector quantity.
So, what does that mean?
00:43:33.000 --> 00:43:38.000
I just mean the x component of
a force is given by integrating
00:43:38.000 --> 00:43:41.000
G m x over rho cubed delta dV.
The y components,
00:43:41.000 --> 00:43:43.000
same thing with y.
The z components,
00:43:43.000 --> 00:43:46.000
same thing with z.
OK, there's no,
00:43:46.000 --> 00:43:51.000
like, you know,
just integrate component by
00:43:51.000 --> 00:43:56.000
component to get each component
of the force.
00:43:56.000 --> 00:44:01.000
So, now we could very well to
this in rectangular coordinates
00:44:01.000 --> 00:44:04.000
if we want.
But the annoying thing is this
00:44:04.000 --> 00:44:06.000
rho cubed.
Rho cubed is going to be x
00:44:06.000 --> 00:44:10.000
squared plus y squared plus z
squared to the three halves.
00:44:10.000 --> 00:44:13.000
That's not going to be a very
pleasant thing to integrate.
00:44:13.000 --> 00:44:24.000
So, it's much better to set up
these integrals in spherical
00:44:24.000 --> 00:44:29.000
coordinates.
And, if we're going to do it in
00:44:29.000 --> 00:44:32.000
spherical coordinates,
then probably we don't want to
00:44:32.000 --> 00:44:34.000
bother too much with x and y
components because those would
00:44:34.000 --> 00:44:38.000
be unpleasant.
It would give us rho sine phi
00:44:38.000 --> 00:44:47.000
cos theta or sine theta.
So, the actual way we will set
00:44:47.000 --> 00:44:57.000
up things, set things up,
is to place the solid so that
00:44:57.000 --> 00:45:04.000
the z axis is an axis of
symmetry.
00:45:04.000 --> 00:45:07.000
And, of course,
that only works if the solid
00:45:07.000 --> 00:45:10.000
has some axis of symmetry.
Like, if you're trying to find
00:45:10.000 --> 00:45:13.000
the gravitational attraction of
the Pyramid of Giza,
00:45:13.000 --> 00:45:16.000
then you won't be able to set
up so that it has rotation
00:45:16.000 --> 00:45:18.000
symmetry.
Well, that's a tough fact of
00:45:18.000 --> 00:45:21.000
life, and you have to actually
do it in x, y,
00:45:21.000 --> 00:45:24.000
z coordinates.
But, if at all possible,
00:45:24.000 --> 00:45:27.000
then you're going to place
things.
00:45:27.000 --> 00:45:30.000
Well, I guess even then,
you could center it on the z
00:45:30.000 --> 00:45:32.000
axis.
But anyway, so you're going to
00:45:32.000 --> 00:45:37.000
mostly place things so that your
solid is actually centered on
00:45:37.000 --> 00:45:41.000
the z-axis.
And, what you gain by that is
00:45:41.000 --> 00:45:45.000
that by symmetry,
the gravitational force will be
00:45:45.000 --> 00:45:52.000
directed along the z axis.
So, you will just have to
00:45:52.000 --> 00:45:58.000
figure out the z component.
So, then the force will be
00:45:58.000 --> 00:46:03.000
actually, you know in advance
that it will be given by zero,
00:46:03.000 --> 00:46:11.000
zero, and some z component.
And then, you just need to
00:46:11.000 --> 00:46:19.000
compute that component.
And, that component will be
00:46:19.000 --> 00:46:27.000
just G times m times triple
integral of z over rho cubed
00:46:27.000 --> 00:46:30.000
delta dV.
OK, so that's the first
00:46:30.000 --> 00:46:35.000
simplification we can try to do.
The second thing is,
00:46:35.000 --> 00:46:38.000
well, we have to choose our
favorite coordinate system to do
00:46:38.000 --> 00:46:45.000
this.
But, I claim that actually
00:46:45.000 --> 00:46:57.000
spherical coordinates are the
best -- -- because let's see
00:46:57.000 --> 00:47:04.000
what happens.
So, G times mass times triple
00:47:04.000 --> 00:47:09.000
integral, well,
a z in spherical coordinates
00:47:09.000 --> 00:47:14.000
becomes rho cosine phi over rho
cubed.
00:47:14.000 --> 00:47:17.000
Density, well,
we can't do anything about
00:47:17.000 --> 00:47:21.000
density.
And then, dV becomes rho
00:47:21.000 --> 00:47:28.000
squared sine phi d rho d phi d
theta.
00:47:28.000 --> 00:47:34.000
Well, so, what happens with
that?
00:47:34.000 --> 00:47:37.000
Well, you see that you have a
rho, a rho squared,
00:47:37.000 --> 00:47:39.000
and a rho cubed that cancel
each other.
00:47:39.000 --> 00:47:42.000
So, in fact,
it simplifies quite a bit if
00:47:42.000 --> 00:47:44.000
you do it in spherical
coordinates.
00:48:08.000 --> 00:48:12.000
OK, so the z component of the
force, sorry,
00:48:12.000 --> 00:48:18.000
I'm putting a z here to remind
you it's the z component.
00:48:18.000 --> 00:48:19.000
That is not a partial
derivative, OK?
00:48:19.000 --> 00:48:27.000
Don't get things mixed up,
just the z component of the
00:48:27.000 --> 00:48:35.000
force becomes Gm triple integral
of delta cos phi sine phi d rho
00:48:35.000 --> 00:48:40.000
d phi d theta.
And, so this thing is not dV,
00:48:40.000 --> 00:48:42.000
of course.
dV is much bigger,
00:48:42.000 --> 00:48:45.000
but we've somehow canceled out
most of dV with stuff that was
00:48:45.000 --> 00:48:49.000
in the integrand.
And see, that's actually
00:48:49.000 --> 00:48:55.000
suddenly much less scary.
OK, so just to give you an
00:48:55.000 --> 00:49:01.000
example of what you can prove it
this way, you can prove Newton's
00:49:01.000 --> 00:49:06.000
theorem, which says the
following thing.
00:49:06.000 --> 00:49:23.000
It says the gravitational
attraction -- -- of a spherical
00:49:23.000 --> 00:49:29.000
planet,
I should say with uniform
00:49:29.000 --> 00:49:32.000
density,
or actually it's enough for the
00:49:32.000 --> 00:49:34.000
density to depend just on
distance to the center.
00:49:34.000 --> 00:49:50.000
But we just simplify the
statement is equal to that of a
00:49:50.000 --> 00:50:05.000
point mass -- -- with the same
total mass at its center.
00:50:05.000 --> 00:50:11.000
OK, so what that means is that,
so the way we would set it up
00:50:11.000 --> 00:50:18.000
is u would be sitting here and
your planet would be over here.
00:50:18.000 --> 00:50:21.000
Or, if you're at the surface of
it, then of course you just put
00:50:21.000 --> 00:50:25.000
it tangent to the xy plane here.
And, you would compute that
00:50:25.000 --> 00:50:27.000
quantity.
Computation is a little bit
00:50:27.000 --> 00:50:30.000
annoying if a sphere is sitting
up there because,
00:50:30.000 --> 00:50:31.000
of course, you have to find
bounds,
00:50:31.000 --> 00:50:33.000
and that's not going to be very
pleasant.
00:50:33.000 --> 00:50:37.000
The case that we actually know
how to do fairly well is if you
00:50:37.000 --> 00:50:39.000
are just at the surface of the
planet.
00:50:39.000 --> 00:50:41.000
But then,
what the theorem says is that
00:50:41.000 --> 00:50:44.000
the force that you're going to
feel is exactly the same as if
00:50:44.000 --> 00:50:48.000
you removed all of the planet
and you just put an equivalent
00:50:48.000 --> 00:50:50.000
point mass here.
So, if the earth collapsed to a
00:50:50.000 --> 00:50:53.000
black hole at the center of the
earth with the same mass,
00:50:53.000 --> 00:50:55.000
well, you wouldn't notice the
difference immediately,
00:50:55.000 --> 00:51:00.000
or, rather, you would,
but at least not in terms of
00:51:00.000 --> 00:51:04.000
your weight.
OK, that's the end for today. | 677.169 | 1 |
Its 9 vertices fall in three parallel planes in sets of 3. The outer planes contain the extreme triangles, while the plane between them intersects with the figure in another triangle with an edge length 1+52{\displaystyle {\frac {1+{\sqrt {5}}}{2}}} times the edge length of the polyhedron. This observation led to a generalization known as the ursatopes, which have vertices falling in three hyperplanes of any dimension; some ursatopes are CRF as the tridiminished icosahedron is. | 677.169 | 1 |
Page 4 ... triangle at all , for the six parts of the spherical triangle are measures of the six parts of the solid angle at O. See fig . A a E 9. If a spherical triangle have one of its angles a right angle , it is called a right - angled triangle ...
Page 10 ... triangle . ON RIGHT - ANGLED SPHERICAL TRIANGLES . 18. The preceding formulÌ will apply to right - angled tri- angles , if we make any one of the angles = 90 ¯ . If A = 90 ¯ we have cos a = cos b cos c ...... ( 1 ) sin b = sin a sin B ...
Page 11 ... right - angled triangles , as well as if he had all the formulÌ by heart . The circular parts of a right - angled spherical triangle are five , namely , the two sides , the complement of the hypothe- nuse , and the complements of the two ...
Page 15 ... triangle BAC ( fig . p . 12 ) right - angled at A , satisfy the equation ; produce BA and BC till they inter- sect in D , then take DA ' : BA , and DC ' = BC , the triangles BAC , DA'C ' will be equal in all respects , then the angle A is a | 677.169 | 1 |
What does Gd and t mean?
Geometric Dimensioning and Tolerancing
GD is an acronym that stands for Geometric Dimensioning and Tolerancing. It is a symbolic language used by designers to communicate manufacturing constraints and tolerances clearly. This information is conveyed in the form of annotations included in the design of the part.
What do GD symbols mean?
Geometric dimensioning and tolerancing (GD) is a system of symbols used on engineering drawings to communicate information from the designer to the manufacturer through engineering drawings. GD tells the manufacturer the degree of accuracy and precision needed for each controlled feature of the part.
What is symmetrical tolerance?
Symmetry Tolerance is a three-dimensional geometric tolerance that controls how much the median points between two features may deviate from a specified center plane or axis. This tolerance is similar to concentricity, and the verification of symmetry tolerance is likewise time-consuming and difficult.
What is symmetry in machining?
Description: GD Symmetry is a 3-Dimensional tolerance that is used to ensure that two features on a part are uniform across a datum plane.
How do you calculate symmetry in GD?
Using a coordinate measuring machine Initially, the CMM is set up to establish the theoretical center plane. Then, both symmetrical sides are measured using the CMM stylus to calculate where the median points fall. The positions of all the median points along the feature's length are compared with the datum plane.
What does t mean in GD?
tangent plane
"T" stands for "tangent plane." How angled a plane in contact with the surface is to the datum plane within the range of specified surface is indicated by parallelism. Unlike parallelism, this specifies the convex of the surface and not the concave. Unequally disposed profile tolerance (ASME only)
What is a symmetrical in math?
What is symmetric in math? In Mathematics, the meaning of symmetry is that one shape is exactly like the other shape when it is moved, rotated, or flipped.
What is the symbol of symmetrical?
The symbol σv describes a symmetry plane vertical to the principal axis, while a symmetry plane running horizontally through the principal axis is denoted as σh. In highly symmetric systems one additional type of symmetry plane occurs that divides the angle between two Cn axes. This type of plane is denoted σd.
How do you describe symmetry?
In geometry, symmetry is defined as a balanced and proportionate similarity that is found in two halves of an object. It means one-half is the mirror image of the other half. The imaginary line or axis along which you can fold a figure to obtain the symmetrical halves is called the line of symmetry.
What is symmetry in engineering?
Symmetrical and synchronized systems are often. used to meet the stability criteria of rotating structures in mechanical engineering. On the other. hand, in telecommunications engineering, since the speed or the amount of data is the same in both. directions, many systems are symmetrical.
Why GD and T is necessary to put in the design as a control?
When the GD system is used properly it reduces the amount of notes, dimensions and tolerances required on a drawing. In addition, by establishing datums, GD provides a precise method for describing a reference coordinate system that can be used during the manufacturing and inspection processes.
What is symmetry explain with example?
Symmetry is an attribute where something is the same on both sides of an axis. An example of symmetry is a circle that is the same on both sides if you fold it along its diameter. noun.
What is the meaning of symmetry in maths?
What is symmetry in drawing?
What is Symmetry in Art? Symmetry in art is when the elements of a painting or drawing balance each other out. This could be the objects themselves, but it can also relate to colors and other compositional techniques.
What is symmetry in GD?
In GD, Symmetry is a 3D tolerance that is used to ensure that two features on a part are symmetrical across a daum plane. Put another way, Symmetry establishes a tolerance zone for the median points of non-cylindrical part features.
What is geometric dimension and tolerance (GD)?
Geometric dimension and tolerance ( GD ) is a type of tolerance that are used along with linear tolerance to define nominal and allowable variations in the part geometry or an assembly. ASME Y14.5-2009 standard has defined GD symbols in detail. This series of articles on geometric dimension and control will help you in following ways.
What is the use of diameter symbol in GD?
Generally diameter symbols is used to make circular or cylindrical tolerance zones. Concentricity tolerance in GD controls the central axis of a cylinder or sphere with respect to the datum plane or an axis. It creates a 3-dimensional cylindrical tolerance zone around the datum axis.
What is symmetry and why is it important?
Put another way, Symmetry establishes a tolerance zone for the median points of non-cylindrical part features. Those median points are the points halfway between the two features and must lie within a certain distance of the datum plane. | 677.169 | 1 |
Meta
Perfect Beauty of math in the Symmetry
数学和完美对称性质
Look at the above figures. Which of them are symmetric? Which of them are not?
For the symmetric figures, what are the lines of symmetry (if applicable)?
In a symmetric figure, we can produce as many isosceles as we like. Do you know the easiest way to find them?
We need to explain on the last figure (the figure at the bottom right). It is consisted of two circular arcs with different radii. The two arcs are glued smoothly (at the gluing point). This last figure is not symmetric.
Not only this figure is NOT symmetric, but also we CANNOT pick out from it a sub-figure that is symmetric, as long as the figure contains both arcs with different radii. But can you argue for this claim?
Sometimes a figure might not be perfect under the given condition, but we are tempted to produce a figure that is more perfect than the one given in the question. For example, we might draw an isosceles when the given one is scalene, and draw a special quadrilateral (e.g. a rectangle) when the given one is just a trapezoid, or even arbitrary quadrilateral. This is the Pitfall! that we shall avoid. | 677.169 | 1 |
hedral Angle
A dihedral angle is the angle between two intersecting planes or half-planes. In chemistry, it is the clockwise angle between half-planes through two sets of three atoms, having two atoms in common. In solid geometry, it is defined as the union of a line and two half-planes that have this line as a common edge. In higher dimensions, a dihedral angle represents the angle between two hyperplanes.
The planes of a flying machine are said to be at positive dihedral angle when both starboard and port main planes (commonly called wings) are upwardly inclined to the lateral axis. When downwardly inclined they are said to be at a negative dihedral angle.
Mathematical background
When the two intersecting planes are described in terms of Cartesian coordinates by the two equations
: a_1 x + b_1 y + c_1 z + d_1 = 0
:a_2 x + b_2 y + c_2 z + d_2 = 0
the dihedral angle, \varphi between them is given by:
:\cos \varphi = \frac
and satisfies 0\le \varphi \le \pi/2.
Alternatively, if Angle
In geometry and trigonometry, a right angle is an angle of exactly 90 degrees or radians corresponding to a quarter turn. If a ray is placed so that its endpoint is on a line and the adjacent angles are equal, then they are right angles. The term is a calque of Latin ''angulus rectus''; here ''rectus'' means "upright", referring to the vertical perpendicular to a horizontal base line.
Closely related and important geometrical concepts are perpendicular lines, meaning lines that form right angles at their point of intersection, and orthogonality, which is the property of forming right angles, usually applied to vectors. The presence of a right angle in a triangle is the defining factor for right triangles, making the right angle basic to trigonometry.
Etymology
The meaning of ''right'' in ''right angle'' possibly refers to the Latin adjective ''rectus'' 'erect, straight, upright, perpendicular'. A Greek equivalent is ''orthos'' 'straight; perpendicular' (see orthog everydayStraight Linetex (geometry)
In geometry, a vertex (in plural form: vertices or vertexes) is a point where two or more curves, lines, or edges meet. As a consequence of this definition, the point where two lines meet to form an angle and the corners of polygons and polyhedra are vertices.
Definition Of an angle
The ''vertex'' of an angle is the point where two rays begin or meet, where two line segments join or meet, where two lines intersect (cross), or any appropriate combination of rays, segments, and lines that result in two straight "sides" meeting at one place.
:(3 vols.): (vol. 1), (vol. 2), (vol. 3).
Of a polytope
A vertex is a corner point of a polygon, polyhedron, or other higher-dimensional polytope, formed by the intersection of edges, faces or facets of the object.
In a polygon, a vertex is called " convex" if the internal angle of the polygon (i.e., the angle formed by the two edges at the vertex with the polygon inside the angle) is less than π radians (180°, two right angles);clus
Proclus Lycius (; 8 February 412 – 17 April 485), called Proclus the Successor ( grc-gre, Πρόκλος ὁ Διάδοχος, ''Próklos ho Diádokhos''), was a Greek Neoplatonist philosopher, one of the last major classical philosophers of late antiquity. He set forth one of the most elaborate and fully developed systems of Neoplatonism and, through later interpreters and translators, exerted an influence on Byzantine philosophy, Early Islamic philosophy, and Scholastic philosophy.
Biography
The primary source for the life of Proclus is the eulogy ''Proclus, or On Happiness'' that was written for him upon his death by his successor, Marinus, Marinus' biography set out to prove that Proclus reached the peak of virtue and attained eudaimonia. There are also a few details about the time in which he lived in the similarly structured ''Life of Isidore'' written by the philosopher Damascius in the following century.
According to Marinus, Proclus was born in 412 AD | 677.169 | 1 |
What is Euclidean: Definition and 211 Discussions
Euclidean space is the fundamental space of classical geometry. Originally, it was the three-dimensional space of Euclidean geometry, but in modern mathematics there are Euclidean spaces of any nonnegative integer dimension, including the three-dimensional space and the Euclidean plane (dimension two). It was introduced by the Ancient Greek mathematician Euclid of Alexandria, and the qualifier Euclidean is used to distinguish it from other spaces that were later discovered in physics and modern mathematics.
Ancient Greek geometers introduced Euclidean space for modeling the physical universe. Their great innovation was to prove all properties of the space as theorems by starting from a few fundamental properties, called postulates, which either were considered as evident (for example, there is exactly one straight line passing through two points), or seemed impossible to prove (parallel postulate).
After the introduction at the end of 19th century of non-Euclidean geometries, the old postulates were re-formalized to define Euclidean spaces through axiomatic theory. Another definition of Euclidean spaces by means of vector spaces and linear algebra has been shown to be equivalent to the axiomatic definition. It is this definition that is more commonly used in modern mathematics, and detailed in this article.In all definitions, Euclidean spaces consist of points, which are defined only by the properties that they must have for forming a Euclidean space.
There is essentially only one Euclidean space of each dimension; that is, all Euclidean spaces of a given dimension are isomorphic. Therefore, in many cases, it is possible to work with a specific Euclidean space, which is generally the real n-space
R
n
,
{\displaystyle \mathbb {R} ^{n},}
equipped with the dot product. An isomorphism from a Euclidean space to
R
n
{\displaystyle \mathbb {R} ^{n}}
associates with each point an n-tuple of real numbers which locate that point in the Euclidean space and are called the Cartesian coordinates of that point.
