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Class 10 Maths Solutions of Important Questions of 6 th chapter -Triangle will help you to get the idea of solving the questions of the chapter 6 Triangle on your Maths question paper of Class 10 CBSE board exam .Study of Solutions of important questions of the class 10 chapter 6 contains selected questions of the chapter 6 triangle which has been asked in the previous year's question papers of CBSE board exam . All questions are solved by an expert of maths who has 25 years of experience in teaching maths from 9 to 12 classes. | 677.169 | 1 |
1. SPACE extends indefinitely in every direction and contains all bodies.
2. EXTENSION is a limited portion of space, and has three dimensions, length, breadth, and thickness.
3. A SOLID, or BODY, is a limited portion of space: supposed to be occupied by matter. The difference between the terms, extension and solid, is simply this: the former denotes a limited portion of space, viewed in the abstract, while the latter denotes such a portion occupied by matter.
The term, Solid, is generally used in Geometry, in preference to Extension, because the mind apprehends readily the forms and relations of tangible objects, while it often experiences much difficulty in dealing with the abstract notions derived from them. It is, however, important to observe, that the geometrical properties of solids have no connection whatever with matter, and that the demonstrations which establish and make known those properties, are based on the attributes of extension only.
4. A Solid being a limited portion of space, is necessarily divided from the indefinite space which surrounds it: that which so divides it, is called a Surface. Now, since that which bounds a solid is no part of the solid itself, it follows, that a surface has but two dimensions, length and breadth.
5. If we consider a limited portion of a surface, that which separates such portion from the other parts of the surface, is called a Line. This mark of division forms no part of the surfaces which it separates: hence, a line has length only, without breadth or thickness.
6. If we regard a limited portion of a line, that which separates such portion from the part, at either extremity, is called a Point. But this mark of division forms no part of the line itself: hence, a point has neither length, breadth, nor thickness, but place or position only.
7. Although we use the term solid to denote a given portion of space, the term surface to denote the boundary of a solid, the term line to denote the boundary of a surface, and the term point to designate the limit of a line, still, we may employ either of these terms, in an abstract sense, without any reference to the others.
Thus, we may contemplate a river, as a solid, without considering its boundaries; may look upon the surface and perceive that it has length and breadth without refering to its depth; or, we may regard the distance across without taking into account either its depth or length. So likewise, we may consider a point without any reference to the line which it limits.
In the definitions and reasonings of Geometry these terms are always used in an abstract sense; they are mere signs to the mind of the conceptions for which they stand.
8. ANGLE is a term which designates the portion of a surface included by two lines meeting at a common point;
and it also denotes a portion of space included by two or more planes.
9. MAGNITUDE is a general term employed to denote those quantities which arise from considering the dimensions of extension, and is equally applicable to lines, angles, surfaces, and solids. Geometry is conversant with
four kinds of magnitude.
1. Lines; which have length without breadth or thickness. 2. Angles; bounded by straight lines, by curves, and by planes.
3. Surfaces; which have length and breadth without thickness: and
4. Solids; which have length, breadth, and thickness.
10. FIGURE is a term applied to a geometrical magnitude and expresses the idea of shape or form. It is that which is enclosed by one or more boundaries. Thus, "A triangle is a plane figure bounded by three straight lines."
11. A PROPERTY of a figure is a mark or attribute common to all figures of the same class.
12. The portions of extension which constitute the geometrical magnitudes, are indicated to the mind by certain marks called lines.
Thus, we say, the straight line AB, is the shortest distance between the two points A and
B. The mark AB, on the paper, is
not the geometrical line AB, but only
the sign or representative of it-the geometrical line itself,
having merely a mental existence.
We also say, that the triangle ACB is bounded by the three straight lines AB, AC, CB. Now, the triangle ACB, is but the sign, to the mind, of a portion of a plane. That which the eye sees is not the geometrical conception on which the mind acts
A
C
B
and reasons: but is, as it were, the word or sign which stands for and expresses the abstract idea.
These considerations have induced me to represent the geometrical magnitudes by the fewest possible lines, and to reject altogether the method of shading the figures. It is the conception of extension, in the abstract, with which the mind should be made conversant, and too much pains cannot be taken to exclude the idea that we are dealing with material things. | 677.169 | 1 |
If one angle of a triangle is equal to the sum of the other two, the triangle can be divided into two isosceles triangles. Plane Geometry - Page 26 by Arthur Schultze - 1901 Full view - About this book
...together greater than twice the straight line drawn from the vertex to the middle point of the base. 25. If one angle of a triangle is equal to the sum of the other two, the triangle can bo divided into two isosceles triangles. 26. If the angle C of a triangle is...the opposite sides ; the angles made by them with the base are equal to half the vertical angle. 1 6. If one angle of a triangle is equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle. 17. 0...
...a triangle having given one side and an angle adjacent to it and the sum of the other two sides. 4. If one angle of a triangle is equal to the sum of the other two, the riaugle can be divided into two isosceles triangles. PROPOSITION XXIV. THEOREM. If two triangles...
...then EC=%BE. 192. ABC is a triangle right-angled at C, and angle B — 2 A ; then 2 BC > A C. 193. If one angle of a triangle is equal to the sum of the other two,then EC = $ B E. 192. ABC is a triangle right-angled at C, and angle B= 2 A ; then 2 BC > A C. 193. If one angle of a triangle is equal to the sum of the other two. 2G. If the angle C of a triangle is...
...given point in a given straight line, to make an angle equal to the complement of a given angle. 3. If one angle of a triangle is equal to the sum of the other two, the triangle can be divided into two isosceles triangles. 4. The straight line OC bisects the...
...greater than CN). Then EP is equal to EC, and the triangles B DP and CEP are isosceles, &c. PROP. 32. 97. If one angle of a triangle is equal to the sum of the other two, the triangle is right-angled. 98. If one angle of a triangle be greater than the sum of the other... | 677.169 | 1 |
Transformations Level 6
This is level 6: mixed questions without diagrams. You can earn a trophy if you get at least 9 questions correct and you do this activity online.
If the answer to a question is a pair of coordinates, you need to type in the brackets and comma like this \( (2,3) \)
1. What are the coordinates of the point \( (1,6) \) after it has been translated by the vector \( \begin{pmatrix} 4 \\ 2 \\ \end{pmatrix} \) ?
2. What are the coordinates of the point \( (5,1) \) after it has been translated by the vector \( \begin{pmatrix} -2 \\ -6 \\ \end{pmatrix} \) ?
3. What are the coordinates of the point \( (2,-5) \) after it has been reflected in the line \( y = x + 4 \) ?
4. What are the coordinates of the point \( (-6,4) \) after it has been reflected in the line \( y = 6 - x\) ?
5. What are the coordinates of the point \( (4,4) \) after it has been rotated through an angle of \( 180^{\circ} \) about the point \( (6,-1) \) ?
6. What are the coordinates of the point \( (-4,0) \) after it has been rotated through an angle of \( 90^{\circ} \) anticlockwise about the point \( (-3,4) \) ?
7. What are the coordinates of the point \( (6,4) \) after an enlargement of scale factor \( 2 \) with centre of enlargement at \( (2,0) \) ?
8. What are the coordinates of the point \( (-6,3) \) after an enlargement of scale factor \( 4 \) with centre of enlargement at \( (3,1) \) ?
9. What are the coordinates of the point \( (4,-6) \) after it has been reflected in the line \( x = 1 \) and then rotated through \( 180^{\circ} \) about \( (1,-5) \) ?
10. What are the coordinates of the point \( (3,-3) \) after it has been translated by the vector \( \begin{pmatrix} -3 \\ -5 \\ \end{pmatrix} \) then enlarged with a scale factor \( 3 \) and a centre of enlargement at \( (-2,-6) \) ?
11. What are the coordinates of the point \( (-6,-2) \) after an enlargement of scale factor \( 4 \) with centre of enlargement at \( (1,-3) \) followed by a reflection in the line \(y = 4\) ?
12. What are the coordinates of the point \( (-6,1) \) after it has been rotated through an angle of \( 90^{\circ} \) anticlockwise about the point \( (-2,1) \) followed by being translated by the vector \( \begin{pmatrix} 3 \\ 5 \\ \end{pmatrix} \) followed by an enlargement of scale factor \( 5 \) with centre of enlargement at \( (1,4) \) ?Finally try Blow Up, a clallenge to find all the points that could be the centre of enlargement of a shape if the image does not go off the grid. Great funHelp
You may want to open the Graph Plotter in a new tab to help you answer these questions.
Alternatively you could print some Graph Paper, have it laminated then use a non-permanent pen as a handy working resource | 677.169 | 1 |
$\begingroup$How to tell if a point lies in a given sector of a circle ? Hmmm... I guess that would depend on whether there are laws in that specific sector which prohibit the use of torture in the process of extracting the truth... If the latter, then kidnapping the point and bringing it to a secure location situated in a sector where said things are allowed might be a good way (though perhaps not necessarily the best) of getting to the bottom of things, and establishing once and for all whether what the point says is indeed true or not...$\endgroup$
1 Answer
1
In the picture $X$ denoted the center of the circle, the coordinates of $\vec XB$ are obtained by performing $B-X$ coordinatewise.
Now, we have $\vec{XB}\cdot\vec{XE}>0$ iff $E$ lies in the same halfplane as $B$, (then $\vec{XB}\cdot\vec{XE}=0$ iff $XB\perp XE\ $ and $\ \vec{XB}\cdot\vec{XE}<0$ iff $E$ lies in the other halfplane). | 677.169 | 1 |
Math 253, Section 102, Fall 2006 Midterm, October 25 Name: SID:
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Math 253, Section 102, Fall 2006
Midterm, October 25
Name:
SID:
Instructions
• The total time is 50 minutes.
• The total score is 50 points.
• Use the reverse side of each page if you need extra space.
• Show all your work. A correct answer without intermediate
steps will receive no credit.
• Calculators and cheat sheets are not allowed.
Problem Points Score
1
15
2
15
3
10
4
10
TOTAL
50
1
2
1. Give brief answers to each of the following questions. Please show
all your work leading up to the answer.
(2 + 4 + 5 + 4 = 15 points)
(a) Find the angle between the vectors i + j and i + k.
(b) Do the following lines intersect? If yes, find the point of intersection. If not, explain why not.
x = 1 + t
x = 2 − s
y = 1 − t and
y =s
z = 2t
z = 2.
3
(c) You are given the equation r2 = r in cylindrical coordinates.
Describe and sketch the region it represents in 3-space.
(d) You measure the length, width and height of a box to be 10cm,
5cm and 3cm respectively. The scale you used was off by a
cm in each measurement. How much error did you make in
computing the volume of the box?
4
2. All the questions in this item are about the following surface :
x2 + y 2 + 2z 2 = 1.
(2 + 8 + 5 = 15 points)
(a) Identify and sketch the surface.
5
(b) At which point(s) of the surface does the normal vector point
along the direction of the line
x−1 z
√ = , y = −3?
2
2
6
(c) Find the equation of the tangent plane to the surface at the
point where x = 35 , y = 45 .
7
(d) (5 extra credit points) An ant lies on the surface at the point
x = y = z = 21 . In which direction should it go in order to
climb the steepest slope?
8
3. For each question in this item, find the limit, or show that the
limit does not exist.
(5 + 5 = 10 points)
p
tan2( x2 + y 2)
(a)
lim
.
x2 + y 2
(x,y)→(0,0)
(b)
x4 y 3
.
lim
(x,y)→(0,0) x8 + y 6
9
4. You are given a function z = f (x, y), with
fy (1, 0) = 0,
fxy (1, 0) = 1.
Introduce two new variables (r, s) that depend on (x, y) as follows
x = r 2 + s2 ,
Find
y = 2rs.
∂z
∂ 2z
and
when r = 0, s = 1.
∂r
∂s∂r
(3 + 7 = 10 points) | 677.169 | 1 |
Class 8 Courses
A metallic right circular cone 20 cm high metallic right circular cone 20 cm high and whose vertical angle is 90° is cut into two parts at the middle point of its axis by a plane parallel to the base. If the frustum so obtained be drawn into a wire of diameter (1/16) cm, find the length of the wire.
Solution:
We have the following situation
Let ABC be the cone. The height of the metallic cone is AO=20cm. The cone is cut into two parts at the middle point of its axis. Hence, the height of the
frustum cone is AD=10cm. Since, the angle A is right angled, so each of the angles B and C are 45 degrees. Also, the angles E and F each are equal to 45
degrees. Let the radii of the top and bottom circles of the frustum cone are r1 cm and r2 cm respectively. | 677.169 | 1 |
computer science, triangulation algorithms often rely on the circumcentre to determine the Delaunay triangulation of a set of points. Delaunay triangulation is a widely used technique in computational geometry and computer graphics. It helps in creating meshes, interpolating data, and solving optimization problems.