I was reading a paper by J.M.C Montanus which was published in <low quality journal reference removed> in which he claims under AEST the new gravitational dynamics and electrodynamics are reformulated in close correspondence with classical physics, and subsequently leads to the correct...
(a) Let be m a line and the only two semiplans determined by m.
(i) Show that: If are points that do not belong to such , so and are in opposite sides of m.
(ii) In the same conditions of the last item, show: and .
(iii) Determine the union result , carefully justifying your answer...
I have the following (small) problem:
Let $ ( , ):V \times V \rightarrow \mathbb{R} $ be a real-valued non-degenerate inner product on the real vector space $V$.
Given, for all $v \in V$ we have $(v,v) \geq 0$
Now prove that if $(x,x)=0$ then $x=0$ for $x \in V$, that is, prove that the inner...
Hi guys,
Hopefully, no geometry-enthusiasts are going to read these next few lines, but if that's the case, be lenient :)
I have always hated high-school geometry, those basic boring theorems about triangles, polygons, circles, and so on, and I have always "skipped" such classes, studying...
I've been given a curve α parametrized by t :
α (t) = (cos(t), t^2, 0)
How would I go about finding the euclidean coordinate functions for this curve? I know how to find euclidean coord. fns. for a vector field, but I am a bit confused here.
(Sorry about the formatting)
Hello
As you know, the geometric definition of the dot product of two vectors is the product of their norms, and the cosine of the angle between them.
(The algebraic one makes it the sum of the product of the components in Cartesian coordinates.)
I have often read that this holds for Euclidean...
Consider a point A outside of a line α. Α and α define a plane.Let us suppose that more than one lines parallels to α are passing through A. Then these lines are also parallels to each other; wrong because they all have common point A.
Before looking at the proof of basic theorems in Euclidean plane geometry, I feel that I should draw pictures or use other physical objects to have some idea why the theorem must be true. After all, I should not just plainly play the "game of logic". And, it is from such observations in real...
The flatness problem (also known as the oldness problem) is a cosmological fine-tuning problem within the Big Bang model of the universe.
The fine-tuning problem of the last century was solved by introducing the theory of inflation which flattens...
I read in my textbook Calculus on Manifolds by Spivak that a set ##A\subset \mathbb{R}^n## is bounded iff there is a closed n-rectangle ##D## such that ##A\subset D##. It should be plain that if I wanted to define unboundedness, I should just say something along the lines of: "A set ##A\subset...
If we suppose the following 8-dimensional manifold given by
$$a_1=cos(x)cos(y)cos(z)$$
$$a_2=cos(x)cos(y)sin(z)$$
$$a_3=cos(x)sin(y)cos(z)$$
$$a_4=cos(x)sin(y)sin(z)$$
$$a_5=sin(x)cos(y)cos(z)$$
$$a_6=sin(x)cos(y)sin(z)$$
$$a_7=sin(x)sin(y)cos(z)$$
$$a_8=sin(x)sin(y)sin(z)$$
Then obviously...
Homework Statement
Let ##f:X\rightarrow Y## with X = Y = ##\mathbb{R}^2## an euclidean topology.
## f(x_1,x_2) =( x^2_1+x_2*sin(x_1),x^3_2-sin(e^{x_1+x_2} ) )##
Is f continuous?
Homework Equations
f is continuous if for every open set U in Y, its pre-image ##f^{-1}(U)## is open in X.
or if...
Homework Statement
I am asked to write an expression for the length of a vector V in terms of its dot product in an arbitrary system in Euclidean space.
Homework EquationsThe Attempt at a Solution
The dot product of a vector a with itself can be given by I a I2. Does that expression only apply...
In the book "The Basics of Abstract Algebra" Bland defines a Euclidean Domain using two conditions as follows:
In the book "Abstract Algebra"by Dummit and Foote we find that a Euclidean Domain is defined using only one of Bland's conditions ... as follows:What are the consequences of these...
Homework Statement
Suppose that m divisions are required to find gcd(a,b). Prove by induction that for m >= 1, a >= F(m+2) and b>= F(m+1) where F(n) is the Fibonacci sequence.
Hint: to find gcd(a,b), after the first division the algorithm computes gcd(b,r).
Homework Equations
Fibonacci...
Homework Statement
Consider the contractions of the 3D Euclidean symmetry while preserving the SO(2) subgroup. In the physics point of view, explain the resulting symmetries G(2) (Galilean symmetry group) and H(3) (Heisenberg-Weyl group for quantum mechanics) and give their Lie algebras...
Good Morning
I am having some trouble categorizing a few concepts (I made the one that is critical to this post to be BOLD)
Remote parallelism: the ability to move coordinate systems and frames around in space.
Euclidean Space
Coordinate systems: Cartesian vs. cylindrical
I am aware that if...Let us look at the topological space R_d x R where R_d is the set of real numbers with the discrete toplogy and R the euclidean topology. This set is not second countable, because R_d has no countable basis.
I am wondering if this space is locally euclidean, and if so, of what dimension? Given...
Hello! (Wave)
I am looking at the following exercise: Let $b=r_0, r_1, r_2, \dots$ be the successive remainders in the Euclidean algorithm applied to $a$ and $b$. Show that after every two steps, the remainder is reduced by at least one half. In other words, verify that $$r_{i+2}< \frac{1}{2}...
Homework Statement
please see the image
Homework Equations
I'm not sure if this is relevant:
##r_2 \leq \frac{1}{2}r_1## ... ##r_n \leq (\frac{1}{2})^nr_1##
The Attempt at a Solution
i have shown that ##r_{i+2} < r_i## by showing the ##r_{i+2} - r_i## is negative, but how do I show that the...
I know this is some kind of exercise problem, but it isnot widely discussed in general general relativity textbook. Sorry to post it here.
I want to calculate the mass and entropy of non-rotating BTZ black hole using Euclidean method. When I calculate the Euclidean action, I always get an...
This is an exercise from Hartman's lecture 6th. Using the Euclidean method to calculate the BTZ black hole mass entropy. The BTZ metric is given by
$$ ds^2=(r^2-8M)d\tau^2 +\frac{dr^2}{r^2-8M}+r^2d\phi^2$$
and ##\tau \sim \tau+\beta, \beta=\frac{\pi}{\sqrt{2M}}##.
Then we calculate the...
I am attempting to understand a question posed to me by an acquaintance, who asked me if I could refer him to literature freely available on the Internet on "self-dual solutions to Maxwell's equations on Euclidean space, or pseudo-Euclidean space, not Minkowski space (where there are none)" and...
I've been trying to understand how we know that the observable universe is flat, and I'm having difficulty finding any sources that explain exactly how the calculations were done. On this WMAP website ( it says:
"A central feature of...
So I was reading this book, "Euclidean and non Euclidean geometries" by Greenberg
I solved the first problems of the first chapter, and I would like to verify my solutions
1. Homework Statement
Homework Equations
[/B]
Um, none that I can think of?
The Attempt at a Solution
(1) Correct...
Homework Statement
Show that the length of a curve γ in ℝn is invariant under euclidean motions. I.e., show that L[Aγ] = L[γ] for Ax = Rx + a
Homework Equations
The length of a curve is given by the arc-length formula: s(t) = ∫γ'(t)dt from t0 to tThe Attempt at a Solution
I would imagine I...
Yesterday I was asking questions from someone and in between his explanations, he said that the Euclidean action in a QFT is actually equal to its Hamiltonian. He had to go so there was no time for me to ask more questions. So I ask here, is it true? I couldn't find anything on google. If its...
So if I pick any 2 points on a 2d manifold, say x1 and x2, then the distance between these two points is a secant line that passes through 3 space that isn't part of the manifold. So no matter what, there doesn't exist an point epsilon, e , where ||e ||>||0 || and ||x2-x1||<|| e ||
No matter...
Hi
I am working on an assignment which is has asked us to derive an expression for a differential number count of supernovae in a euclidean flat non-expanding space.
I am bit perplexed by this question and am wondering whether it is a trick question. We are allowed to do research to find an...
Hi,
When I started learning about GR I wondered if it emerged from SR (which the name suggests) or if the connection between the two is mere technical. GR describes the behaviour of the metric of space-time, which is locally Minkowskian and therefore SR applies.
But is a curvature-based theory...
Hello! (Wave)
I have applied a lot of times the euclidean division of $x^6-1$ with $x^2- \alpha^{a+1} (\alpha+1)x+ \alpha^{2a+3}, a \geq 0$, $\alpha$ a primitive $6$-th root of unity. But I don't get the right result... (Sweating)
We are over $\mathbb{F}_7$.
I got that $x^6-1=(x^2-...
Hello everyone,
I have been reading around that when performing the analytic continuation to Euclidean space (t\to-i\tau) one also has to continue the gauge field (A_t\to iA_4) in order to keep the gauge group compact.
I already knew that the gauge field had to be continued as well but I didn't...
If the Euclidean plane is partitioned into convex sets each of area A in such a way that each contains exactly one vertex of a unit square lattice and this vertex is in its interior, is it true that A must be at least 1/2?
If not what is the greatest lower bound for A?
The analogous greatest... | 677.169 | 1 |
Mensa Triangle Problem - Seeking Answers
In summary, the discussion focused on the mensa triangle problem and how to solve it. The key point is to look at the slopes of the triangles and notice that they are not equal, indicating that the "hypotenuse" of the entire triangle is not a straight line and therefore not a real triangle. The suggested method to solve the problem is to focus on the number of squares in each triangle rather than counting the number of triangles.
Nov 16, 2006
#1
chriswall89@hotmail.
9
0
Ok, I've just signed up to this forum and i see there are a lot of very smart people and i just wondered if anyone knows the answer to the mensa triangle problem. I've sat and stared at it for hours and still i have no answer. Any theories.
Use the search function. There have been several threads about this already.
Bottom line: Look at the slopes, one of the shapes isn't even a real triangle.
For example, the second triangle's green part has a steeper slope than the red part.
EDIT: The green triangle slopes don't equal the red triangle slope.
EDIT2: To see this, line up a ruler to the first triangle, connect the bottom left to the top right. Notice how the actual "triangle" is slightly below this. Do the same to the second one, and you will see the "triangle" is slightly bent upwards. So the change in that area is that extra square.
Last edited: Nov 16, 2006
Nov 16, 2006
#3
chriswall89@hotmail.
9
0
Damn you make it all sound easy but i see what you mean. I've just been counting the number of squares per triangle. Thats where I've been going wrong. Cheers
Oh, that thing! Check the slopes of the hypotenuse of the red right triangle: it has base 8 and height 3; slope 3/8. Now check the slope on the green right triangle: it has base 5 and height 2; slope 2/5. Since those are not the same, that "hypotenuse" of the entire "triangle" is NOT a straight line and it isn't really a triangle at all.
1. What is the Mensa Triangle Problem?
The Mensa Triangle Problem is a famous mathematical puzzle that involves arranging nine numbered cards in a triangle shape in such a way that the sum of the numbers on each side of the triangle is equal.
2. How difficult is the Mensa Triangle Problem?
The difficulty of the Mensa Triangle Problem varies from person to person, but it is generally considered to be a challenging puzzle that requires logical thinking and problem-solving skills.
3. Has the Mensa Triangle Problem been solved?
Yes, the Mensa Triangle Problem has been solved and there are several known solutions. However, there may be other possible solutions that have not been discovered yet.
4. Who created the Mensa Triangle Problem?
The Mensa Triangle Problem was created by a group of mathematicians and puzzle enthusiasts who were members of the organization Mensa, hence the name of the puzzle. The original creators are unknown, but the puzzle has been popularized by Mensa and other puzzle organizations.
5. What is the significance of the Mensa Triangle Problem?
The Mensa Triangle Problem is significant because it showcases the beauty and complexity of mathematics and challenges individuals to think critically and creatively. It is also a popular puzzle that has been enjoyed by many people all over the world. | 677.169 | 1 |
Geometry ActivitiesTeaching geometry? Check out this review of Geometiles, an awesome new math manipulative with endless possibilities. The kind people at Geometiles recently contacted me to ask if I would be willing to write a review in exchange for a free set. I love using manipulatives in the math classroom, so I jumped at the chanceTo give my students a bit more practice with word problems, I typed up these Romeo and Juliet themed pythagorean theorem problems. These are slightly modified from this Weebly site. Free Download of Romeo and Juliet Pythagorean Theorem Questions
I decided my students needed a Pythagorean Triples reference page in their notebooks to go along with the posters I have hanging in my classroom. Here is the page. You will notice that 6, 8, 10 is not included on the list because it is related to 3, 4, 5. Free Download of Pythagorean practice book to give my trigonometry students some much needed review of finding the distance between any two points on the coordinate plane. I'm a rebel and chose not to teach the distance formula since it's way easier to just use the Pythagorean Theorem instead. Instructions for Finding the Distance Between Any …
I used this One Big Square Task with my trigonometry students to review 45-45-90 special right triangles. I found this task in the Open Curriculum Task Database. I loved listening to the conversations my students had as they struggled through this problem. And, I mean "struggle" in the best possible way! Free Download of One The page I'm showing you FIRST is actually what we did LAST. Yes, I'm that teacher who makes my students …
Today I want to share some area and perimeter posters I created for my classroom. I spent another afternoon working in my classroom today, and I feel like I actually made a lot of progress. I still need to start working on actual math lesson plans at some point. My principal e-mailed out the scheduleSince I'm teaching trig this year, I wanted to create some special right triangles posters to jog my students' memories. Special right triangles are a topic they study in geometry but have often forgotten by the time they end up in my trig class. I designed the title to print on 8.5″ x 11″ and …
I created this set of free printable pythagorean triple posters to help decorate my new classroom. I'm teaching Algebra 1, Trig, and Math Concepts (a class for 9th graders not yet ready for Algebra 1) this year. As I was making a list of things my trig students need to review at the beginning Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com. | 677.169 | 1 |
In land surveying, a bearing is the clockwise or counterclockwise angle between north or south and a direction. … In surveying, bearings can be referenced to true north, magnetic north, grid north (the Y axis of a map projection), or a previous map, which is often a historical magnetic north.
Why is it called a protractor?
protractor (n.)
1610s, "one who lengthens (an action)," Modern Latin agent noun from Latin protrahere "to draw forward" (see protraction). Medieval Latin protractor meant "one who calls or drags another into court." The surveying sense of "instrument for measuring and drawing angles on paper" is recorded from 1650s.
How are Clinometers used?
A clinometer is a tool that is used to measure the angle of elevation, or angle from the ground, in a right – angled triangle. You can use a clinometer to measure the height of tall things that you can't possibly reach to the top of, flag poles, buildings, trees.
How is a protractor made?
Fold a triangle with one edge from the upper left corner of the paper to the bottom center of the paper. This will create a 60º angle. Create the same fold from the right side to create a 120º angle. Mark these angles on your protractor.
What are the 16 compass points?
What are 2 uses of a protractor?
We use a protractor to measure angles. Architects and designers use a more precise protractor called an angle protractor tool which gives more accurate measures. A compass helps to construct an angle. Set squares, also known as triangle protractor, are used to draw parallel and perpendicular lines.
Can a person be a protractor?
a person or thing that protracts. (in surveying, mathematics, etc.)
What angle is 45?
A 45-degree angle is exactly half of a 90-degree angle formed between two rays. It is an acute angle and two angles measuring 45 degrees from a right angle or a 90-degree angle. We know that an angle is formed when two rays meet at a vertex.
Why do protractors have two scales?
The two scales make it easy for us to measure angles facing different ways. To measure the size of angle ABC, place the protractor over the angle so that the centre of the protractor is directly over the angle's vertex, B; and the base line of the protractor is along the arm, BA, of the angle.
What is average life of bearing?
Average life –median lives of groups of bearings are averaged–somewhere between 4 and 5 times the L10 life. The constant radial load which a group of bearings can endure for a rating life of 1 million revolutions of the inner ring (stationary load and stationary outer ring).
How do you calculate bearing life?
Bearing Rating Life Calculation
C = Dynamic Capacity (dN or Lbs)
P = Equivalent Bearing Load (N or Lbs)
N = Rotating speed in RPM.
e = 3.0 for ball bearings, 10/3 for roller bearings.
What is E in bearing?
RS – Bearing with rubber seal on one side, one side open. 2 Z / ZZ – Bearing with a metal seal on both sides. Z – Bearing with a metal seal on one side, one side open. E – Reinforced Design. P2 – Highest precision.
What are 3 figure bearings?
Three-figure bearings
Imagine that you are in a boat at sea. You are lost and you cannot see any land.
A bearing is a direction. …
The points of the compass can be written as 3-figure bearings:
In navigation on sea, land or air it is three figure bearings that are used more often than the points of the compass.
What are common applications of bearings?
Bearing Applications
Aviation Cargo Systems.
Aerospace Wing Actuators.
Anemometer.
ATMs & Card Readers.
Bicycles.
Commercial Blenders.
Dental Hand Tools.
Electrical Motors.
What are three things that use ball bearings?
Part of our everyday life, ball bearings are found in things such as blenders and exercise equipment. The list goes on and on. Bicycles, DVD players, water pumps, washing machines and fans are just a few of many day to day products that we use that use ball bearings.
What is a real world example of a plane?
Examples of a plane would be: a desktop, the chalkboard/whiteboard, a piece of paper, a TV screen, window, wall or a door. | 677.169 | 1 |
find the areaof the parallelogram with vertices A(1,2,2), B(1,3,6), C(3,8,6), and D(3,7,3)
First of all, the four vertices don't make a parallelogram.
If it was, then
So the two vectors are not parallel.
12016 Complex Plane that are the Verticesof a Parallelogram Find necessary and sufficient conditions (with proofs) such that the points z1, z2, z3, and z4 in the complex plane are the verticesof a parallelogram.
271892 Four Midpoints of Any Quadrilateral Form a Parallelogram How can one prove that the 4 midpoints of the four sides of any quadrilateral form the verticesof a parallelogram using graph geometry (ie. x1, y1 etc. to denote the four vertices)? | 677.169 | 1 |
Question 1.
Draw rough diagrams to illustrate the following:
(i) open simple curve
(ii) closed simple curve
(iii) open curve that is not simple
(iv) closed curve that is not simple.
Solution:
Question 2.
Consider the given figure and answer the following questions:
(i) Is it a curve?
(ii) Is it a closed curve?
(iii) Is it a polygon?
Solution:
Question 3.
Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior?
Solution:
Question 4.
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them.
Solution:
Question 5.
In context of the given figure:
(i) Is it a simple closed curve?
(ii) Is it a quadrilateral?
(iii) Draw its diagonals and name them.
(iv) State which diagonal lies in the interior and which diagonal lies in the exterior of the quadrilateral.
Solution:
Question 6.
Draw a rough sketch of a quadrilateral KLMN. State,
(i) two pairs of opposite sides
(ii) two pairs of opposite angles
(iii) two pairs of adjacent sides
(iv) two pairs of adjacent angles.
Solution: | 677.169 | 1 |
What is the definition of a line segment in geometry?
A line segment is part of a line that has two endpoints and is finite in length. A ray is a line segment that extends indefinitely in one direction.
What does subset mean in math?
What is a Subset in Maths? Set A is said to be a subset of Set B if all the elements of Set A are also present in Set B. In other words, set A is contained inside Set B. Example: If set A has {X, Y} and set B has {X, Y, Z}, then A is the subset of B because elements of A are also present in set B.
Which is a subset of a line that has two?
line segment
A line segment is a part of a line that has two defined endpoints. A line segment represents a collection of points inside the endpoints and it is named by its endpoints.
What is line segment simple definition?
In geometry, a line segment is bounded by two distinct points on a line. Or we can say a line segment is part of the line that connects two points. A line has no endpoints and extends infinitely in both the direction but a line segment has two fixed or definite endpoints.
How do we name the subsets of a line ray and line segment?
A segment is named by its two endpoints, for example, ¯AB . A ray is a part of a line that has one endpoint and goes on infinitely in only one direction. You cannot measure the length of a ray. A ray is named using its endpoint first, and then any other point on the ray (for example, →BA ).
What is a line ray and segment?
a ray is a line that has one endpoint and on one side it goes on forever. a line goes on forever in both directions. a line segment has two endpoints.
What is a subset of a line that its length is between its and point?
A line segment has two endpoints. It contains these endpoints and all the points of the line between them. You can measure the length of a segment, but not of a line. A segment is named by its two endpoints, for example, ¯AB .
How do you find a line segment in geometry?
You can find the length of a line segment by counting the units that the line segment covers. Counting the units on a graph is like counting the number of blocks traveled between your house and your friend's house. Count the number of units between the two end points.
What is the subset symbol?
⊆
Subset of a Set. A subset is a set whose elements are all members of another set. The symbol "⊆" means "is a subset of". The symbol "⊂" means "is a proper subset of".
What is opposite of subset?
Opposite of an amount or section which, when combined with others, makes up the whole of something. whole. everything. superset.
How do we name the subset of line ray and line segment?
What does a ⊆ B mean?
A set A is a subset of a set B if every element in A is also in B . For example, if A={1,3,5} and B={1,2,3,4,5} , then A is a subset of B , and we write. A⊆B. The line under the sideways ∪ means that A may also be equal to B (that is, they may be identical sets).
When can you say that a subset of a line is a line segment?
A line segment is a subset of a line with two end points. A ray is different from a line because it has a starting point or origin, and extends infinitely from there, whereas a line extends infinitely in opposite directions. | 677.169 | 1 |
Congruent Tangents and Circumscribed Polygons
Although words like "tangent" might sound complex at first, these concepts are actually quite simple when we can visualize how they work. Like many geometric concepts, tangent is something that can be easily illustrated and explained. What are congruent tangents and circumscribed polygons? Let's find out:
What are congruent tangents?
Let's start with congruent tangents. A tangent to a circle is simply a line that touches the edge of the circle without actually cutting through it. "Tangere" is the Latin verb for "touch," which helps us remember how this line works. The Latin word for "cut" is "secare," which is where we get the name "secant." Remember that if you see a line that "cuts through" a circle, you're looking at a secant line and not a tangent line.
If two lines emanate from the same exterior point (A) and are tangent to a circle, then we know that they are always congruent. Congruence is essentially just another word for equality. In this diagram, we can see that line AB is congruent to line AC.
Circumscribed polygons
Now let's add another concept into the mix: Circumscribed polygons. When we circumscribe something, we simply draw another shape around it. The Latin "circum" means "around," while "scribere" is the verb "to cut."
We can circumscribe polygons around a circle. Here's what that looks like:
When a polygon is circumscribed around a circle, we know that the vertices never lie on the circle itself. We also know that each side is tangent to the circle. Another interesting feature is that each side of the polygon touches the circle at its midpoint.
Pair your student with a tutor who understands congruent tangents and circumscribed polygons
Tutoring is an excellent option if your student needs to reinforce concepts like congruent tangents and circumscribed polygons outside of class. This 1-on-1 environment allows your student to work at a pace that suits their ability level. They can also ask as many questions as they like and ask for help if they get stuck. The tutor can then step in and guide the student toward confidence and comprehension. If you'd like to learn more about the benefits of tutoring, don't hesitate to contact our Educational Directors. Remember, Varsity Tutors will pair your student with a suitable tutor. | 677.169 | 1 |
How many triangles #4
1.) How many triangles do you see in this image ?
how many triangles do you see in this image
See AnswerHide Answer
Answer with Solution
Looking at the symmetry, we can divide this triangle in 5 equal parts(square) and we will first count the number of triangles in each part and then number of triangles formed by combining 2 or more parts together. Number of triangles in one square: 8(non overlapping triangles) + 8(overlapping triangles with middle lines a part of the triangle) =16*5 = 80 Number of triangles by taking two squares(one middle and one other) at a time: 4*4(there are four such combinations) = 16 Number of triangles by taking (only) three squares(one middle and two others) at a time: 4(all squares should be a part of the triangle) *4(four such combinations) = 16 Number of triangles by taking 4 squares together: 2 *4(four such combinations) = 8 | 677.169 | 1 |
...be the line so divided in the points C and Dj (fig. Euc-. n. 5) shew that A IP = 4.CD'i 4.AD.DB. 9. If a straight line be bisected and produced to any point, the square on the whole line when thus produced, is equal to the square on the part produced and four times the rectangle contained...
...unequal lines AC, CD, is equal to the rectangle contained by their sum and difference. PROP. VL— THEOREM. If a straight line be bisected, and produced to any point ; then the rectangle contained by the whole line thus produced, and the part of it produced, together
...straight lines AC, CD, is equal to the rectangle contained by their sum and difference. M PROPOSITION 6. THEOREM. If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the...
...straight lines AC, CD, is equal to the rectangle contained by their sum and difference. PROPOSITION 6. THEOREM. If a straight line be bisected, and produced to any point, the rectangle cjntained by the whole line thus produced, and the part of it produced, together with the...
...shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. 10. If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the...
...angles of a triangle can in no case exceed two right angles. 4. Give two proofs of Eucl. I. 47. 5. If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of...
...coefficient of of ( 1 — as) in the expansion of - — — — , prove Cr=BJ.-B,._1. (1—3?)" 9. If a straight line be bisected and produced to any point ; the rectangle contained by the whole line thus produced and the part of it produced together with the square...
...the right angle is equal to the squares described upon the sides which contain the right angle 20 4. If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced, and | 677.169 | 1 |
Of the Construction and Use of the Theodolite.
Fig. A
This Instrument is made of Wood, Brass, or any other solid Matter, commonly circular and about one Foot in Diameter. In the Center of this Instrument is set upright a little Brass Cylinder, or Pivot, about which an Index turns, furnished with two Sights, or a Telescope, having a right Line, called The Fiducial Line, exactly answering to the Center of the aforesaid little Cylinder, whose Top ought to be cut into a Screw, for receiving a Nut to fasten the Index, upon which is fixed a small Compass for finding the Meridian Line.
The Limb of the Theodolite is a Circle of such a Thickness, as to contain about six round Pieces of Pasteboard within it (of which we are going to speak), and of such a Breadth as to receive the Divisions of 360 Degrees, and sometimes of every fifth Minute.
There are several round Pieces of Pasteboard, of the Bigness of the Theodolite, pierced thro' the Middle with a round Hole, exactly to fit the Pivot; so that the Pivot may be put thro' each of the aforesaid Holes in the Pieces of Pasteboard, and the upper Pasteboard may have the Index moving upon it. This upper Pasteboard may be fixed at pleasure, by means of a little Point fastened to the Limb of the Instrument, and entering a little way into the Pasteboard. There is commonly drawn with Ink, upon each of these Pasteboards, a Radius or Semidiameter, serving for a Station-Line.
Figs. D & G
Underneath the Theodolite is fastened a Ball and Socket, represented by the Figure D, which is a Brass Ball enclosed between two Shells of the same Metal, that may be more or less opened by means of a Screw, and a Socket G, in which goes the Head of a three-legged Staff; of which more by and by.
Fig. A, represents the Instrument put together. We now proceed to shew the Construction of the Pieces composing it, in beginning with the Division of it's Limb.
First, Draw upon the Limb two or three concentrick Circles, to contain the Degrees, and the Numbers set at every tenth Degree; then divide one of these Circumferences into four very equal Parts, each of which will be 90 Degrees; and dividing each of these four Parts into 9 more, the Circumference will be divided into every tenth Degree. Again, each of these last Parts being divided by 2, and each of those arising into 5 equal Parts, the whole Circumference will be divided into 360 Degrees. This being done, you must draw the Lines of these Divisions upon their convenient Arcs, by means of a Ruler moving about the Center. Afterwards Numbers must be set to every tenth Degree, beginning from the Fiducial Line, which is that whereon the two fixed Sights or Telescope is fastened.
A Theodolite thus divided is of much greater Use than those whose Limbs are not divided; for it may serve exactly to take the Plots of Places, and measure inaccessible Distances by Trigonometry.
Fig. B
The Figures B represent the Sights which are placed upon different Instruments; that to which is placed the Eye, hath a long strait Slit, which ought to be very perpendicular, made with a fine Saw; and that which is turned towards the Object, hath a square Hole, so large, that the adjacent Parts of a distant Object may be perceived thro' it: And along the Middle of this Hole is drained a very fine Gut, in order to vertically cut Objects, when they are perceived thro' the Slit of the other Sight. But that the Eye may be indifferently placed at any one of the two Sights at pleasure, so that Objects may be as well perceived thro' the Sights on one Side the Instrument, on which they are placed, as on the other; there is made in each Sight a square Hole and a Slit, the Hole in one Sight being below the Slit, and in the other Sight above it, as the little Figures shew. These Sights ought to be exactly placed on the Extremes, and in the fiducial Line, as well of Instruments as Indexes, and are fastened in little square Holes with Nuts underneath, or else by means of Screws, according as the Place they are fastened on requires.
Fig. C
The little Figure C represents the aforesaid Cylinder, or Pivot, with it's Nut, for joining the Index to the Theodolite; those of Semicircles, and other Instruments, are made in the same manner, only they are rivetted underneath.
Figs. D & G
The Figure D represents the Ball and Socket for supporting the Instrument, and is composed of a Brass Ball inclosed between two Shells of the same Metal, which are made very round, with Balls of tempered Steel cut in manner of a File. These Shells are locked more or less by means of a Screw, that so they may press the Ball inclosed between them according to necessity. One of these Shells is soldered to the Socket G, which is a turned Brass Ferrel, in which the Foot of the Instrument is put. Balls and Sockets are made of different Bignesses, according to the Bignesses of Instruments, and are fastened to the Instruments with Screws, in a Plate riveted to the Top of the Ball.
Construction of the Feet for Supporting of Instruments.
We have already mentioned the simple Feet for supporting Surveying-Crosses, which are to be forced into the Ground; but those whose Description we are now going to give, are not to be forced into the Ground, but are opened or shut according as the Inequality of the Ground, the Instrument is to be used upon, requires.
Fig. E
The Foot E is a triangular Plate, in whose Middle is a Piece b, which is to go into the Socket G.
Underneath the aforesaid Plate are fastened three Ferrels, or Sockets, moveable by means of Joints, for receiving three round Staves of such a Length, that the Observer's Eye, when the Instrument is using, may commodiously view Objects thro' the Telescope, or Sights. The Extremities of these Staves are furnished with Ferrels and Iron Points, in order to keep the Instrument firm when it is using.
Fig. F
The Foot F consists of four Staves, about two Foot long, whereof that in the Middle, called the Shank, hath it's Top rounded, that so it may go into the Socket; the rest of this Staff is cut in Figure of a Triangle, that so the three Faces thereof may receive upon them three other Staves, fastened by means of three Screws (all of a piece) and so many Nuts. These three Staves are furnished with Ferrels and Iron Points, being flat within side, and have three Faces without.
When we have a mind to carry this Foot, we re-unite all the Staves together, so that they make, as it were, but one, and by this means are shorter by about the half, than when the Foot is using.
We generally hang to the Middle of each of these Feet a Thread and Plummet, in order to know the Station-Point.
Use of the Theodolite.
Fig. 1
To take the Map of a Country by this Instrument, chuse two high Places, for Example, the Observatory, and the Salt-Petre House, from whence the Country nigh Paris, a Map of which is to be made, may be seen; then mark round the Center of the upper Pasteboard the Name of the Place chosen for the first Station, and having fixed it by means of the Point on the Limb of the Theodolite, put the Index upon it, which sufficiently screw down by means of the Nut and Screw.
Now having placed the Theodolite upon it's Foot, planted at the Observatory, and given it a Situation nearly horizontal, so that it may remain steddy while the Index is moving, observe thro' the Sights the Steeple of the Salt-Petre House, and along the fiducial Line of the Index from the Center draw the Station-Line.
Then turn the Index, and observe some remarkable Object thro' the Sights, as the Steeple of Vaugirard, towards which a Line must be drawn upon the Pasteboard, from the Center, along the fiducial Line of the Index, and along this Line write the Name of the Place viewed thro' the Sights.
Again, direct the Index towards some other Object (as Mont-rouge) and draw a Line towards it from the Center, along the fiducial Line, and upon this Line write the Name of the Place observed. Proceed in the same Manner with all the considerable Places that can be seen from the Observatory.
Now having removed the Theodolite from it's first Station, having well observed it's Place, and transported it to some other designed Place, as to the Salt-Petre House; measure the exact Distance between the two Stations upon level Ground, the Number of Toises of which must be set down upon your Pasteboard, which must now be turned, or taken from under the Index, that so at every different Station, the upper Face of the Pasteboard, upon which the Index is, may be clean: then set down about the Center of this new Pasteboard, the Name of the Place of your second Station, and upon the Base Line the Number of Toises measured, that so you may remember this Line is the same as that on the precedent Pasteboard. The Theodolite being placed here, dispose it so, that placing the fiducial Line of the Index upon the Station Line, you may discover thro' the Sights, the Observatory, which was your first Station.
The Instrument remaining firm in this Situation, turn the Index, and successively. view thro' the Sights the former Objects observed from the Observatory, and draw Lines, as before, upon the Pasteboard, along the Index, from the Center towards the Places viewed, and upon each Line write the correspondent Name of the Place.
If all the Places you have a mind to set down in your Map, cannot be seen from the two precedent Stations, you must chuse a third Place from whence they may be observed, and make as many new Stations, as are necessary for perceiving each remarkable Object, from two Places sufficiently distant from each other.
Now to represent this Map upon a Sheet of Paper, first draw a right Line at pleasure upon it, for a common Base, which divide into the same Number of equal Parts, as you have measured Toises upon the Ground. About one End of this Line, as a Center, describe circular Arcs equal to those drawn upon the first Pasteboard, and upon the other Extreme, Arcs equal to those drawn upon the second Pasteboard, and produce the Lines forming the Arcs 'till they meet each other; then the Points of Concourse, will be the Points of Position of the Places observed.
The aforesaid Places may be laid down upon the Paper easier, by placing the Centers of the Pasteboards upon the Extremities of the common Base, and noting upon the Paper the Ends of the Lines drawn upon the Pasteboard, and then drawing Lines from the Stations thro' those Points 'till they intersect.
By means of this Theodolite may be had in Degrees, or Parts, all the Angles that the Places viewed thro' the Sights or Telescopes, make, with the Places whereat the Instrument is placed.
What we have said, is sufficient to shew the Manner of using the Theodolite in taking the Position of Places, and making of Maps, because the Operations are the same for all different Places; but for it's Uses, with regard to Trigonometry, they are the same as those of the Semi-circle and Quadrant, of which we are going to treat. | 677.169 | 1 |
.. Naked tamzin taber
10 people found it helpful. fichoh. The answers of the geometry exercise are : 1.) x = 25° ; 2.) x = 129°. 3.) x = 73° ; y = 107° ; z = 73°. 4.) s = 167°. 5.) s = 52°. 6.) M1 = …Sep 7, 2023 ... Comments · Angle Addition Postulate explained with examples · Junk Food · Geometry Unit 1 · Angle Relationships · Hon Geom Lesson...Day 6 - Angle Pair Relationships. My students came to school with lots of questions about their homework. So, I made up this page on the fly before we started the new lesson. I think this is the bones …Sep 7, 2023 ... ... : Partitioning a Segment Topic 5: Angle Measures (adjacent angles, vertical angles, complementary angles, supplementary angles, linear pair)4.7/5. Level: Master's, University, College, PHD, High School, Undergraduate, Professional. Min Baths . Any. User ID: 833607 / Mar 30, 2022. ID 3320. 506 ... Latest Topic For Essay Writing In Ielts, Unit 1 Homework 5 Angle Relationships, How To Write Telephone Numbers Argumentative Essay, …A comprehensive study looked into the lives of more than 1,000 of these workers. For little more than pennies per hour, in homes across India, some of the country's most vulnerable...5. Maneuvering the middle llc 2017 worksheets answers. 6. Interpreting political cartoons 11 answer key. 7. Maneuvering the middle linear relationships answer key. 8. Dilations worksheet answer key kuta. Showing 8 worksheets for Maneuvering The Middle Llc 2017.Probability Stat Answers Final. Let x is the amount can win from this game Let probability of failing A as p (A) and probability of failing B as p (B)... God :Supreme Mathematics. 1. Knowledge-the sum of what is known. knowledge is facts, awareness or familiarity gained by doing the knowledge.HW #3 - Angle Relationships Answer Key - Free download as PDF File (.pdf), Text File (.txt) or read online for free. L'Anse Creuse High School - North Geometry Instructor: Lauren Wilson.20 Qs. Complementary and Supplementary Angles. 4.6K plays. 5th - 7th. 15 Qs. 5.2K plays. 9th - 10th. Applying Angle Relationships HW 2 quiz for 8th grade students. Find other quizzes for Mathematics and more on Quizizz for free!Train containers, organized by sections 5: This extra service permits monitoring the writing technique of unit one geometry fundamentals homework 1 reply key massive orders because the paper might be despatched to you for approval in partsdrafts earlier than the ultimate. Unit 1 Geometry Fundamentals Homework 5 Angle …Displaying top 8 worksheets found for - Unit 1 Geometry Basics Homework 3 Angle Relationships. Some of the worksheets for this concept are Gina wilson unit 1 geometery basics, Unit 1 lesson 2 geometry answer key, Geometry unit 1 workbook, Tools of geometry chapter 1 all in one teaching resources, Unit 1 tools of geometry reasoning and proof, …This Geometry Unit Bundle includes a dictionary, guided notes, homework assignments, four quizzes, a study guide, a unit test, and more that cover the following topics: • Naming and Classifying Angles. • Angle Relationships (Vertical, Adjacent, Complementary, and Supplementary Angles) • Parallel Lines Cut by a … Unit 1 Homework 5 Angle Relationships. Niamh Chamberlain. #26 in Global Rating. Calculate the price. Minimum Price. 12 Customer reviews. Pay only for completed parts of your project without paying upfront. View All Writers. Level: College, University, High School, Master's, PHD, Undergraduate ….We are here to help you write a brilliant thesis by the provided requirements and deadline needed. It is safe and simple. Legal. Critical Thinking Essay on Nursing. 4.5-star rating on the Internet. Unit 1 Homework 5 Angle Relationships. 1 (888)814-4206 1 (888)499-5521. Level: College, University, High School, Master's, PHD, Undergraduate ... Unit 1 Homework 5 Angle Relationships. 100% Success rate. 4.9 (2939 reviews) User ID: 123019. PenMyPaper offers you with affordable 'write me an essay service'. We try our best to keep the prices for my essay writing as low as possible so that it does not end up burning a hole in your pocket. The Angle Addition Postulate is a concept in geometry which states that if point B lies in the interior of angle AOC, then the measure of angle AOB + the measure of angle BOC is equal to the measure of angle AOC. To apply this in examples involving vectors, we can consider the analytical methods of vector addition and subtraction.. | 677.169 | 1 |
In quadrilateral ABDC, AB ∥ CD. Which additional piece of information is needed to determine that ABDC is a parallelogram?
answer
AC ≅ BD
question
ABCD is a square.