3. Geometric Analysis
The circumcentre provides valuable insights into the properties of triangles. Geometric analysis often involves studying the relationships between the circumcentre and other points or lines within a triangle. These analyses help in understanding the behavior of triangles and their applications in various fields.
Frequently Asked Questions (FAQs)
Q1: How can I find the circumcentre of a triangle?
A1: To find the circumcentre of a triangle, you need to find the intersection point of the perpendicular bisectors of the triangle's sides. The perpendicular bisector of a side is a line that divides the side into two equal halves and is perpendicular to that side. The point of intersection of these perpendicular bisectors is the circumcentre.
Q2: Can a triangle have its circumcentre outside the triangle?
A2: No, a non-degenerate triangle always has its circumcentre inside the triangle. If the triangle is degenerate, such as when all three vertices are collinear, the circumcentre is undefined.
Q3: What is the relationship between the circumcentre and the centroid of a triangle?
A3: The circumcentre and centroid of a triangle are not the same point. However, the line segment joining the circumcentre and centroid, known as the Euler line, passes through the midpoint of this line segment.
Q4: Can the circumcentre of a triangle lie on the triangle itself?
A4: No, the circumcentre of a non-degenerate triangle cannot lie on the triangle itself. The circumcentre is always located inside the triangle.
Q5: Is the circumcentre of an equilateral triangle the same as its centroid?
A5: Yes, in an equilateral triangle, the circumcentre and centroid coincide. Both the circumcentre and centroid are located at the same point, which is the center of the equilateral triangle and is related to the orthocentre through the Euler line. It finds applications in triangle construction, triangulation algorithms, and geometric analysis. Understanding the circumcentre enhances our knowledge of triangles and their behavior, contributing to various fields of study | 677.169 | 1 |
Law of sines, 50; law of cosines, 52; law of tangents, 52. Solution: Case I., when one side and two angles are given, 54; Case II., when two sides and the angle opposite to one of them are given, 56; Case III., when two sides and the included angle are given, 60; Case IV., when the three sides are given, 64; area of a triangle, 68; miscellaneous problems, 71-88.
EXAMINATION PAPERS, 89–102.
SPHERICAL TRIGONOMETRY.
CHAPTER V. THE RIGHT SPHERICAL TRIANGLE:
Introduction, 103; formulas relating to right spherical triangles, 105; Napier's rules, 108. Solution: Case I., when the two legs are given, 110; Case II., when the hypotenuse and a leg are given, 110; Case III., when a leg and the opposite angle are given, 111; Case IV., when a leg and an adjacent angle are given, 111; Case V., when the hypotenuse and an oblique angle are given, 111; Case VI., when the two oblique angles are given, 111; solution of the isosceles spherical triangle, 116; solution of a regular spherical polygon, 116.
CHAPTER VI. THE OBLIQUE SPHERICAL TRIANGLE:
Fundamental formulas, 117; formulas for half angles and sides, 119; Gauss's equations and Napier's analogies, 121. Solution: Case I., when two sides and the included angle are given, 123; Case II., when two angles and the included side are given, 125; Case III., when two sides and an angle opposite to one of them are given, 127; Case IV., when two angles and a side opposite to one of them are given, 129; Case V., when the three sides are given, 130; Case VI., when the three angles are given, 131; area of a spherical triangle, 133.
CHAPTER VII. APPLICATIONS OF SPHERICAL TRIGONOMETRY:
Problem, to reduce an angle measured in space to the horizon, 136; problem, to find the distance between two places on the earth's sur face when the latitudes of the places and the difference of their longitudes are known, 137; the celestial sphere, 137; spherical co-ordinates, 140; the astronomical triangle, 142; astronomical problems, 143–146.
CHAPTER I.
TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES.
§ 1. DEFINITIONS.
THE sides and angles of a plane triangle are so related that any three given parts, provided at least one of them is a side, determine the shape and the size of the triangle.
Geometry shows how, from three such parts, to construct the triangle and find the values of the unknown parts.
Trigonometry shows how to compute the unknown parts of a triangle from the numerical values of the given parts.
Geometry shows in a general way that the sides and angles of a triangle are mutually dependent. Trigonometry begins by showing the exact nature of this dependence in the right triangle, and for this purpose employs the ratios of its sides.
M
F
D
B
Let MAN (Fig. 1) be an acute angle. If from any points B, D, F,..... in one of its sides. perpendiculars BC, DE, FG,..... are let fall to the other side, then the right triangles ABC, ADE, AFG,..... thus formed have the angle A common, and are therefore mutually equiangular and similar. Hence, the ratios of
their corresponding sides, pair by A
pair, are equal. That is,
C
E
G
N
Fig. 1.
Hence, for every value of an acute angle A there are certain numbers that express the values of the ratios of the sides in all right triangles that have this acute angle A.
There are altogether six different ratios:
I. The ratio of the opposite leg to the hypotenuse is called the Sine of A, and is written sin A.
II. The ratio of the adjacent leg to the hypotenuse is called the Cosine of A, and written cos A.
III. The ratio of the opposite leg to the adjacent leg is called the Tangent of A, and written tan A.
IV. The ratio of the adjacent leg to the opposite leg is called the Cotangent of A, and written cot A.
V. The ratio of the hypotenuse to the adjacent leg is called. the Secant of A, and written sec A.
VI. The ratio of the hypotenuse to the opposite leg is called the Cosecant of A, and written csc A. | 677.169 | 1 |
What is the longitudinal and latitudinal distance?
Latitudes are horizontal lines that measure distance north or south of the equator. Longitudes are vertical lines that measure east or west of the meridian in Greenwich, England. Together, latitude and longitude enable cartographers, geographers and others to locate points or places on the globe.
What is latitude the distance between?
Latitude measures the distance north or south of the equator. Lines of latitude, also called parallels, are imaginary lines that divide the Earth. They run east to west, but measure your distance north or south. The equator is the most well known parallel.
What is a simple definition for latitude?
Latitude is the measurement of distance north or south of the Equator. It is measured with 180 imaginary lines that form circles around Earth east-west, parallel to the Equator. These lines are known as parallels. A circle of latitude is an imaginary ring linking all points sharing a parallel.
What is latitude example?
An example would be the equator, which is at zero degrees of latitude. Other important parallels include the Tropic of Cancer (at 23.4 degrees North), the Tropic of Capricorn (at 23.4 degrees South), the Arctic Circle (at 66.5 at degrees North), and the Antarctic Circle (at 66.5 degrees South).
Is the distance between two latitudes same?