What is the measure of angle BAC?
answer
45°
question
Complete the proof to show that ABCD is a parallelogram.
The slope of BC is .
The slope of AD is .
The slope of CD is .
The slope of BA is .
BC ∥ AD and CD ∥ BA because the __________________________________. Therefore, ABCD is a parallelogram because both pairs of opposite sides are parallel.
answer
slopes of opposite sides are equal
question
In the diagram, SR = 4√2 and QR = √10. What is the perimeter of parallelogram PQRS?
answer
8√2 + 2√10 units
question
In parallelogram WXYZ, what is CY?
answer
15
question
Figure CDEF is a parallelogram.
What is the value of r?
answer
5
question
The perimeter of a square is 56 cm. What is the approximate length of its diagonal?
answer
19.8 cm
question
Which statement proves that △XYZ is an isosceles right triangle?
answer
The slope of XZ is 3/4, the slope of XY is -4/3, and XZ = XY = 5.
question
What are the missing angle measures in parallelogram RSTU?
answer
m∠R = 110°, m∠T = 110°, m∠U = 70°
question
What is the perimeter of rectangle EFGH?
answer
2√10 + 2√29 units
question
A partial proof was constructed given that MNOP is a parallelogram.
By the definition of a parallelogram,
MN ∥ PO and MP ∥ NO.
Using MP as a transversal, ∠M and ∠P are same-side interior angles, so they are supplementary.
Using NO as a transversal, ∠N and ∠O are same-side interior angles, so they are supplementary.
Using OP as a transversal, ∠O and ∠P are same-side interior angles, so they are supplementary.
Therefore, __________ and _________ because they are supplements of the same angle.
Which statements should fill in the blanks in the last line of the proof?
answer
∠M ≅ ∠O; ∠N ≅ ∠P
question
Rhombus LMNO is shown with its diagonals.
Angle MNO measures 112°. What is the measure of angle LMN?
answer
68°
question
LMNP is a parallelogram.
What additional information would prove that LMNP is a rectangle?
answer
LP ⊥ PN
question
A rectangle has a width of 9 units and a length of 40 units. What is the length of a diagonal?
answer
41 units
question
Figure ABCD is a parallelogram.
What is the perimeter of ABCD?
answer
44 units
question
The vertices of a quadrilateral in the coordinate plane are known. How can the perimeter of the figure be found?
answer
Use the distance formula to find the length of each side, and then add the lengths.
question
Which statements are true of all squares? Check all that apply.
answer
The diagonals are perpendicular.
The diagonals are congruent to each other.
The diagonals bisect the vertex angles.
The diagonals bisect each other.
question
The perimeter of square JKLM is 48 units.
What is the value of x?
answer
9
question
A quilt piece is designed with four congruent triangles to form a rhombus so that one of the diagonals is equal to the side length of the rhombus.
Which measures are true for the quilt piece? Check all that apply.
answer
a = 60°
The perimeter of the rhombus is 16 inches.
The length of the longer diagonal is approximately 7 inches | 677.169 | 1 |
Closure
Tangents to Circles in Real Life
Imagine a superhero joining the Olympics to throw a hammer. An athlete would typically spin counterclockwise three or four (rarely five) times, then release the hammer. As viewed from above, the hammer travels on a path that is tangent to the circle created when the athlete spins. The diagram below shows the path of the superhero's hammer throw. See how the super hero fairs! Note, it is a not-to-scale drawing.
On August 30,1986, in a packed stadium full of fans, Yuriy Sedykh set the world record with a throw of 86.74 meters. Perhaps a throw farther than the superhero! | 677.169 | 1 |
How to Memorize the Trigonometric Functions of the Common Angles (and the quadrantal angles)
In the video below, you will see how to memorize the value of the trig functions for all the common angles, including the quadrantal angles like 0° and 180°. This is based on this huge table of values for the trig functions (you can find a blank copy there too). For the process we will look at, memorize may be the wrong word since it is more about understanding how the values work with each other. Even so, after some practice, you won't need the table to be able to answer the question "what is \(\sin\left(\dfrac{\pi}{3}\right)\)" because you will have the idea of the table in your head!
You may notice that this is, in a way, very similar to learning/memorizing the unit circle. Even so, you would still use the unit circle after memorizing this. For example, if you wanted to know the \(\sin\left(\dfrac{2\pi}{3}\right)\), you would still use the value you have in this table as a reference angle and then the sign would come from the quadrant. The unit circle already has that "built in", but has more angles laid out than this table. | 677.169 | 1 |
The Pythagorean Theorem Makes Construction and GPS Possible
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Pythagoras, an ancient Greek thinker — equal parts philosopher, mathematician and mystical cult leader — lived from 570 to 490 B.C.E and is credited with devising one of the most famous theorems of all time. Wikimedia Commons (CC By-SA 4.0)(CC By-SA 3.0)/HowStuffWorks
Key Takeaways
The Pythagorean theorem is crucial in various fields, including construction, manufacturing and navigation, enabling precise measurements and the creation of right angles for large structures.
It underpins our entire system of measurement, allowing for accurate navigation by pilots and ships, and making GPS measurements possible through the calculation of distances and angles.
Beyond navigation, the theorem is essential in geometry, physics, geology, engineering and even practical applications by carpenters and machinists.
OK, time for a pop quiz. You've got a right-angled triangle — that is, one where two of the sides come together to form a 90-degree angle. You know the length of those two sides. How do you figure out the length of the remaining side?
The Pythagorean theorem states that with a right-angled triangle, the sum of the squares of the two sides that form the right angle is equal to the square of the third, longer side, which is called the hypotenuse. As a result, you can determine the length of the hypotenuse with the equation a2 + b2 = c2, in which a and b represent the two sides of the right angle and c is the long side.
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Who Was Pythagoras?
A pretty slick trick, huh? But the man whom this math trick is named for is nearly as fascinating. Pythagoras, an ancient Greek thinker who was born on the island of Samos and lived from 570 to 490 B.C.E, was kind of a trippy character — equal parts philosopher, mathematician and mystical cult leader. In his lifetime, Pythagoras wasn't known as much for solving for the length of the hypotenuse as he was for his belief in reincarnation and adherence to an ascetic lifestyle that emphasized a strict vegetarian diet, adherence to religious rituals and plenty of self-discipline that he taught to his followers.
Pythagoras biographer Christoph Riedweg describes him as a tall, handsome and charismatic figure, whose aura was enhanced by his eccentric attire — a white robe, trousers and a golden wreath on his head. Odd rumors swirled around him — that he could perform miracles, that he had a golden artificial leg concealed beneath his clothes and that he possessed the power to be in two places at one time.
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Pythagoras founded a school near what is now the port city of Crotone in southern Italy, which was named the Semicircle of Pythagoras. Followers, who were sworn to a code of secrecy, learned to contemplate numbers in a fashion similar to the Jewish mysticism of Kaballah. In Pythagoras' philosophy, each number had a divine meaning, and their combination revealed a greater truth.
With a hyperbolic reputation like that, it's little wonder that Pythagoras was credited with devising one of the most famous theorems of all time, even though he wasn't actually the first to come up with the concept. Chinese and Babylonian mathematicians beat him to it by a millennium.
"What we have is evidence they knew the Pythagorean relationship through specific examples," writes G. Donald Allen, a math professor and director of the Center for Technology-Mediated Instruction in Mathematics at Texas A&M University, in an email. "An entire Babylonian tablet was found that shows various triples of numbers that meet the condition: a2 + b2 = c2."
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How Is the Pythagorean Theorem Useful Today?
The Pythagorean theorem isn't just an intriguing mathematical exercise. It's utilized in a wide range of fields, from construction and manufacturing to navigation.
As Allen explains, one of the classic uses of the Pythagorean theorem is in laying the foundations of buildings. "You see, to make a rectangular foundation for, say, a temple, you need to make right angles. But how can you do that? By eyeballing it? This wouldn't work for a large structure. But, when you have the length and width, you can use the Pythagorean theorem to make a precise right angle to any precision."
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Beyond that, "This theorem and those related to it have given us our entire system of measurement," Allen says. "It allows pilots to navigate in windy skies, and ships to set their course. All GPS measurements are possible because of this theorem."
In navigation, the Pythagorean theorem provides a ship's navigator with a way of calculating the distance to a point in the ocean that's, say, 300 miles north and 400 miles west (480 kilometers north and 640 kilometers west). It's also useful to cartographers, who use it to calculate the steepness of hills and mountains.
"This theorem is important in all of geometry, including solid geometry," Allen continues. "It is also foundational in other branches of mathematics, much of physics, geology, all of mechanical and aeronautical engineering. Carpenters use it and so do machinists. When you have angles, and you need measurements, you need this theorem."
Now That's Theoretical
One of the formative experiences in the life of Albert Einstein was writing his own mathematical proof of the Pythagorean theorem at age 12. Einstein's fascination with geometry eventually played a role in his development of the theories of special and general relativity.
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Cite This!
Please copy/paste the following text to properly cite this HowStuffWorks.com article: | 677.169 | 1 |
Notice that two angles form a straight angle when together. Web complementary and supplementary angles. Web complementary angles are two angles with a sum of 90 ∘. Supplementary angles find the value worksheet. If two angles add up to 90°, they are _____ angles. A common case is when they form a right angle. Get an grip on geometry with this helpful worksheet!
Train students to keenly observe the right angles and straight angles and equate them to find the. Since the sum of two complementary angles is 90°, one of them cannot be an obtuse angle. Angles in a straight line. Web complementary and supplementary angles (visual) google classroom. Web complementary and supplementary angles at a glance.
Supplementary Angles Worksheet
Angles in a straight line. If two angles add up to 180o they are _____ angles. Get an grip on geometry with this helpful worksheet! Supplementary angles are pairs of. B c a o 149.
Complementary And Supplementary Angles Worksheet Answers
Angles are not necessarily drawn to scale. If two angles add up to 90°, they are _____ angles. Supplementary angles are pairs of. One angle of a pair of complementary angles is given. B c.
Supplementary Angles (A)
Find the complement or supplement of the indicated. Which one might that be? This means that the angles are supplementary and have a sum of \ (180^\circ\). Supplementary angles are pairs of. Web finding unknown.
12 Best Images of Supplementary Angles Worksheet Supplementary Angles
Complementary angle are usually part of a right angle and add up to 90 degrees. Web complementary and supplementary angles at a glance. Web complementary angles are two angles with a sum of 90 ∘..
Complementary and supplementary angle plansres
Supplementary angles find the value worksheet. If two angles add up to 180o they are _____ angles. Web complementary and supplementary angles worksheets. A common case is when they form a right angle. Angles in.
Complementary And Supplementary Angles Worksheet Kuta —
If two angles add up to 180o they are _____ angles. Web complementary and supplementary angles. Since the sum of two complementary angles is 90°, one of them cannot be an obtuse angle. What is.
Complementary Angles And Supplementary Angles Worksheet - Train students to keenly observe the right angles and straight angles and equate them to find the. Web finding unknown angles in a complementary or supplementary pair. Web supplementary angles are angles that when added equal 180 degrees. Angles in a straight line. Two right angles are supplementary as their sum is 180°. Notice that two angles form a straight angle when together.
Glue each angle next to it next to its complement. Web finding unknown angles in a complementary or supplementary pair. (answers on the second page.) complementary angles worksheet (find unknown angles) online. Angles are not necessarily drawn to scale. Web complementary and supplementary angles.
Web complementary and supplementary angles. Since the sum of two complementary angles is 90°, one of them cannot be an obtuse angle. Cut out the angle tiles at the bottom of the page. Your student will learn the difference between complementary and.
Web Complementary And Supplementary Angles.
Train students to keenly observe the right angles and straight angles and equate them to find the. Web supplementary angles are angles that when added equal 180 degrees. Two right angles are supplementary as their sum is 180°. Web complementary and supplementary angles (visual) google classroom.
Find The Complement Or Supplement Of The Indicated.
[show me] supplementary angles are two angles with a sum of 180. Supplementary angles are pairs of. Complementary angles are pairs of angles which sum to 90° (a right angle). Get an grip on geometry with this helpful worksheet!
Complementary Angles Find The Value Worksheet.
If two angles add up to 180o they are _____ angles. Angles are not necessarily drawn to scale. Supplementary angles find the value worksheet. Your student will learn the difference between complementary and.
Angles In A Straight Line.
Web finding unknown angles in a complementary or supplementary pair. If two angles add up to 90°, they are _____ angles. Web complementary and supplementary angles at a glance. Through artistic, interactive guided notes,. | 677.169 | 1 |
Difference between a point, a line segment, a ray and a line
On the screen, we can see a point A in blue color, a line segment BC in black color, a ray DE in green color and a line FG in red color. Move these points A, B, C, D, E, F and G and see how the position of the point, the line segment, the ray and the line gets changed.
1. What difference do you observe between a point, a line segment, a ray and a line? | 677.169 | 1 |
Question 2.
For two vectors to be equal, they should have the
(A) same magnitude
(B) same direction
(C) same magnitude and direction
(D) same magnitude but opposite direction
Answer:
(C) same magnitude and direction
Question 4.
Find a vector which is parallel to \(\overrightarrow{\mathrm{v}}\) = \(\hat{i}\) – 2\(\hat{j}\) and has a magnitude 10.
Answer:
Substituting for wx in (i) using equation (ii),
Using equation (ii),
Answer:
Required vector is \(\frac{10}{\sqrt{5}} \hat{\mathbf{i}}\) – \(\frac{20}{\sqrt{5}} \hat{\mathbf{j}}\)
Alternate method:
When two vectors are parallel, one vector is scalar multiple of another,
i.e., if \(\overrightarrow{\mathrm{v}}\) and \(\overrightarrow{\mathrm{w}}\) are parallel then, \(\overrightarrow{\mathrm{w}}\) = n\(\overrightarrow{\mathrm{v}}\) where, n is scalar.
Question 4.
Given \(\bar{v}_{1}\) = 5\(\hat{i}\) + 2\(\hat{j}\) and \(\bar{v}_{2}\) = a\(\hat{i}\) – 6\(\hat{j}\) are perpendicular to each other, determine the value of a.
Solution:
As \(\bar{v}_{1}\) and \(\bar{v}_{2}\) are perpendicular to each other, θ = 90°
Answer:
Value of a is \(\frac{12}{5}\). | 677.169 | 1 |
...formulas can be expressed in words : In any triangle, the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides multiplied by the cosine of their included angle. NOTE. In Fig. 49 a, A is acute and I. 47, II. 13, we see that in the triangle ABC, if the angle...
...side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it. JD Hyp. In Aabc, p is the projection of 6 upon c, and the angle opposite a is obtuse. To prove a 2...
...side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it. Hyp. In A abc, p is the projection of b upon c, and the angle opposite a is obtuse. To prove a2 = 62...
...opposite the obtuse angle is equivalent to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it. B Let ABC be an obtuse-angled A, and CD be the projection of BC on AC (prolonged). To Prove AB2 = BC'2...
...side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it. c Hyp. In Aabc, p is the projection of 6 upon c, and the angle opposite a is obtuse. To prove a2 —...
...side opposite an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of these sides and the projection of the other side upon it. 7. Prove : The area of a regular polygon is equal to one-half the product of its perimeter and apothem....
...opposite an acute angle is equivalent to the sum of the squares of the other two sides, diminished by twice the product of one of these sides and the projection of the oiher side upon it. b B Let ABC be a A in which BC lies opposite an acute angle, and AD is the projection i. 47, II. 13, we see that in the triangle ABC, if the angle...
...about the triangle ABC. Law of cosines. In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides into the cosine of their included angle. In figures 35 regard AD, DB, and AB as directed... | 677.169 | 1 |
The Pythagoras theorem which is also referred to as the Pythagorean theorem explains the relationship between the three sides of a right-angled triangle. According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides of a triangle. Let us learn more about the Pythagoras theorem, the Pythagoras theorem formula, and the proof of Pythagoras theorem along with examples.
What is the Pythagoras Theorem?
The Pythagoras theorem states that if a triangle is a right-angled triangle, then the square of the hypotenuse is equal to the sum of the squares of the other two sides. Observe the following triangle ABC, in which we have BC2 = AB2 + AC2. Here, AB is the base, AC is the altitude (height), and BC is the hypotenuse. It is to be noted that the hypotenuse is the longest side of a right-angled triangle.
Pythagoras Theorem Equation
The Pythagoras theorem equation is expressed as, c2 = a2 + b2, where 'c' = hypotenuse of the right triangle and 'a' and 'b' are the other two legs. Hence, any triangle with one angle equal to 90 degrees produces a Pythagoras triangle and the Pythagoras equation can be applied in the triangle.
History of Pythagoras Theorem
Pythagoras theorem was introduced by the Greek Mathematician Pythagoras of Samos. He was an ancient Greek philosopher who formed a group of mathematicians who worked religiously on numbers and lived like monks. Although Pythagoras introduced the theorem, there is evidence that proves that it existed in other civilizations too, 1000 years before Pythagoras was born. The oldest known evidence is seen between the 20th to the 16th century B.C in the Old Babylonian Period.
Pythagoras Theorem Formula
The Pythagorean theorem formula states that in a right triangle ABC, the square of the hypotenuse is equal to the sum of the squares of the other two legs. If AB and AC are the sides and BC is the hypotenuse of the triangle, then: BC2 = AB2 + AC2. In this case, AB is the base, AC is the altitude or the height, and BC is the hypotenuse.
Another way to understand the Pythagorean theorem formula is using the following figure which shows that the area of the square formed by the longest side of the right triangle (the hypotenuse) is equal to the sum of the area of the squares formed by the other two sides of the right triangle.
In a right-angled triangle, the Pythagoras Theorem Formula is expressed as:
c2 = a2 + b2
Where,
'c' = hypotenuse of the right triangle
'a' and 'b' are the other two legs.
Pythagoras Theorem Proof
The Pythagoras theorem can be proved in many ways. Some of the most common and widely used methods are the algebraic method and the similar triangles method. Let us have a look at both these methods individually in order to understand the proof of this theorem.
Proof of Pythagorean Theorem Formula using the Algebraic Method
The proof of the Pythagoras theorem can be derived using the algebraic method. For example, let us use the values a, b, and c as shown in the following figure and follow the steps given below:
Step 1: This method is also known as the 'proof by rearrangement'. Take 4 congruent right-angled triangles, with side lengths 'a' and 'b', and hypotenuse length 'c'. Arrange them in such a way that the hypotenuses of all the triangles form a tilted square. It can be seen that in the square PQRS, the length of the sides is 'a + b'. The four right triangles have 'b' as the base, 'a' as the height and, 'c' as the hypotenuse.
Step 2: The 4 triangles form the inner square WXYZ as shown, with 'c' as the four sides.
Step 3: The area of the square WXYZ by arranging the four triangles is c2.
Pythagorean Theorem Formula Proof using Similar Triangles
Two triangles are said to be similar if their corresponding angles are of equal measure and their corresponding sides are in the same ratio. Also, if the angles are of the same measure, then by using the sine law, we can say that the corresponding sides will also be in the same ratio. Hence, corresponding angles in similar triangles lead us to equal ratios of side lengths.
Pythagoras Theorem Triangles
Right triangles follow the rule of the Pythagoras theorem and they are called Pythagoras theorem triangles. The three sides of such a triangle are collectively called Pythagoras triples. All the Pythagoras theorem triangles follow the Pythagoras theorem which says that the square of the hypotenuse is equal to the sum of the two sides of the right-angled triangle. This can be expressed as c2 = a2 + b2; where 'c' is the hypotenuse and 'a' and 'b' are the two legs of the triangle.
Pythagoras Theorem Squares
As per the Pythagorean theorem, the area of the square which is built upon the hypotenuse of a right triangle is equal to the sum of the area of the squares built upon the other two sides. These squares are known as Pythagoras squares.
Applications of Pythagoras Theorem
The applications of the Pythagoras theorem can be seen in our day-to-day life. Here are some of the applications of the Pythagoras theorem.
Engineering and Construction fields
Most architects use the technique of the Pythagorean theorem to find the unknown dimensions. When the length or breadth is known it is very easy to calculate the diameter of a particular sector. It is mainly used in two dimensions in engineering fields.
Face recognition in security cameras
The face recognition feature in security cameras uses the concept of the Pythagorean theorem, that is, the distance between the security camera and the location of the person is noted and well-projected through the lens using the concept.
Woodwork and interior designing
The Pythagoras concept is applied in interior designing and the architecture of houses and buildings.
Navigation
People traveling in the sea use this technique to find the shortest distance and route to proceed to their concerned places.
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Pythagorean Theorem Examples
Example 1: The hypotenuse of a right-angled triangle is 16 units and one of the sides of the triangle is 8 units. Find the measure of the third side using the Pythagoras theorem formula.
Solution:
Given: Hypotenuse = 16 units
Let us consider the given side of a triangle as the perpendicular height = 8 units
On substituting the given dimensions to the Pythagoras theorem formula
Hypotenuse2 = Base2 + Height2
162 = B2 + 82
B2 = 256 – 64
B = √192 = 13.856 units
Therefore, the measure of the third side of the triangle is 13.856 units.
Example 2: Julie wanted to wash her building window which is 12 feet off the ground. She has a ladder that is 13 feet long. How far should she place the base of the ladder away from the building?
Solution:
We can visualize this scenario as a right triangle. We need to find the base of the right triangle formed. We know that, Hypotenuse2 = Base2 + Height2. Thus, we can say that b2 = 132 – 122 where 'b' is the distance of the base of the ladder from the feet of the wall of the building. So, b2 = 132 – 122 can be solved as, b2 = 169 – 144 = 25. This means, b = √25 = 5. Hence, we get 'b' = 5.
Therefore, the base of the ladder is 5 feet away from the building.
Example 3: Use the Pythagoras theorem to find the hypotenuse of the triangle in which the sides are 8 units and 6 units respectively.
Practice Questions on Pythagoras Theorem
FAQs on Pythagoras Theorem
What is the Pythagoras Theorem in Math?
The Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem can be expressed as, c2 = a2 + b2; where 'c' is the hypotenuse and 'a' and 'b' are the two legs of the triangle. These triangles are also known as Pythagoras theorem triangles.
What is the Converse of Pythagoras Theorem?
The converse of Pythagoras theorem is: If the sum of the squares of any two sides of a triangle is equal to the square to the third (largest) side, then it is said to be a right-angled triangle.
What is the Use of the Pythagorean Theorem Formula?
The Pythagoras theorem works only for right-angled triangles. When any two values are known, we can apply the Pythagoras theorem and calculate the unknown sides of the triangle. There are other real-life applications of the Pythagoras theorem like in the field of navigation, engineering and architecture.
What is the use of the Pythagoras Theorem?
The Pythagorean theorem is used in various fields. A few of its applications are given below.
Architecture, construction and Navigation industries.
For computing the distance between points on the plane.
For calculating the perimeter, the surface area, the volume of geometrical shapes, and so on.
Can the Pythagorean Theorem Formula be Applied to any Triangle?
No, the Pythagorean theorem can only be applied to a right-angled triangle since the Pythagorean theorem expresses the relationship between the sides of the triangle where the square of the two legs is equal to the square of the third side which is the hypotenuse.
How to Work Out the Pythagoras Theorem?
Pythagoras theorem can be used to find the unknown side of a right-angled triangle. For example, if two legs of a right-angled triangle are given as 4 units and 6 units, then the hypotenuse (the third side) can be calculated using the formula, c2 = a2 + b2; where 'c' is the hypotenuse and 'a' and 'b' are the two legs. Substituting the values in the formula, c2 = a2 + b2 = c2 = 42 + 62 = 16 + 36 = √52 = 7.2 units.
What is the Formula of Pythagoras Theorem?
The formula of Pythagoras theorem is expressed as, Hypotenuse2 = Base2 + Height2. This is also written as, c2 = a2 + b2; where 'c' is the hypotenuse and 'a' and 'b' are the two legs of the right-angled triangle. Using the Pythagoras theorem formula, any unknown side of a right-angled can be calculated if the other two sides are given.
Why is the Pythagoras Theorem Important?
The Pythagoras theorem is important because it helps in calculating the unknown side of a right-angled triangle. It has other real-life applications in the field of architecture and engineering, navigation, and so on.
How is Pythagoras Theorem used in Navigation?
The Pythagoras theorem is commonly used in air navigation and ship navigation. The Pythagoras theorem provides a way to the ship's navigator to calculate the distance to a point in the ocean, for example, if the distance between two points is given as 600 km north and 800 km west, the required distance can be calculated using the Pythagoras theorem.
When is Pythagoras Theorem used?
Pythagoras theorem is used when any two sides of a right-angled triangle are given and the third side needs to be calculated. For example, if the perpendicular and base of a right-angled triangle are given as 12 units and 5 units respectively, and we need to find the third side (the hypotenuse) we can calculate it using the theorem which says hypotenuse2 = perpendicular2 + base2. After substituting the values in the equation we get hypotenuse2 = 122 + 52 = 144 + 25 = 169. So, hypotenuse = √169 = 13 units.
What is the Pythagoras Property of Triangles?
The Pythagoras property of triangles is another term for the Pythagoras theorem. According to the Pythagoras property, in a right-angled triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides. This theorem is expressed as, c2 = a2 + b2; where 'c' is the hypotenuse and 'a' and 'b' are the two legs of the triangle. | 677.169 | 1 |
A Holistic Analysis Of Pythagoras Theorem Formula
Pythagoras Theorem In the discipline of mathematics, the Pythagoras theorem holds immense significance and had unfolded different mysteries and areas of research in the triangle geometry. As the name signifies, the theorem was found by the Greek mathematician Pythagoras. The mathematician was born in the year 569 BC on the island of Samos, Greece. He has interjected an immense contribution in the field of mathematical subdivisions like polyhedral, areas, angles, polygons, triangles and other polygons. Not only limiting his contribution to academic mathematics, but he has also set many inventions and theories in musical mathematics. Among all these advances, his Pythagorean theorem is considered as the crucial point in the evolution of the present-day discipline of mathematics. Although he had made this much of advances in the field of mathematics, not many pieces of evidence are found regarding the implication of this theorem by the Babylonian mathematicians. Hence it could be deduced that his theorems were not well accepted by his contemporary colleagues. Because of his inventions, Pythagoras of Samos is nicknamed as the first pure mathematician. His theorem has been later researched by many of the specialists and currently, it has around 400 of its variations. There are many versions of the Pythagoras theorem published by different researchers, and to eliminate any confusion or anomaly we have provided a generic theorem in the below section
In a right-angled triangle, the total sum of the individual squares of the base and altitude would be equal to the square of the length of the hypotenuse.
Let us denote the base of the right-angled triangle as letter 'b', the altitude of the triangle as 'a' and hypotenuse of the same triangle as the letter 'h', then as per the Pythagoras theorem
h2 = b2 + a2
The numerical triplets which would satisfy this formula are termed as Pythagorean triples. The numerical triplets like (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), etc. comes under the category of Pythagorean triplets. The multiples of these basic triplets also behave like a normal triplet, and if considered (a, b, c) as a basic Pythagorean triplet then the multiple of these triplets (ka, kb, kc) also comes under the category of the Pythagorean triplet. It should be noted that the constant 'k' in this context is a non-negative integer.
Evolution of the Pythagoras theorem As mentioned earlier, the Pythagoras theorem formula is considered as one of the most renowned and path clearing concept in the history of mathematics. This theorem is so significant that even if today a student studying in high school is asked to state a theorem, it is most probable that he would state this theorem. Though the suspicions are being raised by the historians that there is a possibility that this theorem was being discovered by mathematicians and scholars in earlier times at different places.
If going down to the ancient history of India, the text Baudhayana Sulba Sutra, an ancient document discusses Mn of the Pythagorean triplets which were derived algebraically. These calculations were pertaining to the right-angled isosceles triangles. By utilizing the methodology of area computation, the numerical proof has been derived for the Pythagoras theorem in the document of Apastamba Sulba Sutra. These texts were confirmed to be drafted around the year of 600 BC. In the Chinese documents like Zhou Bi Suan Jing, the Pythagoras theorem is being and explained in a very efficient way.
If you search the mathematical research papers, you could come across derivations and explanation for the Pythagoras theorem. Scholars have stated that these numbers of derivations sum up to around 100. The oldest explanation is contemporary to the age of Euclid and the latest one dates to the 21st century. Below is provided one of the most accepted derivations for the Pythagoras theorem formula.
Proof: Consider a square in which four right-angled triangles with sides a, b, and c are arranged inwards along its perimeter so that the hypotenuse, c of the right-angled triangles turn out to be the side of the square. There would be a void in between the arrangements of the right-angled triangle which is in the shape of a square with side b-a. From the below diagram it would be evident that the area of each triangle would be equal to ½ ab.
Algebraically, the area of the whole square considering the dimensions of the triangles and inner square would be
(b-a)2 + (4/2) ab = a2 + b2
It is also clear that the actual length of the square is c and hence they are of the square would be c2
Hence c2 = a2 + b2
Augmented studies on this theorem
As per the principles of mathematics and its subsidiaries, a comprehensive theorem is the one that brings the way to the innovation of new theorems. The Pythagoras theorem had even led to the generation of a new subtopic in mathematics, trigonometry. The scope of mathematics is very large and had been the basis of many scientifically and mathematically diverse theorems.
A set of numerical triplets would be termed as the primitive Pythagorean triplet if their GCD is 1. To prove this statement, let us look at the theorem and proof given below.
Theorem 1
Consider two variables, v and u which are relatively prime positive integers and along with it they are reasonably odd by satisfying the condition u > v.
Let us take into account another set of variables, a, b, and c, where a = 2uv, b = u2 – v2 , and c = u2 + v2 , where it is already provided that gcd (a, b, c) = 1 and c2 = a2 + b2 .
If referred to one of the prominent books, Arithmetica written by Diophantus, very elucidated and authentic proof for this theorem could be observed. This proof had led to one of the prominent and path-breaking discoveries in the discipline of number theory which was later termed as Fermat's last theorem. Below is provided the general and simplified statement of Fermat's last theorem.
Theorem 2: Fermat's Last Theorem: If taken into account the following theorem there would be no solution for the algebraic triplets (X, Y, Z) or integers.
XP + YP + ZP = 0
Provided that the XYZ ≠ 0 and P ≥ 3 is a prime.
The French mathematician Pierre de Fermat had mentioned in his theorem that the segregation of the cube is not feasible and any variable having the power greater than two could not be segregated into equal parts of the same degree.
In the modern era, the Pythagorean theorem is not only used by the mathematicians but also the businessmen and artists. Though the Pythagoras theorem formula is still considered to be the most prominent, relevant, powerful, complex, intellectual, and perplexed ideology in mathematics.
Let us have a look at some of the theorems derived from the Pythagoras theorem formula
Distance Formula As per this theory, the formula of the distance could be mentioned as
D = d=∆x2 + ∆y2
Where the variable x = x1 – x2 and y = y1 – y2
If the provided distance formula is squared on either side, then you would come up with the Pythagoras theorem formula, which is
D2 = x2 + y2
Cosine Law
The cosine theorem is the most prominent and crucial theorem which was derived out of the Pythagoras theorem formula. By implying the law of cosine, anyone could find the third side if the other two sides are given along with the angle entailed between them. Below is provided the formula to be followed in the cosine law.
c2= a2+ b2– 2ab cos γ
where the variables a, b, and c denote the length of the sides of the provided triangle and y is the value of the angle created in between the two sides i.e. between a and b hence opposite to the side c.
The cosine would consequentially turn out to be zero if the angle y opposite to the side c and in between the sides a and b to be a right angle.
This would satisfy the Pythagoras theorem formula c2 = a2 + b2
Below are given some examples of problems.
Problem 1
Consider a right triangle angled with height 5 m and base 12 m. Calculate the length if hypotenuse of the same right-angled triangle.
Problem 3
Imagine a right-angled Δ ABC with its hypotenuse of length 50 m and length of the base 30 m. Calculate the altitude or the perpendicular height of Δ ABC.
Answer
While implying the Pythagoras theorem in this context, the below formula could be implemented
C2 = A2 + B2
To deduct the perpendicular height from the above equation, we could imply that
A2 = C2 – B2
A = √ (C2 – B2)
Hence if implied the values in this expression
A = √ (502 – 302)
= √ 900 = 30
Problem 4
By taking into account the length of all the sides of the triangle, determine whether it is a right-angled triangle.
AC = 75, BA = 45, CB = 55.
Answer:
To determine whether the triangle is right-angled, you should prove that the given values of the sides should satisfy the Pythagoras theorem. If considered the sides of a right-angled triangle, the side with the greatest sides should be the hypotenuse. Hence the greatest side would be AC, having the length 75. Let us take the side BA with length 45 as the perpendicular height of the triangle ABC | 677.169 | 1 |
Isosceles triangles are a fundamental concept in geometry, captivating mathematicians and learners alike with their unique properties and applications.In this extensive exploration, we'll unravel the mysteries surrounding these intriguing polygons, covering everything from their definition to practical examples and FAQs.
Definition of Isosceles Triangle
An isosceles triangle is a triangle with two sides of equal length. This characteristic sets it apart from other types of triangles, making it a fascinating subject of study in geometry. The equal sides are known as legs, while the remaining side is called the base
Angles of Isosceles Triangle
In an isosceles triangle, the angles opposite the equal sides are congruent.These angles are typically referred to as the base angles, while the angle formed by the two equal sides is known as the vertex angle.
Simplifying Right Triangle Trigonometry: Tips and Tricks Do you ever listen to right-angle triangles? If yes, then you
Practice Questions
Calculate the area of an isosceles triangle with a base of 10 units and a height of 8 units.
Determine the perimeter of an isosceles triangle with two equal sides measuring 12 units each and a base of 9 units.
Conclusion:
In this way, you've got a good grounding of this enigmatic phenomenon that tires you to know the inner working of it and applying it at your playthroughs. Keep reading, and you will eventually understand it is fun to get lost in the mysterious depths of geometry! | 677.169 | 1 |
How do you measure and classify angles?
When two rays meet at a point In plane geometry, they form an angle at that common endpoint, called the vertex of the angle. Angles lie in a plane, but this plane does not have to be a Euclidean plane. There are a variety of different types of angles. These types include straight, right, acute or obtuse. These are a portion of the total circle that 360˚. A straight angle is the same as half the circle and is 180° whereas a right angle is a quarter of a circle and is 90°.
All these can be measured using a protractor in the geometry box, that is shaped like a capital D.
You will definitely need a protractor for these worksheets and lessons. The lesson walks you step-by-step through measuring the angles by matching them arms up on the protractor. Once you know the measure of the angle we classify the angles. Straight angles are basically straight lines and have a measure of 180 degrees. Right angles measure 90 degrees. Acute angles are less than 90 degrees. Obtuse angles are the goldilocks of angles; they measure between 90 and 180 degrees. Reflex angles get a bit extreme and measure between 180 and 360 degrees. These worksheets explain how to classify angles as straight, right, acute, obtuse, or reflex. The different angle classifications are defined for students. | 677.169 | 1 |
Trapezoid Sentence Examples
Three races were sailed on a trapezoid course with different conditions in all of them.
10
4
Vanessa Minnillow has a 4 carat diamond ring set in platinum with trapezoid diamonds from Nick Lachey.
15
9
The demihexagon is an isosceles trapezium (trapezoid in American usage) in which three of the sides have the same length.
7
2
Central leg allows for trapezoid tables to be used on both sides.
5
1
The new lights are equally imposing, with a trapezoid shape incorporating a subtle circular highlight that cuts into the bumper.
3
1
The handlebar, trapezoid headlamp, front mudguard and trimmed-down rear lids were all new.
3
1
The head when viewed anteriorly is narrowly trapezoid in shape.
3
1
The second row consists of a broad and flat magnum, supporting the great third metacarpal, having to its radial side the trapezoid, and to its ulnar side the unciform, which are both small, and articulate inferiorally with the rudimentary second and fourth metacarpals.
3
2
The ring is a 16.5 carat emerald cut white diamond center stone with two 2-carat side trapezoid white diamonds set in a platinum band for a total of nearly 20.5 carats.
The trapezium is also sometimes called a " trapezoid," but it will be convenient to reserve this term for a different figure (§ 24).
14
2010
19
The trapezoid and magnum of the carpus, and the cuboid and navicular of the tarsus are distinct. | 677.169 | 1 |
Dentro del libro
Resultados 1-5 de 18
Página 6 ... construct a figure , or to solve a question . A Theorem ( from theoreema , a subject of contemplation ) , is the assertion of a geometrical truth , and requires demonstration . The Data ( from datum , a thing granted ) , are the things ...
Página 13 ... constructed : as the Sextant , an arc of 60 degrees ; the Quadrant , an arc of 90 degrees ; and the Theodolite , a whole circle of 360 degrees , divided into two semicircles , of 180 degrees each . The Theodolite ( a word of uncertain ...
Página 18 ... construct a Scale of Equal Parts ; Take AB of indefinite length towards B , and let C be the given st . line or part , that is to be cut off from AB : from AB cut off a part equal to C , as AE ; then from EB another part equal to C , as ...
Página 19 ... construct figures that are a true index of the positions and real distances of cities , mountains and seas , and in some respects of the constellations of heaven . PROP . 4. - THEOR .- ( Important . ) If two triangles have two sides of | 677.169 | 1 |
Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It has numerous applications in various fields, including physics, engineering, and computer science. One of the fundamental concepts in trigonometry is the cos(a-b) formula, which allows us to find the cosine of the difference between two angles. In this article, we will explore the cos(a-b) formula in detail, understand its derivation, and examine its practical applications.
What is the Cos(a-b) Formula?
The cos(a-b) formula is a trigonometric identity that expresses the cosine of the difference between two angles, a and b, in terms of the cosines and sines of those angles. It is derived from the more general trigonometric identity known as the cosine of the sum of two angles, cos(a+b).
The cos(a-b) formula is given by:
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
This formula allows us to find the cosine of the difference between two angles without directly calculating the individual cosines of those angles. Instead, it relies on the known values of the cosines and sines of the angles involved.
Derivation of the Cos(a-b) Formula
The derivation of the cos(a-b) formula involves manipulating the cosine of the sum of two angles, cos(a+b), using trigonometric identities. Let's go through the derivation step by step:
Start with the trigonometric identity for the cosine of the sum of two angles:
cos(a+b) = cos(a)cos(b) – sin(a)sin(b)
Replace b with -b:
cos(a+(-b)) = cos(a)cos(-b) – sin(a)sin(-b)
Use the trigonometric identity for the cosine of a negative angle:
cos(-b) = cos(b)
Use the trigonometric identity for the sine of a negative angle:
sin(-b) = -sin(b)
Substitute the values from steps 3 and 4 into step 2:
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
Thus, we have derived the cos(a-b) formula from the cosine of the sum of two angles identity.
Applications of the Cos(a-b) Formula
The cos(a-b) formula finds applications in various fields, including physics, engineering, and computer science. Let's explore some practical examples where this formula is used:
1. Navigation and GPS Systems
In navigation and GPS systems, the cos(a-b) formula is used to calculate the distance and direction between two points on the Earth's surface. By knowing the latitude and longitude of two locations, we can determine the angle between them and use the cos(a-b) formula to find the cosine of that angle. This information is crucial for determining the shortest distance and the direction to travel.