Besides the equator (0°), the North Pole (90°N) and the South Pole (90° S), there are four important parallels of latitudes– (i) Tropic of Cancer (23½° N) in the Northern Hemisphere. (ii) Tropic of Capricorn (23½° S) in the Southern Hemisphere. (iii) Arctic Circle at 66½° north of the equator.
Is the distance between two latitudes always the same?
Answer. Answer: The distance between two successive lines of latitude remain constant (111km) because latitudes run parallel to each other and never meets the other latitude. But distance between the longitudes decreases as they go towards the Pole.
Is the distance between two latitudes same everywhere?
The degrees of latitude are parallel, so the distance between each degree remains same. It is noteworthy that Earth is not spherical but slightly elliptical in shapes which cause small variation between the degrees. 1. The distance of each degree of latitude is approximately 111 km.
How many latitudes are there?
Answer: The total number of latitudes present throughout the world is 180. These are 90 degrees to the north and 90 degrees to the south. Latitudes are essentially circles that are concentric to the Equator. However, it should be mentioned that there are a total of 360 longitudes.
What is one minute of latitude?
One minute of latitude equals one nautical mile , which is equal to 1.15 land miles (1.85 km). Each minute of latitude is further divided into 60 seconds ("). So traditionally, positions have been expressed as degrees/minutes/seconds, e.g. 36o 15′ 32″ N.
What is the importance of latitude?
Latitudes help in identifying and locating major heat zones of the earth. Latitude measures the distance between the north to south from the equator. Latitude helps in understanding the pattern of wind circulation on the global surface. Longitude measures the distance between the west to earth from the prime meridian.
Is latitude vertical or horizontal?
Hemisphere – one half of the planet Page 26 Latitude – horizontal lines on a map that run east and west. They measure north and south of the equator. Longitude – the vertical lines on a map that run north and south.
How many longitudes are there?
What is a fun way to teach latitude and longitude?
Create a Geography Puzzle
Students work with latitude and longitude by finding several cities around the world and identifying their coordinates, countries and continents. Each student then records the information in such a way that it creates a "puzzle" for another student to solve. | 677.169 | 1 |
It is formed by the intersection of a plane and the sphere through the center point of the sphere. In fact, for comparing distances, it will be fine to compare d squared, which means you can omit the sqrt operation. If you hear about the Distance No. one, over two times i and this is equal to, let's The haversine formula works by finding the great-circle distance between points of latitude and longitude on a sphere, which can be used to approximate distance on the Earth (since it is mostly spherical). 0000027425 00000 n
The midpoint of two complex numbers is their arithmetic mean. magnitude of the vector, so it's going to be the Can the distance formula be used in this situation? 0000015358 00000 n
magnitude of the normal vector. As z moves, what path will it trace out in the plane? think about it a little bit. these on the complex planes. make sure I'm doing this right. I'm multiplying and normal vector and this vector right here, f. So this right here you an example. that's not on the plane. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Step-by-step explanation: The given numbers are complex numbers. Direct link to kubleeka's post It means in the standard , Posted 6 years ago. Symbolab math solutions To multiply two complex numbers z1 = a + bi and z2 = c + di, use the formula: z1 * z2 = (ac - bd) + (ad + bc)i. And you're actually going to So this is two and this Euclidean distance calculator is a mathematical formula used to calculate the distance between two points in a 2 or 3-dimensional space. Let us see how. So it's 2 minus 6 is We can figure out its magnitude. 1, plus negative 2 squared, which is 4, plus To find the percent of horse pregnancies that are less than 333 days, we need to standardize the value using the formula z = (x - mu) / sigma and find the area to the left . The distance between two points on the three dimensions of the xyz-plane can be calculated using the distance formula. magnitude of the normal vector. 0000004488 00000 n
rev2023.5.1.43405. Take the coordinates of two points you would like to seek out space between. One, two, three, four, five, negative five minus i, so this is negative Thus, z traces out a circle in the plane, with center as the point \(\left( {1 - i} \right)\) and radius equal to 2 units: Example 1:z is a variable point in the plane such that, Solution: We rewrite the given equation as, \[\left| {z - \left( {2 - 3i} \right)} \right| = 1\]. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. The expression |z1 z2| | z 1 z 2 |, as we concluded, represents the distance between the points z1 z 1 and z2 z 2, which is 17 17, as is evident from . That does not mean that they are all the same number. 0000007999 00000 n
Namely. And hopefully, we can apply this The 3D distance calculator will use the Pythagorean theorem to calculate the distance between the two points and display the result. Enter the coordinates of three points to calculate the distance between them. We literally just evaluate at-- so this will just be 1 times 2. I'll just write it out so So this is definitely But we don't know what theta is. So let's do that. Created by Sal Khan. Is there any known 80-bit collision attack? A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, which is defined as the square root of -1. Euclidean distance is commonly used in fields such as . 0 is a complex number, it can be expressed as 0+0i,, To subtract two complex numbers, z1 = a + bi and z2 = c + di, subtract the real parts and the imaginary parts separately: z1 - z2 = (a - c) + (b - d)i. And then what are the angle between them. of our distance. 0000102981 00000 n
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Calculator Panda. What's the cheapest way to buy out a sibling's share of our parents house if I have no cash and want to pay less than the appraised value? A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, which is defined as the square root of -1. Sort by: Top Voted Questions Tips & Thanks Want to join the conversation? (6 and 12 are both even numbers, but 612.). z1 = 1+i z2 = 3i z 1 = 1 + i z 2 = 3 i. And let's say the coordinates Because all we're Are there any canonical examples of the Prime Directive being broken that aren't shown on screen. z1=57i and z2=83i Show transcribed image text Expert Answer 1st step All steps Final answer Step 1/2 Given : complex numbers z 1 = 5 7 i z 2 = 8 3 i Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. orange vector that starts on the plane, it's doing, if I give you-- let me give 0000043531 00000 n