2. Robotics and Automation
In robotics and automation, the cos(a-b) formula is used to calculate the angles and positions of robotic arms and manipulators. By knowing the lengths of the arm segments and the desired position, the cos(a-b) formula can be used to determine the required joint angles to reach that position accurately.
3. Signal Processing
In signal processing, the cos(a-b) formula is used in various applications, such as audio and image compression. By representing signals in the frequency domain using Fourier transforms, the cos(a-b) formula helps in analyzing and manipulating the signals efficiently.
Examples of Using the Cos(a-b) Formula
Let's consider a few examples to illustrate the practical use of the cos(a-b) formula:
Example 1: Finding the Cosine of the Difference between Two Angles
Suppose we want to find the cosine of the difference between two angles, a = 30 degrees and b = 45 degrees. Using the cos(a-b) formula, we can calculate:
cos(30-45) = cos(30)cos(45) + sin(30)sin(45)
Using the known values of cos(30) = √3/2, cos(45) = √2/2, sin(30) = 1/2, and sin(45) = √2/2, we can substitute these values into the formula:
cos(30-45) = (√3/2)(√2/2) + (1/2)(√2/2)
Simplifying the expression, we get:
cos(30-45) = (√6 + √2)/4 ≈ 0.612
Therefore, the cosine of the difference between 30 degrees and 45 degrees is approximately 0.612.
Example 2: Calculating the Distance between Two Points
Suppose we have two points on the Earth's surface with latitude and longitude coordinates: Point A (40.7128° N, 74.0060° W) and Point B (34.0522° N, 118.2437° W). We want to calculate the distance between these two points using the cos(a-b) formula.
First, we convert the latitude and longitude coordinates from degrees to radians. Using the conversion factor π/180, we get:
Latitude of Point A in radians: 40.7128° × π/180 ≈ 0.710
Longitude of Point A in radians: -74.0060° × π/180 ≈ -1.291
Latitude of Point B in radians: 34.0522° × π/180 ≈ 0.594
Longitude of Point B in radians: -118.2437° × π/180 ≈ -2.063
Next, we can use the Haversine formula, which involves the cos(a-b) formula, to calculate the distance between the two points | 677.169 | 1 |
thefifthpubhouseandcafe
How do I solve this complicated problem?
Accepted Solution
A:
Interesting problem! In order to solve this one, we'll know some information about the measure of these triangles' angles. First, remember that triangles have the very unique property that *equal angles sweep out equal lengths*. The other side of that is that if two sides of a triangle are congruent, we know the angles opposite them are congruent, too.
Since all the sides of triangle ABE are congruent, all its angles are congruent, and since the angles of a triangle add to 180, this means that each of the angles in ABE must be 60 degrees (since 60 + 60 + 60 = 180).
The other triangle of notice in the problem is triangle ABC. Notice that, since it's formed from a square, it's a *right* triangle. This means that one of its angles must be 90. Sharing two sides with the square ABCD also means that two of the triangle's sides are congruent, which *also* means the two non-right angles must be congruent. This forces them to be 45 degrees (since 90 + 45 + 45 = 180).
Now, take a look at angle EAC. That small angle is what we end up with if we take angle EAB (an angle of triangle ABE) and subtract angle CAB (part of triangle ABC) from it. EAB measures 60 degrees, and CAB measures 45, so angle EAB must be 60 - 45 = 15 degrees.
At this point, we're almost done. I want to bring your attention to the point where line segments EB and AC intersect; let's call that point F. If we want to find the measure of x, we can look at the angle right across from it, angle AFE. Notice that this angle is one of three in the triangle AFE. We've already found the other two, actually: angle AEF is 60 degrees, and angle EAF is 15, so to find AFE, we can simply subtract those two from 180 to find a measure of
180 - 60 - 15 = 180 - 75 = 105 degrees.
And since AFE and x are vertical angles, we know x must also be 105 degrees. | 677.169 | 1 |
arc 9
In this construction, dividing AB into 6 equal parts, two circles are drawn. They define an equilateral triangle in the middle. Extending one side of this triangle we can construct the two parts of the arc. | 677.169 | 1 |
Law of Sines Example Problem
The law of sines is a useful rule showing a relationship between an angle of a triangle and the length of the side opposite of the angle.
The law is expressed by the formula
The sine of the angle divided by the length of the opposite side is the same for every angle and its opposing side of the triangle.
Law of Sines – How does it work?
It is easy to show how this law works. First, let's take the triangle from above and drop a vertical line to the side marked c .
This cuts the triangle into two right triangles which share a common side marked h.
The sine of an angle in a right triangle is the ratio of the length of the side opposite of the angle to the length of the hypotenuse of the right triangle. In other words:
Take the right triangle including the angle A . The length of the side opposite of A is h and the hypotenuse is equal to b .
Solve this for h and get
h = b sin A
Do the same thing for the right triangle including angle B . This time, the length of the side opposite of B is still h but the hypotenuse is equal to a .
h = a sin B
Since both of these equations are equal to h, they are equal to each other.
b sin A = a sin B
We can rewrite this to get the same letters on the same side of the equation to get
You can repeat process for every angle and get the same result. The overall result will be the law of sines.
Question: Use the law of sines to find the length of the side x.
Solution: The unknown side x is opposite the 46.5° angle and the side with length 7 is opposite the 39.4° angle. Plug these values into the Law of Sines equation.
Solve for x
7 sin(46.5°) = x sin(39.4°)
7 (0.725) = x (0.635)
5.078 = x (0.635)
Answer: The unknown side is equal to 8.
Bonus: If you wanted to find the missing angle and length of the last side of the triangle, remember that all three angles of a triangle all add up to 180°.
180° = 46.5° + 39.4° + C C = 94.1°
Use this angle in the law of sines the same way as above with either of the other angles and get a length of side c equal to 11.
Potential Issue of the Law of Sines
One potential problem to keep in mind using the law of sines is the possibility of two answers for an angle variable. This tends to appear when you are given two side values and an acute angle not between the two sides.
These two triangles are an example of this problem. The two sides are 100 and 75 in length and the 40° angle is not between these two sides. Notice how the side with length 75 could swing to hit a second place along the bottom side. Both of these angles will give a valid answer using the law of sines.
Fortunately, these two angle solutions add up to 180°. This is because the triangle formed by the two 75 sides is an isosceles triangle (triangle with two equal sides). The angles between the sides and their shared side will also be equal to each other. This means the angle on the other side of the angle θ will be the same as angle φ. The two angles added together make a straight line, or 180°.
Law of Sines Example Problem 2
Question : What are the two possible angles of a triangle with sides of 100 and 75 with a 40° as marked in the triangles above?
Solution : Use the law of sines formula where the 75 length is opposite of 40°, and 100 is opposite of θ.
sin θ = 0.857
θ + φ = 180°
φ = 180° – θ
φ = 180° – 58.97° φ = 121.03°
Answer : The two possible angles for this triangle is 58.97° and 121.03°.
Science Notes Trigonometry Help
Law of Cosines Example Problems
Right Triangles – Trigonometry Basics
Right Triangle Trigonometry and SOHCAHTOA
SOHCAHTOA Example Problem – Trigonometry Help
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Trig Identities Study Sheet PDF
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Great law of sines problems
These law of sines problems below will show you how to use the law of sines to solve some real life problems. You will need to use the sine formula shown below to solve these problems.
Law of sines
The ratio of the sine of an angle of a scalene triangle to the side opposite that angle is the same for all angles and sides in the triangle.
sin A / a = sin B / b = sin C / c
Law of sines problems
Solving an angle-side-angle (asa) triangle with the law of sines.
Two How far is the fire from station A?
Solution The biggest trick is this problem is to understand the meaning of N52°E and N36°W.
N52°E means 52 degrees east of north.
N36°W means 36 degrees west of north.
Here is what the graph will look like after you draw it.
Notice that to find 38°, you need to subtract 52° from 90°. By the same token, to find 54°, you need to subtract 36° from 90°.
Notice also that the angle opposite 15 is missing, so we need to find it.
38 + 54 + missing angle = 180
92 + missing angle = 180, so missing angle = 88
Now, we can use the sine rule to find the distance the fire is from station A.
Let x be the distance from the fire to station A.
Solving an angle-angle-side (AAS) triangle with the law of sines
The leaning tower of pisa is inclined 5.5 degrees from the vertical. At a distance of 100 meters from the wall of the tower, the angle of elevation to the top is 30.5 degrees. Use the law of sines to estimate the height of the leaning tower (height as shown in the image below with the green line)
The trickiest thing here is making the graph. We show it below. Notice that the height is shown with a green line.
Let us find angle y and angle z
angle y + 5.5 = 90, so angle y = 84.5
Notice that knowing the value of the angle y is not enough to find the unknown height of the leaning tower of pisa. You must know the value of the angle z!
84.5 + 30.5 + angle z =180
115 + angle z = 180
angle z = 65
Now, we can use law of sines.
Solving a side-side-angle (SSA) triangle with the ambiguous case of the law of sines
A homeowner is trying to build a raised garden bed that has a triangular shape. His neighbor gave him two pieces of lumber with lengths 20 feet and 8 feet and he puts them together to begin his triangle. He is planning to go to the store to buy the third piece of lumber to finish the triangle. He wants to build the triangular garden so that third piece of lumber will make an angle of 40 degrees with the piece that is 20 feet long. Will it be possible for him to make the garden? If so, how long should the third piece of lumber be?
If the homeowner could build such a garden, it will look like the figure shown above. An angle A would then be formed between the third piece of lumber and the piece that is 8 feet long.
Let us use the law of sines to see if this is possible.
sin (40 degrees) / 8 = sin (A) / 20
0.642 / 8 = sin (A) / 20
0.08025 = sin (A) / 20
Multiply both sides of the equation by 20
0.08025(20) = [sin (A) / 20](20)
0.08025(20) = sin (A)
1.605 = sin(A)
The sine of an angle must be smaller than 1. Since Angle A cannot be formed, it will not be possible to make such a garden. Since the angle cannot be formed, the piece that is 8 feet long will never meet with the third piece. As a result, he will never know the length of the third piece.
If the homeowner knows how to use the ambiguous case of law of sines , then he will do that first before driving to the store!
You cannot use the law of sines to solve word problems if the triangle you end up with is an SSS triangle or an SAS triangle. You must use the law of cosines instead,
Proof of the law of sines
Law of cosines
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The Law of Sines
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10.1 Non-right Triangles: Law of Sines
Learning objectives.
In this section, you willlique .
While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.
Law of Sines
Given a triangle with angles and opposite sides labeled as in Figure 6 , solve an oblique triangle, use any pair of applicable ratios.
Solving for Two Unknown Sides and Angle of an AAS Triangle
Solve the triangle shown in Figure 7 to the nearest tenth.
The three angles must add up to 180 degrees. From this, we can determine that
To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° α = 50° and its corresponding side a = 10. a = 10. We can use the following proportion from the Law of Sines to find the length of c . c .
Similarly, to solve for b , b , we set up another proportion.
Therefore, the complete set of angles and sides is
Solve the triangle shown in Figure 8 to the nearest tenth.Possible Outcomes for SSA Triangles
Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides a a and b b and known angle α . α .
Solving an Oblique SSA Triangle
Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth.
Use the Law of Sines to find angle β β and angle γ , γ , and then side c . c . Solving for β , β , we have the proportion
However, in the diagram, angle β β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β ? β ? Let's investigate further. Dropping a perpendicular from γ γ and viewing the triangle from a right angle perspective, we have Figure 11 . It appears that there may be a second triangle that will fit the given criteria.
The angle supplementary to β β is approximately equal to 49.9°, which means that β = 180° − 49.9° = 130.1° . β = 180° − 49.9° = 130.1° . (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ , γ , we have
We can then use these measurements to solve the other triangle. Since γ ′ γ ′ is supplementary to the sum of α ′ α ′ and β ′ , β ′ , we have
Now we need to find c c and c ′ . c ′ .
To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12 .
However, we were looking for the values for the triangle with an obtuse angle β . β . We can see them in the first triangle (a) in Figure 12 .
Given α = 80° , a = 120 , α = 80° , a = 120 , and b = 121 , b = 121 , find the missing side and angles. If there is more than one possible solution, show both.
Solving for the Unknown Sides and Angles of a SSA Triangle
In the triangle shown in Figure 13 , solve for the unknown side and angles. Round your answers to the nearest tenth.
In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85° , γ = 85° , and its corresponding side c = 12 , c = 12 , and we know side b = 9. b = 9. We will use this proportion to solve for β . β .
To find β , β , apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β . β . It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.
In this case, if we subtract β β from 180°, we find that there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7° . β = 180° − 48.3° ≈ 131.7° . To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives
which is impossible, and so β ≈ 48.3° . β ≈ 48.3° .
To find the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7° . α = 180° − 85° − 48.3° ≈ 46.7° . Now, only side a a is needed. Use the Law of Sines to solve for a a by one of the proportions.
The complete set of solutions for the given triangle is
Given α = 80° , a = 100 , b = 10 , α = 80° , a = 100 , b = 10 , find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.
Finding the Triangles That Meet the Given Criteria
Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.
Using the given information, we can solve for the angle opposite the side of length 10. See Figure 14 .
We can stop here without finding the value of α . α . Because the range of the sine function is [ − 1 , 1 ] , [ − 1 , 1 ] , it is impossible for the sine value to be 1.915. In fact, inputting sin − 1 ( 1.915 ) sin − 1 ( 1.915 ) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions. Area = 1 2 b h , Area = 1 2 b h , where b b is base and h h is height. For oblique triangles, we must find h h before we can use the area formula. Observing the two triangles in Figure 15 , one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sin α = opposite hypotenuse sin α = opposite hypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sin α = h c sin α = h c or c sin α = h . c sin α = h . However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b b to form a right triangle. The angle used in calculation is α ′ , α ′ , or 180 − α . 180 − α .
Area of an Oblique Triangle
The formula for the area of an oblique triangle is given by
This is equivalent to one-half of the product of two sides and the sine of their included angle.Finding an Altitude
Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16 . Round the altitude to the nearest tenth of a mile.
To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a , a , and then use right triangle relationships to find the height of the aircraft, h . h . distance from one station to the aircraft is about 14.98 miles.
Now that we know a , a , we can use right triangle relationships to solve for h . h .
The aircraft is at an altitude of approximately 3.9 miles B , B , is 62°, and the distance between the viewing points of the two end zones is 145 yards.
Access these online resources for additional instruction and practice with trigonometric applications.
Law of Sines: The Basics
Law of Sines: The Ambiguous Case
10.1 Section Exercises
Describe the altitude of a triangle.
Compare right triangles and oblique triangles.
When can you use the Law of Sines to find a missing angle?
In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?
What type of triangle results in an ambiguous case?
For the following exercises, assume α α is opposite side a , β a , β is opposite side b , b , and γ γ is opposite side c . c . Solve each triangle, if possible. Round each answer to the nearest tenth.
α = 43° , γ = 69° , a = 20 α = 43° , γ = 69° , a = 20
α = 35° , γ = 73° , c = 20 α = 35° , γ = 73° , c = 20
α = 60° , α = 60° , β = 60° , β = 60° , γ = 60° γ = 60°
a = 4 , a = 4 , α = 60° , α = 60° , β = 100° β = 100°
b = 10 , b = 10 , β = 95° , γ = 30° β = 95° , γ = 30°
For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A A is opposite side a , a , angle B B is opposite side b , b , and angle C C is opposite side c . c .
For the following exercises, assume α α is opposite side a , β a , β is opposite side b , b , and γ γ is opposite side c . c . Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.
α = 119° , a = 14 , b = 26 α = 119° , a = 14 , b = 26
γ = 113° , b = 10 , c = 32 γ = 113° , b = 10 , c = 32
b = 3.5 , b = 3.5 , c = 5.3 , c = 5.3 , γ = 80° γ = 80°
a = 12 , a = 12 , c = 17 , c = 17 , α = 35° α = 35°
a = 20.5 , a = 20.5 , b = 35.0 , b = 35.0 , β = 25° β = 25°
a = 7 , a = 7 , c = 9 , c = 9 , α = 43° α = 43°
a = 7 , b = 3 , β = 24° a = 7 , b = 3 , β = 24°
b = 13 , c = 5 , γ = 10° b = 13 , c = 5 , γ = 10°
a = 2.3 , c = 1.8 , γ = 28° a = 2.3 , c = 1.8 , γ = 28°
β = 119° , b = 8.2 , a = 11.3 β = 119° , b = 8.2 , a = 11.3
For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.
In Figure 26 , A B C D A B C D is not a parallelogram. ∠ m ∠ m is obtuse. Solve both triangles. Round each answer to the nearest tenth.
Real-World Applications
A pole leans away from the sun at an angle of 7° 7° to the vertical, as shown in Figure 27 . When the elevation of the sun is 55° , 55° , the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.
To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 28 . Determine the distance of the boat from station A A and the distance of the boat from shore. Round your answers to the nearest whole foot.
Figure 29 shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A A and B , B , which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A A and B B are measured to be 83.9° 83.9° and 86.2° , 86.2° , respectively. How far is the satellite from station A A and how high is the satellite above the ground? Round answers to the nearest whole mile.
A communications tower is located at the top of a steep hill, as shown in Figure 30 . The angle of inclination of the hill is 67° . 67° . A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is 16°. 16°. Find the length of the cable required for the guy wire to the nearest whole meter.
The roof of a house is at a 20° 20° angle. An 8-foot solar panel is to be mounted on the roof and should be angled 38° 38° relative to the horizontal for optimal results. (See Figure 31 ). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth.
Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer's line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° 37° and 44° , 44° , as shown in Figure 32 . Find the distance of the plane from point A A to the nearest tenth of a kilometer.
A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33 . Find the distance of the plane from point A A to the nearest tenth of a kilometer.
In Assuming that the street is level, estimate the height of the building to the nearest foot.
InPoints A A and B B are on opposite sides of a lake. Point C C is 97 meters from A . A . The measure of angle B A C B A C is determined to be 101°, and the measure of angle A C B A C B is determined to be 53°. What is the distance from A A to B , B , rounded to the nearest whole meter?
A man and a woman standing 3 1 2 3 1 2TwoAThree cities, A , B , A , B , and C , C , are located so that city A A is due east of city B . B . If city C C is located 35° west of north from city B B and is 100 miles from city A A and 70 miles from city B , B , how far is city A A from city B ? B ? Round the distance to the nearest tenth of a mile.
TwoBrian's house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in Figure 34 .
TheA yield sign measures 30 inches on all three sides. What is the area of the sign?
Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 35 Jay Abramson
Publisher/website: OpenStax
Book title: Algebra and Trigonometry
Publication date: Feb 13, 2015
Location: Houston, Texas
Book URL:
4.1.7: Trigonometry Word Problems
Angle of Depression and Angle of Elevation
One application of the trigonometric ratios is to find lengths that you cannot measure. Very frequently, angles of depression and elevation are used in these types of problems.
Angle of Depression: The angle measured down from the horizon or a horizontal line.
Angle of Elevation: The angle measured up from the horizon or a horizontal line.
What if you placed a ladder 10 feet from a haymow whose floor is 20 feet from the ground? How tall would the ladder need to be to reach the haymow's floor if it forms a \(30^{\circ}\) angle with the ground?
Example \(\PageIndex{1}\)
A math student is standing 25 feet from the base of the Washington Monument. The angle of elevation from her horizontal line of sight is \(87.4^{\circ}\). If her "eye height" is 5 ft, how tall is the monument?
Example \(\PageIndex{3}\)
Elise is standing on top of a 50 foot building and sees her friend, Molly. If Molly is 30 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise's eye height is 4.5 feet.
Because of parallel lines, the angle of depression is equal to the angle at Molly, or \(x^{\circ}\). We can use the inverse tangent ratio.
\(\tan^{−1} \left(\dfrac{54.5}{30}\right)=61.2^{\circ}=x\)
Example \(\PageIndex{4}\)
Mark is flying a kite and realizes that 300 feet of string are out. The angle of the string with the ground is \(42.5^{\circ}\). How high is Mark's kite above the ground?
It might help to draw a picture. Then write and solve a trig equation.
Use what you know about right triangles to solve for the missing angle. If needed, draw a picture. Round all answers to the nearest tenth of a degree.
A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building?
Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation?
A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck?
Standing 100 feet from the base of a building, Sam measures the angle to the top of the building from his eye height to be \(50^{\circ}\). If his eyes are 6 feet above the ground, how tall is the building?
Over 4 miles (horizontal), a road rises 200 feet (vertical). What is the angle of elevation?
A 90 foot building casts an 110 foot shadow. What is the angle that the sun hits the building?
Luke is flying a kite and realizes that 400 feet of string are out. The angle of the string with the ground is \(50^{\circ}\). How high is Luke's kite above the ground?
An 18 foot ladder rests against a wall. The base of the ladder is 10 feet from the wall. What angle does the ladder make with the ground?
Review (Answers)
To see the Review answers, open this PDF file and look for section 8.9.
Additional Resources
Interactive element.
Video: Trigonometry Word Problems Principles - Basic
Activities: Trigonometry Word Problems Discussion Questions
Practice: Trigonometry Word Problems
Real World: Measuring Mountains2: The Law of Sines
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Carl Stitz & Jeff Zeager
Lakeland Community College & Lorain County Community College
Trigonometry literally means 'measuring triangles' and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to 'solve' triangles – that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 10.2 , 10.3 and 10.6 , we've had some experience solving right triangles. The following example reviews what we know.
Example 11.2.1
Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree.
For definitiveness, we label the triangle below.
To find the length of the missing side \(a\), we use the Pythagorean Theorem to get \(a^{2}+4^{2}=7^{2}\) which then yields \(a=\sqrt{33}\) units. Now that all three sides of the triangle are known, there are several ways we can find \(\alpha\) using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to \(\alpha\). According to Theorem 10.4 , \(\cos (\alpha)=\frac{4}{7}\). Since \(\alpha\) is an acute angle, \(\alpha=\arccos \left(\frac{4}{7}\right)\) radians. Converting to degrees, we find \(\alpha \approx 55.15^{\circ}\). Now that we have the measure of angle \(\alpha\), we could find the measure of angle \(\beta\) using the fact that \(\alpha\) and \(\beta\) are complements so \(\alpha+\beta=90^{\circ}\). Once again, we opt to use the data given to us in the problem. According to Theorem 10.4 , we have that \(\sin (\beta)=\frac{4}{7}\) so \(\beta=\arcsin \left(\frac{4}{7}\right)\) radians and we have \(\beta \approx 34.85^{\circ}\).
A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle 1 and the corresponding lowercase English letter represents the side 2 opposite that angle. Thus, \(a\) is the side opposite \(\alpha, b\) is the side opposite \(\beta\) and \(c\) is the side opposite \(\gamma\). Taken together, the pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it minimizes the chances of propagated error. 3 Third, since many of the applications which require solving triangles 'in the wild' rely on degree measure, we shall adopt this convention for the time being. 4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn't a right triangle? In certain cases, we can use the Law of Sines to help.
Theorem 11.2. The Law of Sines
Given a triangle with angle-side opposite pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\), the following ratios hold
The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle \(\triangle A B C\) below, all of whose angles are acute, with angle-side opposite pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\). If we drop an altitude from vertex \(B\), we divide the triangle into two right triangles: \(\triangle A B Q\) and \(\triangle B C Q\). If we call the length of the altitude \(h\) (for height), we get from Theorem 10.4 that \(\sin (\alpha)=\frac{h}{c}\) and \(\sin (\gamma)=\frac{h}{a}\) so that \(h=c \sin (\alpha)=a \sin (\gamma)\). After some rearrangement of the last equation, we get \(\frac{\sin (\alpha)}{a}=\frac{\sin (\gamma)}{c}\). If we drop an altitude from vertex \(A\), we can proceed as above using the triangles \(\triangle A B Q\) and \(\triangle A C Q\) to get \(\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\), completing the proof for this case.
For our next case consider the triangle \(\triangle A B C\) below with obtuse angle \(\alpha\). Extending an altitude from vertex \(A\) gives two right triangles, as in the previous case: \(\triangle A B Q\) and \(\triangle A C Q\). Proceeding as before, we get \(h=b \sin (\gamma)\) and \(h=c \sin (\beta)\) so that \(\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\).
Dropping an altitude from vertex B also generates two right triangles, \(\triangle A B Q\) and \(\triangle B C Q\). We know that \(\sin \left(\alpha^{\prime}\right)=\frac{h^{\prime}}{c}\) so that \(h^{\prime}=c \sin \left(\alpha^{\prime}\right)\). Since \(\alpha^{\prime}=180^{\circ}-\alpha, \sin \left(\alpha^{\prime}\right)=\sin (\alpha)\), so in fact, we have \(h^{\prime}=c \sin (\alpha)\). Proceeding to \(\triangle B C Q\), we get \(\sin (\gamma)=\frac{h^{\prime}}{a} \text { so } h^{\prime}=a \sin (\gamma)\). Putting this together with the previous equation, we get \(\frac{\sin (\gamma)}{c}=\frac{\sin (\alpha)}{a}\), and we are finished with this case.
The remaining case is when \(\triangle A B C\) is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines.
Example 11.2.2
Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.
\(\alpha=120^{\circ}, a=7 \text { units, } \beta=45^{\circ}\)
\(\alpha=85^{\circ}, \beta=30^{\circ}, c=5.25 \text { units }\)
\(\alpha=30^{\circ}, a=1 \text { units, } c=4 \text { units }\)
\(\alpha=30^{\circ}, a=2 \text { units, } c=4 \text { units }\)
\(\alpha=30^{\circ}, a=3 \text { units, } c=4 \text { units }\)
\(\alpha=30^{\circ}, a=4 \text { units, } c=4 \text { units }\)
Knowing an angle-side opposite pair, namely \(\alpha\) and \(a\), we may proceed in using the Law of Sines. Since \(\beta=45^{\circ}\), we use \(\frac{b}{\sin \left(45^{\circ}\right)}=\frac{7}{\sin \left(120^{\circ}\right)}\) so \(b=\frac{7 \sin \left(45^{\circ}\right)}{\sin \left(120^{\circ}\right)}=\frac{7 \sqrt{6}}{3} \approx 5.72\) units. Now that we have two angle-side pairs, it is time to find the third. To find \(\gamma\), we use the fact that the sum of the measures of the angles in a triangle is \(180^{\circ}\). Hence, \(\gamma=180^{\circ}-120^{\circ}-45^{\circ}=15^{\circ}\). To find \(c\), we have no choice but to used the derived value \(\gamma=15^{\circ}\), yet we can minimize the propagation of error here by using the given angle-side opposite pair \((\alpha, a)\). The Law of Sines gives us \(\frac{c}{\sin \left(15^{\circ}\right)}=\frac{7}{\sin \left(120^{\circ}\right)}\) so that \(c=\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)} \approx 2.09\) units. 5
Since we are given \((\alpha, a)\) and \(c\), we use the Law of Sines to find the measure of \(\gamma\). We start with \(\frac{\sin (\gamma)}{4}=\frac{\sin \left(30^{\circ}\right)}{1}\) and get \(\sin (\gamma)=4 \sin \left(30^{\circ}\right)=2\). Since the range of the sine function is [−1, 1], there is no real number with \(\sin (\gamma)=2\). Geometrically, we see that side \(a\) is just too short to make a triangle. The next three examples keep the same values for the measure of \(\alpha\) and the length of \(c\) while varying the length of \(a\). We will discuss this case in more detail after we see what happens in those examples.
Some remarks about Example 11.2.2 are in order. We first note that if we are given the measures of two of the angles in a triangle, say \(\alpha\) and \(\beta\), the measure of the third angle \(\gamma\) uniquely determined using the equation \(\gamma=180^{\circ}-\alpha-\beta\). Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the 'Angle-Angle-Side' (AAS) case. 8 In number 2, the given side is adjacent to both angles which means we are in the so-called 'Angle-Side-Angle' (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the 'Angle-Side-Side' (ASS) case. 9 In number 3, the length of the one given side \(a\) was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, \(a\) was long enough, but not too long, so that two triangles were possible; and in number 6, side \(a\) was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem.
Theorem 11.3
Suppose \((\alpha, a)\) and \((\gamma, c)\) are intended to be angle-side pairs in a triangle where \(\alpha\), \(a\) and \(c\) are given. Let \(h=c \sin (\alpha)\).
If \(a<h\), then no triangle exists which satisfies the given criteria.
If \(a=h\), then \(\gamma=90^{\circ}\) so exactly one (right) triangle exists which satisfies the criteria.
If \(h<a<c\), then two distinct triangles exist which satisfy the given criteria.
If \(a \geq c\), then \(\gamma\) is acute and exactly one triangle exists which satisfies the given criteria
Theorem 11.3 is proved on a case-by-case basis. If \(a<h\), then \(a<c \sin (\alpha)\). If a triangle were to exist, the Law of Sines would have \(\frac{\sin (\gamma)}{c}=\frac{\sin (\alpha)}{a}\) so that \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}>\frac{a}{a}=1\), which is impossible. In the figure below, we see geometrically why this is the case.
Simply put, if \(a<h\) the side \(a\) is too short to connect to form a triangle. This means if \(a \geq h\), we are always guaranteed to have at least one triangle, and the remaining parts of the theorem tell us what kind and how many triangles to expect in each case. If \(a = h\), then \(a=c \sin (\alpha)\) and the Law of Sines gives \(\frac{\sin (\alpha)}{a}=\frac{\sin (\gamma)}{c}\) so that \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}=\frac{a}{a}=1\). Here, \(\gamma=90^{\circ}\) as required. Moving along, now suppose \(h<a<c\). As before, the Law of Sines 10 gives \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}\). Since \(h<a\), \(c \sin (\alpha)<a\) or \(\frac{c \sin (\alpha)}{a}<1\) which means there are two solutions to \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}\): an acute angle which we'll call \(\gamma_{0}\), and its supplement, \(180^{\circ}-\gamma_{0}\). We need to argue that each of these angles 'fit' into a triangle with \(\alpha\). Since \((\alpha, a)\) and \(\left(\gamma_{0}, c\right)\) are angle-side opposite pairs, the assumption \(c>a\) a in this case gives us \(\gamma_{0}>\alpha\). Since \(\gamma_{0}\) is acute, we must have that \(\alpha\) is acute as well. This means one triangle can contain both \(\alpha\) and \(\gamma_{0}\), giving us one of the triangles promised in the theorem. If we manipulate the inequality \(\gamma_{0}>\alpha\) a bit, we have \(180^{\circ}-\gamma_{0}<180^{\circ}-\alpha\) which gives \(\left(180^{\circ}-\gamma_{0}\right)+\alpha<180^{\circ}\). This proves a triangle can contain both of the angles \(\alpha\) and \(\left(180^{\circ}-\gamma_{0}\right)\), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume \(a \geq c\). Then \(\alpha \geq \gamma\), which forces \(\gamma\) to be an acute angle. Hence, we get only one triangle in this case, completing the proof.
Example 11.2.3
Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is \(30^{\circ}\) and at the second point the angle is \(45^{\circ}\). Assuming a straight coastline, find the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point?
We sketch the problem below with the first observation point labeled as \(P\) and the second as \(Q\). In order to use the Law of Sines to find the distance \(d\) from \(Q\) to the island, we first need to find the measure of \(\beta\) which is the angle opposite the side of length 5 miles. To that end, we note that the angles \(\gamma\) and \(45^{\circ}\) are supplemental, so that \(\gamma=180^{\circ}-45^{\circ}=135^{\circ}\). We can now find \(\beta=180^{\circ}-30^{\circ}-\gamma=180^{\circ}-30^{\circ}-135^{\circ}=15^{\circ}\). By the Law of Sines, we have \(\frac{d}{\sin \left(30^{\circ}\right)}=\frac{5}{\sin \left(15^{\circ}\right)}\) which gives \(d=\frac{5 \sin \left(30^{\circ}\right)}{\sin \left(15^{\circ}\right)} \approx 9.66\) miles. Next, to find the point on the coast closest to the island, which we've labeled as \(C\), we need to find the perpendicular distance from the island to the coast. 11
Let \(x\) denote the distance from the second observation point \(Q\) to the point \(C\) and let \(y\) denote the distance from \(C\) to the island. Using Theorem 10.4 , we get \(\sin \left(45^{\circ}\right)=\frac{y}{d}\). After some rearranging, we find \(y=d \sin \left(45^{\circ}\right) \approx 9.66\left(\frac{\sqrt{2}}{2}\right) \approx 6.83\) miles. Hence, the island is approximately 6.83 miles from the coast. To find the distance from \(Q\) to \(C\), we note that \(\beta=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}\) so by symmetry, 12 we get \(x=y \approx 6.83\) miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point.
We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise.
Theorem 11.4
Suppose \((\alpha, a),(\beta, b) \text { and }(\gamma, c)\) are the angle-side opposite pairs of a triangle. Then the area \(A\) enclosed by the triangle is given by
Example 11.2.4
Find the area of the triangle in Example 11.2.2 number 1.
From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose \(A=\frac{1}{2} a c \sin (\beta)\) from Theorem 11.4 because it uses the most pieces of given information. We are given \(a=7\) and \(\beta=45^{\circ}\), and we calculated \(c=\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\). Using these values, we find \(A=\frac{1}{2}(7)\left(\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\right) \sin \left(45^{\circ}\right)=\approx 5.18\) square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4 .
11.2.1 Exercises
In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\) are angle-side opposite pairs.
\(\alpha=13^{\circ}, \beta=17^{\circ}, a=5\)
\(\alpha=73.2^{\circ}, \beta=54.1^{\circ}, a=117\)
\(\alpha=95^{\circ}, \beta=85^{\circ}, a=33.33\)
\(\alpha=95^{\circ}, \beta=62^{\circ}, a=33.33\)
\(\alpha=117^{\circ}, a=35, b=42\)
\(\alpha=117^{\circ}, a=45, b=42\)
\(\alpha=68.7^{\circ}, a=88, b=92\)
\(\alpha=42^{\circ}, a=17, b=23.5\)
\(\alpha=68.7^{\circ}, a=70, b=90\)
\(\alpha=30^{\circ}, a=7, b=14\)
\(\alpha=42^{\circ}, a=39, b=23.5\)
\(\gamma=53^{\circ}, \alpha=53^{\circ}, c=28.01\)
\(\alpha=6^{\circ}, a=57, b=100\)
\(\gamma=74.6^{\circ}, c=3, a=3.05\)
\(\beta=102^{\circ}, b=16.75, c=13\)
\(\beta=102^{\circ}, b=16.75, c=18\)
\(\beta=102^{\circ}, \gamma=35^{\circ}, b=16.75\)
\(\beta=29.13^{\circ}, \gamma=83.95^{\circ}, b=314.15\)
\(\gamma=120^{\circ}, \beta=61^{\circ}, c=4\)
\(\alpha=50^{\circ}, a=25, b=12.5\)
Find the area of the triangles given in Exercises 1, 12 and 20 above.
(Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6 ) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we first have you change road grades into angles and then use the Law of Sines in an application.
Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a \(7 \%\) grade means that the road (hypotenuse) makes about a \(4^{\circ}\) angle with the horizontal. (It will not be exactly \(4^{\circ}\), but it's pretty close.)
What grade is given by a \(9.65^{\circ}\) angle made by the road and the horizontal? 13
Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road. 14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is \(6^{\circ}\). Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a \(94^{\circ}\) angle with the road.)
(Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, \(\mathrm{N} 40^{\circ} \mathrm{E}\) (read "\(40^{\circ}\) east of north") is a bearing which is rotated clockwise \(40^{\circ}\) from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of \(\theta=50^{\circ}\). Similarly, \(\mathrm{S} 50^{\circ} \mathrm{W}\) would point into Quadrant III along the terminal side of \(\theta=220^{\circ}\) because we started out pointing due south (along \(\theta=270^{\circ}\)) and rotated clockwise \(50^{\circ}\) back to \(220^{\circ}\). Counter-clockwise rotations would be found in the bearings \(\mathrm{N} 60^{\circ} \mathrm{W}\) (which is on the terminal side of \(\theta=150^{\circ}\)) and \(\mathrm{S} 27^{\circ} \mathrm{E}\) (which lies along the terminal side of \(\theta=297^{\circ}\)). These four bearings are drawn in the plane below.
The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as 'due north', 'due south', 'due east' and 'due west', respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.)
The Colonel spots a campfire at a of bearing \(\mathrm{N} 42^{\circ} \mathrm{E}\) from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be \(\mathrm{N} 20^{\circ} \mathrm{W}\) from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot.
A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn't reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of \(\mathrm{N} 53^{\circ} \mathrm{W}\) which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of \(\mathrm{S}6 5^{\circ} \mathrm{E}\) will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead?
The captain of the SS Bigfoot sees a signal flare at a bearing of \(\mathrm{N} 15^{\circ} \mathrm{E}\) from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of \(\mathrm{N} 75^{\circ} \mathrm{W}\). If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is \(\mathrm{N} 50^{\circ} \mathrm{E}\), find the distances from the flare to each vessel, rounded to the nearest tenth of a mile.
Carl spies a potential Sasquatch nest at a bearing of \(\mathrm{N} 10^{\circ} \mathrm{E}\) and radios Jeff, who is at a bearing of \(\mathrm{N} 50^{\circ} \mathrm{E}\) from Carl's position. From Jeff's position, the nest is at a bearing of \(\mathrm{S} 70^{\circ} \mathrm{W}\). If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot.
A hiker determines the bearing to a lodge from her current position is \(\mathrm{S} 40^{\circ} \mathrm{W}\). She proceeds to hike 2 miles at a bearing of \(\mathrm{S} 20^{\circ} \mathrm{E}\) at which point she determines the bearing to the lodge is \(\mathrm{S}{75}{ }^{\circ} \mathrm{W}\). How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile.
A watchtower spots a ship off shore at a bearing of \(\mathrm{N} 70^{\circ} \mathrm{E}\). A second tower, which is 50 miles from the first at a bearing of \(\mathrm{S} 80^{\circ} \mathrm{E}\) from the first tower, determines the bearing to the ship to be \(\mathrm{N} 25^{\circ} \mathrm{W}\). How far is the boat from the second tower? Round your answer to the nearest tenth of a mile.
Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be \(75^{\circ}\) and radios Sally immediately to find the angle of inclination from her position to the craft is \(50^{\circ}\). How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.)
The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is \(55^{\circ}\). From a point five stories below the original observer, the angle of inclination to the gargoyle is \(20^{\circ}\). Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.)
Prove that the Law of Sines holds when \(\triangle A B C\) is a right triangle.
Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides.
Discuss with your classmates why the Law of Sines cannot be used to find the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.)
the information yields no triangle
the information yields exactly one right triangle
the information yields two distinct triangles
the information yields exactly one obtuse triangle
Explain why you cannot choose \(a\) in such a way as to have \(\alpha=30^{\circ}, b=10\) and your choice of \(a\) yield only one triangle where that unique triangle has three acute angles.