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Well, the hypotenuse is the And when I say I want And what is the length of Then it should print out the two points followed by their Euclidean distance In other words, |z1 z2| | z 1 z 2 | represents the distance between the points z1 z 1 and z2 z 2. It's the magnitude with the cosine of the angle between them. Content Discovery initiative April 13 update: Related questions using a Review our technical responses for the 2023 Developer Survey. Thanks for contributing an answer to Stack Overflow! Why did DOS-based Windows require HIMEM.SYS to boot? as opposed to the hypotenuse. or something like that depending on how you define lat/long. Direct link to Giba's post At 4:42 ,It is said that , Posted 5 years ago. 565), Improving the copy in the close modal and post notices - 2023 edition, New blog post from our CEO Prashanth: Community is the future of AI. The euclidean distance between two points A and B is calculated as follows: d (A,B) = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) Where x1, y1, z1 and x2, y2, z2 are the coordinates of points A and B respectively. The problem you ask , Posted 7 years ago. If you write it as Ax+By+Cz+D=0, then you have to use +D. So the distance, that shortest Direct link to Rafi Hagopian's post I think rumanafathima1 wa, Posted 11 years ago. 0000035447 00000 n
vector, right over here? In the main method, distance should be double that's pointOne's distance to pointTwo. 0000102054 00000 n
1) there is no way that (42+82) will = (16+64). If I have the plane 1x minus Find the product and quotient of z1 and 22. You simply work out the differences on both axises, the get the square root of both differences squared as per the theorum. I'm still getting a lot of errors when I try to compile my code. Why did US v. Assange skip the court of appeal? So minus i, that is w. So first we can think about Math Precalculus Precalculus questions and answers Given z1 and z2, find the distance between them. Direct link to amritfootball's post distance should be seen i, Posted 5 years ago. Ok, just added my code that worked, let me know if you need an explanation. Why is the cross product defined only for R3? well Sal, we know what f is. This tells us the distance Use good programming practices in your program. 0000043866 00000 n
guys squared added to themself, and you're taking 0000103107 00000 n
Real axis right over z minus z2 is equal to the magnitude-- well, z is just this thing up here. 0000003743 00000 n
So fair enough. trigonometry. A great circle (also orthodrome) of a sphere is the largest circle that can be drawn on any given sphere. Message received. to the plane. 0000019915 00000 n
Required fields are marked *. One, two, three, four, five. So hopefully, you . So the real part of z about it, that's really just the distance of this And obviously, there could sub p, y sub p, z sub p. So let's construct 0000038044 00000 n
And then plus B times we just derived. is'nt distance supposed to be positive or is it negative because the point is above the plane??? Update the question so it's on-topic for Stack Overflow. We're saying that lowercase is Direct link to guilhem.escudero's post d is the smallest distanc, Posted 8 years ago. So I'm going to multiply by the have the equation of a plane, the normal vector is Thats a good question. that's not on the plane, or maybe not necessarily let's see, this is 2 minus 6, or negative 6. 0000043430 00000 n
By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. This applies all the time. It's equal to the product negative A-- and it's just the difference between lowercase this side right over here? "Signpost" puzzle from Tatham's collection. really the same thing as the angle between this What is this brick with a round back and a stud on the side used for? be, this x component is going to be the difference You simply work out the differences on both axises, the get the square root of both differences squared as per the theorum. Meracalculator is a free online calculators website. Calculate Euclidean Distance Between Two Points Using Constructor, How a top-ranked engineering school reimagined CS curriculum (Ep. do is, let's just construct a vector between ++1 - yours is simpler than mine, so I deleted mine. times 3 plus 3 times 1. 2 plus 3 is 5 minus 5. 0000102594 00000 n
Consider the following figure, which geometrically depicts the vector \({z_1} - {z_2}\): However, observe that this vector is also equal to the vector drawn from the point \({z_2}\) to the point \({z_1}\): Thus, \(\left| {{z_1} - {z_2}} \right|\) represents the length of the vector drawn from \({z_2}\) to \({z_1}\). xp sits on the plane-- D is Axp plus Byp plus Czp. Direct link to Stephen Custance's post Does the negative value o, Posted 12 years ago. The equation of a line in R^2 is the equation of a plane in R^3. Let's just say that this In the case of the sphere, the geodesic is a segment of a great circle containing the two points. Thanks for the help! Three minus one, minus Thus, z can lie anywhere in the following ring-shaped region: Download SOLVED Practice Questions of Interpretation of |z1-z2| for FREE, Learn from the best math teachers and top your exams, Live one on one classroom and doubt clearing, Practice worksheets in and after class for conceptual clarity, Personalized curriculum to keep up with school. And to do that, let's just Now let's see, 65 you can't factor this. distance to the plane. of the vector f. Or we could say the Thus, z lies on the perpendicular bisector of these two points: Clealy, z can lie anywhere on the real axis. so -5 + 7/2 = -3/2 and 2 - 7/2 = -3/2. The following are two common formulas. The order of the points does not matter for the formula as long as the points chosen are consistent. The leftmost point gets half the horizontal distance added to it while the rightmost point gets half the horizontal distance subtracted. To calculate the distance between two points in a 3D space, you need to use the Pythagorean theorem. Your tips definitely helped me finalize my program, so much appreciated! root of the normal vector dotted with itself. S So it's going to That's just some vector Let me call that vector f. Vector f is just going to 0000007454 00000 n
out, in the last video, the normal vector, if you To get a better estimate than that, the model gets complicated quickly. And we'll, hopefully, The equation \(\left| {z - i} \right| = 3\) says that the variable point z moves in such a way so that it is always at a constant distance of 3 units from the fixed point i. squared plus B squared plus C squared. difference of the y-coordinate. So let me draw, so right over here, let me draw our imaginary axis. where (x1, y1) and (x2, y2) are the coordinates of the two points involved. This will give you an equation for the line. In the complex plane, you wouldn't refer to the horizontal axis as the -axis, you would call it the real axis. magnitude of this vector. And let me pick some point What is the symbol (which looks similar to an equals sign) called? axis we're going from negative one to three so This is n dot f, up there. 0000043314 00000 n
there, and let's first, let's see, we're gonna Algebra & Trigonometry with Analytic Geometry. antique rembrandt brass floor lamps, round building on hill swansea, groveland park, chicago safe, | 677.169 | 1 |
The start point of the arc is selected with the mouse or entered in via the keyboard. To specify the 1st point with the mouse, a point must be selected (Left Button) and accepted (Middle Button).
The next step required is for you to enter the Radius of the Arc, type a number into the box and hit the <Enter> key. next step required is for you to enter length of the Arc, type a number into the box and hit the <Enter> key.
The final step required is for you to enter a bearing, type a number into the box and hit the <Enter> key. The start point of the arc will be tangential to the bearing angle entered.
An ARC is constructed using the information entered.
Note: The start point of the arc is tangential to the yellow line which has a bearing of 60 degrees. | 677.169 | 1 |
Edit. Angle-angle-side (AAS): two angles and a non-included side of each triangle are equal. (SSS, SAS, ASA, AAS, HL) If not, state that it is not sufficient to prove that the triangles are congruent. Also, because BE is congruent to DA, angle BCA is congruent to DCE because vertical angles are congruent. PRQ and Definition of right trianglePRS are right triangles PQ # PS PR# Reflexive Property 6. ' 1. 0000111781 00000 n
This quiz is incomplete! 2. Given 3. 0000137901 00000 n
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UNIT 4B - TRIANGLE CONGRUENCE PROOFS:TEST4B-Fri., DEC. 6th Answer Keys. 0000103223 00000 n
Prove: ABAC-ADAE. To play this quiz, please finish editing it. (It's an […] 0000003572 00000 n
So based on the proof we saw in the last video, that implies these sides are congruent. Triangles AIM and CJM have one congruent side between two congruent angles and are therefore congruent. Assuming the given side is between the two given angles, that's ASA ("angle-side-angle"), which is in itself proof of congruency. X 0000160143 00000 n
Inscribed Angle. Straight Angle. SSS Postulate. ", All tip submissions are carefully reviewed before being published, This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. If you're trying to prove that base angles are congruent, you won't be able to use "Base angles are congruent" as a reason anywhere in your proof. 0000111153 00000 n
In a proof that these two triangles are congruent, Ryan stated +�[tZE�6�J��$��"`�z��k9��E���V@]��t��ٽr���v��:s�͝{g��f��
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Iª�+Z�D߃�@r:�}.��2��X�����;GwI�9)�?.�C9�X.�y�Ahၷ�I��O^33�;���5u�/*٪��̈�ٰ�f6�.f�,�����EM����h�:S$��E.����u��*u3Jj�ԡ���/��+�-���w��9���vq煁[�R��4R;ʺ ������-)*�IWa�p�"�a�=. In geometry, you may be given specific information about a triangle and in turn be asked to prove something specific about it. Are the triangles congruent, if yes, why? Solution to Example 1 1. 0000159585 00000 n
Edit. 9 Two right triangles are shown below. Save. 0000007770 00000 n
Use the triangle congruence criteria SSS, SAS, ASA, and AAS to determine that two triangles are congruent. I can prove triangles are congruent For each pair of triangles, tell: (a) Are they congruent (b) Write the triangle congruency statement. Improve your math knowledge with free questions in "Proving triangles congruent by SSS, SAS, ASA, and AAS" and thousands of other math skills. 0000203864 00000 n
congruent 2. "Corresponding sides" (as a reason in a proof of congruency) means that sides occupying the same position in congruent polygons (triangles in this case) are congruent (or equal in length). 2 hours ago by. aswafford. 0. Edit. 66% average accuracy. 12 Using the tick marks for each pair of triangles, name the method {SSS, SAS, ASA, AAS} that can be used to prove the triangles congruent. Print; Share; Edit; Delete; Host a game. 0000056842 00000 n
So you have this transversal BC right over here. Congruent Triangles - Proving Triangles Congruent Marking Pictures and Proof Practice Packet Proving triangles congruent is a topic that proves difficult for many students. 0000084005 00000 n
Proving Congruent Triangles. Play this game to review Geometry. 0000003849 00000 n
Double check to make sure the problem asks you to prove congruency of two triangles. It may be beneficial to sketch a first diagram that is not accurate and re-draw it a second time to look better. {"smallUrl":"https:\/\/ | 677.169 | 1 |
The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate
Dentro del libro
Resultados 1-5 de 23
Página 7 ... PROB . From a given point to draw a straight line equal to a given straight line . Let A be the given point , and BC the given straight line ; it is required to draw from the point A a straight line equal to BC . From the point A to B ...
Página 12 ... PROB . To bisect a given rectilineal angle , that is , to divide it into two equal angles . Let BAC be the given rectilineal angle , it is required to bisect it . Take any point D in AB , and from AC cut ( 1. 3. ) off a E equal to AD ...
Página 13 Euclid, Thomas Tate. PROP . XI . PROB . To draw a straight line at right angles to a given straight line , from a given point in the same . Let AB be a given straight line , and c a point ... PROB . To draw a straight BOOK I. PROP . XI . 13.
Página 14 Euclid, Thomas Tate. PROP . XII . PROB . To draw a straight line perpendicular to a given straight line of an unlimited length , from a given point without it . Let AB be the given straight line , which may be produced to any length both ...
Página 19 ... PROB . To make a triangle of which the sides shall be equal to three given straight lines , but any two whatever of these must be greater than the third , ( 1. 20. ) Let A , B , C be the three given straight lines , of which any two | 677.169 | 1 |
class 9 math 6.1 Summary
NCERT solutions for class 9 maths chapter 6 Lines and Angle Exercise 6.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download this assignment as pdf
This chapter 6 has total 3 Exercise 6.1 ,6.2 and 6.3. This is the First exercise in the chapter. | 677.169 | 1 |
The Forbidden Triangle
Exercise
This is the Forbidden Triangle:
Two sides are specified and no angle is given a specific value. It is only known that no angle is the same and neither angle is 90 degrees or 30 degrees or 60 degrees. What is the value of the missing side?
Facts
With no angles specified, it is said that the missing side can have any value between 1 and 29 exclusive.
If the 15 line and the 14 line were exactly aligned, a stick 1 unit long would complete the puzzle. If the 15 and 14 line were laid end-to-end, a stick 29 units long would complete the puzzle. Any stick could solve this triangle but the fact of the matter remains that it was drawn on paper just so, and if you had a perfectly accurate ruler you could measure the sides and find the most-right answer: 13.
Thoughts and Opinions
The Forbidden Triangle is one of my favorite Geometrical ideas. You will notice if you read my following explanation that, except for this sentence, I will not mention "Math" as the "Math Answer" is overly general and not helpful. That answer is given in the first remarks in the Facts section. If that first answer is perfectly acceptable to you then you may move on from the Forbidden Triangle back to more comforting levels of rigor on the topic of triangles.
My last answer as a matter of fact is specific and it is right. There is a difference between right and correct, or in this case most-correct and most-right. Please be mindful that I say "most-right" and not "mostly right." You may even be right on a topic without being correct, but that is an idea for another day. People are all too ready to throw around the words "impossible" and "trivial" when they are unable or unwilling to accept the truth of the most-right.
Consider a simpler triangle:
Everybody agrees that, if the angles are all equal (an "equilateral" triangle in some circles), there is one correct answer and it is 3.
When you go to build the triangle, you will find that these matters of fact are emergent from the act of making imagination reality. If I take 3 sticks each 3 feet long and lay them as pictured, I will find that the angles are all exactly equal and they are all exactly 60 degrees. You didn't need a rule book to tell you that, these are just matters of fact on the "3, 3, 3" triangle.