Use the cases and diagrams in the proof of the Law of Sines ( Theorem 11.2 ) to prove the area formulas given in Theorem 11.4 . Why do those formulas yield square units when four quantities are being multiplied together?
The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles.
The HMS Sasquatch is about 2.9 miles from the flare.
Jeff is about 371 feet from the nest.
She is about 3.02 miles from the lodge
The boat is about 25.1 miles from the second tower.
The UFO is hovering about 9539 feet above the ground.
The gargoyle is about 27 feet from the observer on the lower floor.
The gargoyle is about 25 feet from the other building.
1 as well as the measure of said angle
2 as well as the length of said side
3 Your Science teachers should thank us for this.
4 Don't worry! Radians will be back before you know it!
5 The exact value of \(\sin \left(15^{\circ}\right)\) could be found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus "exact" here means \(\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\).
8 If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisfies the given criteria.
9 In more reputable books, this is called the 'Side-Side-Angle' or SSA case.
10 Remember, we have already argued that a triangle exists in this case!
11 Do you see why \(C\) must lie to the right of \(Q\)?
12 Or by Theorem 10.4 again . . .
13 I have friends who live in Pacifica, CA and their road is actually this steep. It's not a nice road to drive.
14 The word 'plumb' here means that the tree is perpendicular to the horizontal.
15 See Example 10.1.1 in Section 10.1 for a review of the DMS system.
The Law of Sines
The Law of Sines (or Sine Rule ) is very useful for solving triangles:
a sin A = b sin B = c sin C
It works for any triangle:
And it says that:
When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B , and also equal to side c divided by the sine of angle C
Well, let's do the calculations for a triangle I prepared earlier:
a sin A = 8 sin(62.2°) = 8 0.885... = 9.04...
b sin B = 5 sin(33.5°) = 5 0.552... = 9.06...
c sin C = 9 sin(84.3°) = 9 0.995... = 9.04...
The answers are almost the same! (They would be exactly the same if we used perfect accuracy).
So now you can see that:
Is This Magic?
Not really, look at this general triangle and imagine it is two right-angled triangles sharing the side h :
The sine of an angle is the opposite divided by the hypotenuse, so:
a sin(B) and b sin(A) both equal h , so we get:
a sin(B) = b sin(A)
Which can be rearranged to:
a sin A = b sin B
We can follow similar steps to include c/sin(C)
How Do We Use It?
Let us see an example:
Example: Calculate side "c"
Now we use our algebra skills to rearrange and solve:
Finding an Unknown Angle
In the previous example we found an unknown side ...
... but we can also use the Law of Sines to find an unknown angle .
In this case it is best to turn the fractions upside down ( sin A/a instead of a/sin A , etc):
sin A a = sin B b = sin C c
Example: Calculate angle B
Sometimes there are two answers .
There is one very tricky thing we have to look out for:
Two possible answers.
This only happens in the " Two Sides and an Angle not between " case, and even then not always, but we have to watch out for it.
Just think "could I swing that side the other way to also make a correct answer?"
Example: Calculate angle R
The first thing to notice is that this triangle has different labels: PQR instead of ABC. But that's OK. We just use P,Q and R instead of A, B and C in The Law of Sines.
But wait! There's another angle that also has a sine equal to 0.9215...
The calculator won't tell you this but sin(112.9°) is also equal to 0.9215...
So, how do we discover the value 112.9°?
Easy ... take 67.1° away from 180°, like this:
180° − 67.1° = 112.9°
So there are two possible answers for R: 67.1° and 112.9° :
Both are possible! Each one has the 39° angle, and sides of 41 and 28.
So, always check to see whether the alternative answer makes sense.
... sometimes it will (like above) and there are two solutions
... sometimes it won't (like below) and there is one solution
For example this triangle from before.
As you can see, we can try swinging the "5.5" line around, but no other solution makes sense.
So this has only one solution.
Chapter 8: Applications of Trigonometry Functions
Section 8.2: the law of sines, learning outcomes.
By the end of this section, you will be able toliques.
Using the right triangle relationships, we know that [latex]\sin \alpha =\frac{h}{b}[/latex] and [latex]\sin \beta =\frac{h}{a}[/latex]. Solving both equations for [latex]h[/latex] gives two different expressions for [latex]h[/latex].
We then set the expressions equal to each other.
Similarly, we can compare the other ratios.
Collectively, these relationships are called the Law of Sines .
Note the standard way of labeling triangles: angle [latex]\alpha [/latex] (alpha) is opposite side [latex]a[/latex]; angle [latex]\beta [/latex] (beta) is opposite side [latex]b[/latex]; and angle [latex]\gamma [/latex] (gamma) is opposite side [latex]c[/latex]. See Figure 6.
While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.
A General Note: Law of Sines
Given a triangle with angles and opposite sides labeled as in Figure 6, find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\alpha =50^\circ [/latex] and its corresponding side [latex]a=10[/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[/latex].A General Note: Possible Outcomes for SSA Triangles
Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides [latex]a[/latex] and [latex]b[/latex] and known angle [latex]\alpha [/latex].
Example 2: Solving an Oblique SSA Triangle
Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth.
Use the Law of Sines to find angle [latex]\beta [/latex] and angle [latex]\gamma [/latex], and then side [latex]c[/latex]. Solving for [latex]\beta [/latex], we have the proportion
However, in the diagram, angle [latex]\beta [/latex] appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of [latex]\beta ?[/latex] Let's investigate further. Dropping a perpendicular from [latex]\gamma [/latex] and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.
The angle supplementary to [latex]\beta [/latex] is approximately equal to 49.9°, which means that [latex]\beta =180^\circ -49.9^\circ =130.1^\circ [/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\gamma [/latex], we have
Example 3: Solving for the Unknown Sides and Angles of a SSA Triangle
In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth.
In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle [latex]\gamma =85^\circ [/latex], and its corresponding side [latex]c=12[/latex], and we know side [latex]b=9[/latex]. We will use this proportion to solve for [latex]\beta [/latex].
To find [latex]\beta [/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\beta [/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.
In this case, if we subtract [latex]\beta [/latex] from 180°, we find that there may be a second possible solution. Thus, [latex]\beta =180^\circ -48.3^\circ \approx 131.7^\circ [/latex]. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives
To find the remaining missing values, we calculate [latex]\alpha =180^\circ -85^\circ -48.3^\circ \approx 46.7^\circ [/latex]. Now, only side [latex]a[/latex] is needed. Use the Law of Sines to solve for [latex]a[/latex] by one of the proportions.
We can stop here without finding the value of [latex]\alpha [/latex]. Because the range of the sine function is [latex]\left[-1,1\right][/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\sin }^{-1}\left(1.915\right)[/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.
Determine the number of triangles possible given [latex]a=31,b=26,\beta =48^\circ [/latex]. [latex]\text{Area}=\frac{1}{2}bh[/latex], where [latex]b[/latex] is base and [latex]h[/latex] is height. For oblique triangles, we must find [latex]h[/latex] before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property [latex]\sin \alpha =\frac{\text{opposite}}{\text{hypotenuse}}[/latex] to write an equation for area in oblique triangles. In the acute triangle, we have [latex]\sin \alpha =\frac{h}{c}[/latex] or [latex]c\sin \alpha =h[/latex]. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base [latex]b[/latex] to form a right triangle. The angle used in calculation is [latex]{\alpha }^{\prime }[/latex], or [latex]180-\alpha [/latex].
Find the area of the triangle given [latex]\beta =42^\circ ,a=7.2\text{ft},c=3.4\text{ft}[/latex]. Round the area to the nearest tenth.
about 8.2 square feetExample 6: Finding an Altitude
Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.
To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[/latex]. [latex]B[/latex], is 62°, and the distance between the viewing points of the two end zones is 145 yards.
Key Equations
Key concepts.
The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution.
The ambiguous case arises when an oblique triangle can have different outcomes.
There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution.
The Law of Sines can be used to solve triangles with given criteria.
The general area formula for triangles translates to oblique triangles by first finding the appropriate height value.
There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation.
Section 8.2 Homework Exercises
2. Compare right triangles and oblique triangles.
3. When can you use the Law of Sines to find a missing angle?
4. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?
5. What type of triangle results in an ambiguous case? Solve each triangle, if possible. Round each answer to the nearest tenth.
For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle [latex]A[/latex] is opposite side [latex]a[/latex], angle [latex]B[/latex] is opposite side [latex]b[/latex], and angle [latex]C[/latex] is opposite side [latex]c[/latex]. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.
14. [latex]\alpha =119^\circ ,a=14,b=26[/latex]
15. [latex]\gamma =113^\circ ,b=10,c=32[/latex]
16. [latex]b=3.5,c=5.3,\gamma =80^\circ [/latex]
17. [latex]a=12,c=17,\alpha =35^\circ [/latex]
18. [latex]a=20.5,b=35.0,\beta =25^\circ [/latex]
19. [latex]a=7,c=9,\alpha =43^\circ [/latex]
20. [latex]a=7,b=3,\beta =24^\circ [/latex]
21. [latex]b=13,c=5,\gamma =10^\circ [/latex]
22. [latex]a=2.3,c=1.8,\gamma =28^\circ [/latex]
23. [latex]\beta =119^\circ ,b=8.2,a=11.3[/latex]
For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.
For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth.
27. [latex]a=5,c=6,\beta =35^\circ [/latex]
28. [latex]b=11,c=8,\alpha =28^\circ [/latex]
29. [latex]a=32,b=24,\gamma =75^\circ [/latex]
30. [latex]a=7.2,b=4.5,\gamma =43^\circ [/latex]
For the following exercises, find the length of side [latex]x[/latex]. Round to the nearest tenth.
For the following exercises, find the measure of angle [latex]x[/latex], if possible. Round to the nearest tenth.
For the following exercises, find the area of each triangle. Round each answer to the nearest tenth.
66. In Assuming that the street is level, estimate the height of the building to the nearest foot.
67. In68. Points [latex]A[/latex] and [latex]B[/latex] are on opposite sides of a lake. Point [latex]C[/latex] is 97 meters from [latex]A[/latex]. The measure of angle [latex]BAC[/latex] is determined to be 101°, and the measure of angle [latex]ACB[/latex] is determined to be 53°. What is the distance from [latex]A[/latex] to [latex]B[/latex], rounded to the nearest whole meter?
69. A man and a woman standing [latex]3\frac{1}{2}[/latex]70. Two71. A72. Three cities, [latex]A,B[/latex], and [latex]C[/latex], are located so that city [latex]A[/latex] is due east of city [latex]B[/latex]. If city [latex]C[/latex] is located 35° west of north from city [latex]B[/latex] and is 100 miles from city [latex]A[/latex] and 70 miles from city [latex]B[/latex], how far is city [latex]A[/latex] from city [latex]B?[/latex] Round the distance to the nearest tenth of a mile.
73. Two75. The76. A yield sign measures 30 inches on all three sides. What is the area of the sign?
COMMENTS
Law of Sines Example Problem
Solution: The unknown side x is opposite the 46.5° angle and the side with length 7 is opposite the 39.4° angle. Plug these values into the Law of Sines equation. Solve for x. 7 sin (46.5°) = x sin (39.4°) 7 (0.725) = x (0.635) 5.078 = x (0.635) x = 8. Answer: The unknown side is equal to 8. Bonus: If you wanted to find the missing angle ...
Law of Sines
When solving problems using the Law of Sines, there are usually three (3) cases that we are going to deal with. But the general idea is that if any two angles and one side of an oblique triangle are given then it can easily be solved by the Law of Sines.. Case 1: Solving an SAA (Side-Angle-Angle) Triangle In an SAA Triangle, we are given two angles of a triangle and a side opposite to one of ...
Law of Sines
We can use simple trigonometry in right triangle to find that The same holds for and , thus establishing the identity. Method 2. This method only works to prove the regular (and not extended) Law of Sines. The formula for the area of a triangle is . Since it doesn't matter which sides are chosen as , , and , the following equality holds:
Law of Sines Problems
Law of sines problems Solving an angle-side-angle (ASA) triangle with the law of sines. Problem #1. Two
3.1: The Law of Sines
Law of Sines. 1 If the angles of a triangle are A, B, and C, and the opposite sides are respectively a, b, and c, then. sin A a = sin B b = sin C c. or equivalently, a sin A = b sin B = c sin C. 2 We can use the Law of Sines to find an unknown side in an oblique triangle.
Laws of sines and cosines review (article)
I can't find anything here about ambiguous triangles. What if a question asks you to solve from a description where two triangles exist? Like "Determine the unknown side and angles in each triangle, if two solutions are possible, give both: In triangle ABC, <C = 31, a = 5.6, and c = 3.9."
Law of Sines
This trigonometric law lets you solve problems involving any kind of triangle that you come across. As long as your shape is a triangle, you can use the law of sines to help you solve the problem ...
The Law of Sines (with videos, worksheets, solutions, activities)
The ratio of the sine of an angle to the side opposite it is equal for all three angles of a triangle. We can use the law of sines for solving for a missing length or angle of a triangle is by using the law of sines. The law of sines works for any triangle, not just right triangles. The law of sines is also called the sine rule. Law of Sines ...
3.2.6.5: Non-right Triangles10.1 Non-right Triangles:Solving Applied Problems using Law of Sines
Law of Sines. states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side. oblique triangle. any triangle that is not a right triangle.
4.1.7: Trigonometry Word Problems
Then write and solve a trig equation. \(\begin{aligned} \sin 42.5^{\circ} &=\dfrac{x}{300}\\ 300\cdot \sin 42.5^{\circ} &=x \\ x&\approx 202.7\end{aligned}\) ... law of sines: The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of ...
11.2: The Law of Sines
The remaining case is when \(\triangle A B C\) is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.2.6 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair.
Problem Set: Law of Sines
For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 24. Find angle \displaystyle A A when \displaystyle a=24,b=5,B=22^\circ a = 24, b = 5, B = 22∘. 25.
Law of Sines
The law of sines is an equation that allows us to relate the sines of an angle to their respective opposite sides. The law of sines is applied to find the measures of an angle or the length of a side in a triangle. To use the law of sines, we need to know the measures of two angles and the length of an opposite side or the lengths of two sides ...
3.1 Law of Sines5.2: The Law of Sines
In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. ... (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or ...
Solving Real-World Problems Involving the Law of Sines
Question Video: Solving Real-World Problems Involving the Law of Sines. James, Anthony, and Jennifer stand at three points, 𝐴, 𝐵, and 𝐶, respectively. Suppose that 𝑚∠𝐴𝐵𝐶 = 48°, 𝑚∠𝐵𝐴𝐶 = 54°, and James is exactly 12 feet away from Anthony. Find the distance between Anthony and Jennifer, to two decimal places.
The Law of Sines
Law of Sines (Sine Law)
@MathTeacherGon will demonstrate how to use the law of sines in solving problems in oblique triangles.The Six Trigonometric Ratios of Right Triangle
Section 8.2: TheHow to Solve a Word Problem Using the Law of Sines
Step 1: Draw a picture representing the situation in the word problem. Label all known sides and angles. Step 2: Set up an equation using the law of sines to solve for the unknown value. The law ... | 677.169 | 1 |
What is the difference between circle and ellipse?
The simple answer is that an ellipse is a squashed circle.A more precise answer is that an ellipse is the locus (a collection) of points such that the sum of their distances from two fixed points (called foci) remains a constant. A circle is the locus of points that are all the same distance from a fixed point. If the two foci are moved closer together, the ellipse becomes more and more like a circle and finally, when they coincide, the ellipse becomes a circle. So, a circle is a special case of an ellipse. | 677.169 | 1 |
How do you Find the Degree of a Vector?
From the perspective of the people who usually have to deal with the questions of solving vectors, it's important to know about the degree of a vector. The people who have extended knowledge about solving questions of the vector will not take enough time to find out the degree of the vector. In some situations, you may need to find out the magnitude and angle of the vector instead of its components. By finding out the angle of a vector, it might be easy for you to find out the degree of the given vector shortly.
Introduction
If you want to find out the magnitude, you can use the Pythagorean Theorem. In addition, if you want to find the angle you can use the inverse tangent function which can be inverse cosine or inverse sine. When you have details about the horizontal and vertical components of a vector, you will never face the problem to find out the magnitude of the vector. In this situation, you just need to find out the hypotenuse of a Right Triangle. How do you find the degree of a vector? This is something called basic about the magnitude and angle you need to know.
How you can find the degree of a vector?
First of all, you need to assume that you are given the coordinates of the end of the vector and want to find its magnitude v and the angle theta. Maybe, you have extended knowledge about trigonometry you should know that.
Y/X= Φ Sinθ/Φ Cosθ= tanθ
V= under root x2+ under root y2
It simply means that if you have a vector given by the available coordinates, its magnitude and angle could easily be found.
This is an expression that you actually need in order to find out the degree of a factor. From now, the task of understanding how do you find the degree of a vector might be easy for you and there is not a single doubt about the same concept.
In this given expression, tan theta is the tangent of the angle which means theta = tan inverse (y/x). Next, you can use the Pythagorean Theorem to find out the hypotenuse the magnitude us of the triangle which can be formed by the X, Y, and v.
In easy words, it can be said that the degree of a vector can be expressed as the angle theta. When you want to check the degree of a vector, you can simply use the formula means theta = tan inverse (y/x) without asking anyone else.
Now, you have collected information about the possible ways that you can use for finding out the degree of a given vector in very short amount of time. Make sure that you will not commit a single mistake during the same procedure to have the rest of the benefits.
The ones, who will follow the steps of finding a degree of the given vector carefully without committing a mistake, may not face any kind of problem to complete the question and find out the degree. | 677.169 | 1 |
equal the sum of the diagonals of the given quadrilateral. G. 4. Two triangles are similar, if they have an angle of the one equal to an angle of the other and the sides including those angles proportional. 5. In any triangle, if a straight line is drawn.) 220. Prove, geometrically, that the square described upon the sum of be the two triangles having the...
...triangle AGH, therefore the triangle ABC is similar to the triangle DEF. PROPOSITION 18. If two triangles have an angle of the one equal to an angle of the other, and the sides containing those angles proportionals, the triangles shall be similar. Fig. 25. Let ABC. ) 220. Prove, geometrically, that the square described upon the sum b« the two triangles having the...
...be right, the remaining angles will be right angles. FIRST BOOK. COR. 2. — If two parallelograms have an angle of the one equal to an angle of the other, the remaining angles will be also equal ; for the angles which are opposite to these equal angles are...
...diagonals proportionals, each to each, prove that they are similar. THEOREM XIII. If two triangles have an angle of the one equal to an angle of the other, and the sides about the equal angles proportional, they shall be similar. Let ABC and PQR be two triangles,...
...DEH are equiangular (I. 35), and similar (20) ; therefore : EF D THEOREM X. 231 Two triangles having an angle of the one equal to an angle of the other, and the sides including these angles proportional, are similar. In the triangles ABC,DEF let tiifl...
...(I. 35^, and similar (20) ; therefore BG:EH—AB:DE=AC:DF=BC:EF THEOREM X. 23, Two triangles having an angle of the one equal to an angle of the other, and the sides including these angles proportional, are similar. E D In the triangles ABC, DEF let t!:e... | 677.169 | 1 |
Almost all parallelograms have two pairs of parallel sides. A
square, rectangle and parallelogram can all be considered true
parallelogram by definition of parallel sides. However, in a
rhombus, although each sides' lengths are equal, they cannot be
consided parallel sides because the angle at which they are formed
at each point isn't equal. | 677.169 | 1 |
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