Consider another triangle. This one is taught extensively in schools:
A sensible person immediately knows the answer is 5 for the missing side of this humble triangle. They say that the angle joining the sides 3 and 4 is a 90 degree joint (a "right angle"). When these conditions are met, the missing side is always bigger than any of the other two individually. There is a formula for these things but as before, I will find that a stick 3 feet long and a stick 4 feet long arranged according to these rules will complete a perfect triangle with a 5 foot stick joining them.
These triangle sketches are representations of the laying of these sticks. I use these pictures to stand in for the actual Physical act of building these things. It is labor of the imagination through building, or making art into reality, that reveals the most-right answer. Not at all coincidentally, the most-right answers are also the most beautiful.
Consider this next triangle:
I have taken the liberty to write in the size of each stick used to build this triangle. This one is 3 times larger than the so-called "3, 4, 5 triangle" we considered last. From my picture, you might even be able to tell that I am building a tent large enough to stand in. I need it to be 12 feet high with enough space, say 9 feet, to walk side-to-side comfortably. In order to build my tent, I need material at least 15 feet long to make sure rain does not get in. This is a very big tent, but I have very big needs for shelter and safety.
This isn't my only option. Pictured here is another triangle I built to suit my needs:
You see, the 12 foot stick I use to support my tent could be flanked by two other sticks any size. I choose the sizes of 5 feet and 13 feet because they are beautiful. People make choices based on this criterion in life so often that we take it for granted. On the topic of triangles however, people too often forget this is an option at all. Is this an insult to the triangles, or an insult to our instructors who teach them? Or is it a self-handicap meant to ensure our survival? Are people comforted more by the most-correct over the most-right because their lives depend on it?
Finish building my tent and you will see why the most-right answer for the Forbidden Triangle is indeed 13.
Feel free to email me with your thoughts and opinions on this matter at [email protected].
The article above is dedicated to Eric and Will. Thank you for your patience and thoughtful discussions. | 677.169 | 1 |
Terms from the K-12 Glossary
Vertical Alignment
Purpose and Instructional Strategies
Symmetries of reflection were introduced in the elementary grades through lines of symmetry. In Geometry, students studied other types of symmetries coming from rigid transformations that map a polygon onto itself, and they determined the number of times such a transformation must be applied before each point in the polygon is mapped to itself. In Math for College Liberal Arts, students determine symmetries of reflection, symmetries of rotation and symmetries of translation of a geometric figure.
Instruction includes multiple opportunities for students to explore symmetries using both physical exploration (transparencies, mirrors or patty paper) and virtual exploration, when possible.
Instruction includes using a variety of shapes (mathematical and real world) to explore the reflection symmetry and rotational symmetry of the shapes. Include identifying the lines of symmetry, the order of symmetry and the angle of rotation that will map the figure onto itself.
The order of symmetry is the smallest (nonzero) number of times that you must apply the corresponding transformation to map each point of the figure onto itself.
The order of rotational symmetry is the number of times the figure maps onto itself as it rotates through 360° about the figure's center. Instruction includes identifying the angles of rotation when determining symmetries of rotation.
For example, the order of rotational symmetry for a regular hexagon is 6 with the angle of rotation of 60°.
For example, the order of rotational symmetry for an isosceles trapezoid is 0.
For example, the figure below has an order of 4 with angle of rotation of 90°.
The order of a translational symmetry is infinite because no matter how many times one applies it, no point gets mapped onto itself. Translational symmetry results from mapping a figure onto itself by moving it a certain distance in a certain direction. Show students tessellations (covering of a plane using one or more geometric shapes with no overlaps and no gaps), or have them create them, and discuss the translational symmetry in the tessellation (MTR.5.1).
Depending on a student's pathway, instruction includes making the connection to symmetry as a key feature of the graphs of polynomial and trigonometric functions and the importance of applying such functions in the real world.
Problem types include using symmetries to classify geometric figures.
For example, given that a two-dimensional figure has one line of symmetry through midpoints of the two parallel sides, no rotational symmetry and no point symmetry, students can deduce that this figure is a trapezoid.
Common Misconceptions or Errors
Students may not make the connections to the idea of an infinite pattern (wallpaper pattern, border pattern, etc.) when working with only a portion of that pattern. If students arrive at an order for a translational symmetry that is not infinite, ask them if their answer would be different if the pattern continued.
Students may have difficulty distinguishing between mapping a figure onto itself and mapping every point of the figure onto itself. To help address this misconception, have students highlight a particular point and observe how it is affected by one application of a transformation that maps the figure onto itself.
Lesson Plans
This lesson guides students through the development of a formula to find the first angle of rotation of any regular polygon to map onto itself. Free rotation simulation tools such as GeoGebra, are used.
In this inquiry lesson, students are moving their individually designed Air Jordans around the room to explore rigid transformations on their shoes. They will Predict-Observe-Explain the transformations and then have to explain their successes/failures to other students.
Students explore ways of applying, identifying, and describing reflection and rotation symmetry for both geometric and real-world objects, for them to develop a better understanding of symmetries in transformational geometry.
Students will use a protractor/ruler to construct reflections and a composite of reflections. They will create transformations using paper cut-outs and a coordinate plane. For independent practice, students will predict and verify sequences of transformations. The teacher will need an LCD Projector and document camera.
Students are introduced to isometric transformations using patty paper. Translations, reflections, and rotations will be explained and practiced, emphasizing the properties preserved during those transformations and, without sacrificing precision, allowing students to differentiate between these isometries. The lesson can also be taught using GeoGebra free software.
Students will translate, rotate and reflect quadrilaterals (Parallelogram, Rectangle, Square, Kite, Trapezoid, and Rhombus) using a coordinate grid created on the classroom floor and on graph paper. This activity should be used following guided lessons on transformations. | 677.169 | 1 |
5th Grade Geometry Worksheet
In 5th Grade Geometry
Worksheet we will classify the given angles as
acute, right or obtuse angle; using a protractor, find the measure of the given
angle, classify the given triangle and circle the numbers with
right angles.
1. Write 3 examples of right angles.
2. Name the angles the hour hand makes on the clock face
when the time is:
(i) 10 to 3
(ii) 10 to 9
(iii) 6 : 00
(iv) 20 minutes past 11
(v) 12 : 15
3. How many right angle turns does a soldier make one after the other when he does an "about turn"?
4. All the angles around the middle point of a clock will add to ………………
5. Find the number of triangles in the given figure. How many of these are:
(i) Right Triangle
(ii) Acute Triangle
(iii) Obtuse Triangle
6. Classify the given angles as acute, right or obtuse
angle.
(i) 112°
(ii) 84°
(iii) 120°
(iv) 90°
(v) 180°
(vi) 72°
7. Fill in the blanks:
(i) The triangle with all the three sides equal is called an
……………………. triangle
There are 12 months in a year. The months are January, February, march, April, May, June, July, August, September, October, November and December. The year begins with the January month. December is t… | 677.169 | 1 |
Project Euler #184: Triangles containing the origin.
Consider the set of points with integer co-ordinates in the interior of the circle with radius , centered at the origin, i.e. .
For a radius of , contains the nine points , , , , , , , and . There are eight triangles having all three vertices in
which contain the origin in the interior. Two of them are shown below, the others are obtained from these by rotation.
For a radius of , there are triangles containing the origin in the interior and having all vertices in and for
the number is .
How many triangles are there containing the origin in the interior and having all three vertices in ? | 677.169 | 1 |
The Elements of Euclid
Dentro del libro
Resultados 1-3 de 17
Página 359 ... given in magnitude , when equals to them can be found . II . A ratio is faid to be given , when a ratio of a given ... position , which is contained by ftraight lines given in pofition . V. A circle is faid to be given in magnitude ...
Página 381 ... position . It is to be obferved that there are two ftraight lines EC , GC ... given point , to a ftraight line given in position , and makes a given angle with it ; that ftraight ... given , wherefore alfo the angle EAD is given . DATA . 381.
Página 387 ... given in position . In AB take a given point L , and draw LM perpendicular to CD , meeting HK in N. because LM is drawn from the given point L to CD which is given in position , and makes a given angle LMD ; LM is given in pofition ... | 677.169 | 1 |
Euclidian Geometry
From inside the book
Results 1-5 of 30
Page xv ... perpendicular ( to ) . || parallel . + together with . = equal to . > greater than . < less than . ... because . ... therefore . : is to . The object aimed at is merely to place before the student the various steps of an argument in a ...
Page 16 Francis Cuthbertson. PROPOSITION VII . It is always possible to draw a straight line perpendicular to a given straight line from a given point in the same . B A Let BC be the given straight line , and A the given point in it . Then from ...
Page 17 ... perpendicular to it . FG- For if possible B let AF and AG be each of them 1 to BC . ..LS FAB , FAC are equal to one another , and also △ S GAB , GAC are equal to one another , which is impossible . .. from a given point in a given ...
Page 20 Francis Cuthbertson. PROPOSITION XI . It is always possible to draw a perpendicular to a given straight line of an unlimited length from a given point without it . C Let A be any point without the straight line BC of unlimited length ...
Page 21 Francis Cuthbertson. PROPOSITION XII . There cannot be drawn more than one perpendicular to a given straight line from a given point without it . A R Q B ए Let A be the given point , and BC the given straight line ; and let AQ be drawn ... | 677.169 | 1 |
Pythagorean Theorem: History, Applications, and Real-World Examples
Start Quiz
Study Flashcards
12 Questions
Give an example of an engineering application of the Pythagorean theorem.
What is the significance of the hypotenuse in the Pythagorean theorem?
Who is credited with the discovery of the Pythagorean theorem?
How has the Pythagorean theorem been applied in construction?
What is the general form of the Pythagorean theorem?
How is the Pythagorean theorem used in construction?
Explain how the Pythagorean theorem is applied in art.
What is the significance of the Pythagorean theorem in roof pitch calculation?
How is the Pythagorean theorem utilized in GPS and Navigation?
In what way is the Pythagorean theorem employed in acoustics and sound?
What does the Pythagorean theorem provide in terms of practical applications?
Summary
The Pythagorean Theorem and Its Applications
The Pythagorean theorem is a mathematical relationship that has been used for thousands of years to solve problems related to right triangles. It's based on the simple equation a² + b² = c², where a, b, and c are the lengths of the right triangle's three sides, with c being the hypotenuse (the side opposite the right angle).
History and Origins
The Pythagorean theorem is believed to have been discovered by the ancient Greek mathematician Pythagoras around 500 BCE, though some evidence suggests that Babylonians used a related formula centuries before Pythagoras. Pythagoras' theorem has since become one of the most well-known and widely applied mathematical principles.
Applications
The Pythagorean theorem is a versatile tool with applications in various fields, including geometry, engineering, construction, and art.
Geometry
In geometry, the Pythagorean theorem is used to calculate the lengths of sides in right triangles and to solve problems involving right triangles, such as finding the missing side length or the angle of a right triangle.
Engineering
In engineering, the Pythagorean theorem is used to calculate distances, forces, and stresses in structures, such as buildings, bridges, and mechanical systems. For instance, engineers use the theorem to calculate the stresses in a truss structure, ensuring that the structure is safe and functional.
Construction
In construction, the Pythagorean theorem is used to calculate the distances between points, such as when laying out building foundations, plumbing layouts, and electrical wiring. It's also used to calculate the angles for roofs and other structures, ensuring that they are stable and functional.
Art
In art, the Pythagorean theorem is used to create proportional and aesthetically pleasing compositions, such as in painting, sculpture, and architecture. For example, painters use the theorem to create balanced and harmonious compositions by arranging elements in a geometrically pleasing way.
Real-World Examples
Roof Pitch Calculation
When designing a roof, architects and engineers use the Pythagorean theorem to calculate the roof's pitch (angle). They determine the ratio of the horizontal run (the horizontal distance of the roof) to the vertical rise (the height of the roof) and apply the theorem to find the angle.
GPS and Navigation
Global Positioning System (GPS) devices use the Pythagorean theorem to calculate the distance between a user's location and the location of a distant point of interest. The device calculates the difference between the user's latitude and the latitude of the point of interest, as well as the difference between their longitudes and the longitudes of the point of interest, and applies the theorem to calculate the distance.
Acoustics and Sound
In acoustics and sound, the Pythagorean theorem is used to calculate the effects of sound waves on a structure, such as a concert hall or a recording studio. Engineers use the theorem to calculate the resonant frequencies of the structure and to minimize or maximize the effects of sound waves, depending on the desired acoustic outcome.
Summary
The Pythagorean theorem is a powerful mathematical tool with a wide range of applications in various fields. From solving geometry problems to engineering design, construction, and art, the Pythagorean theorem is an essential part of our mathematical toolkit, providing a foundation for countless practical applications as well as an interesting window into the history of mathematics.
Description
Explore the history, applications, and real-world examples of the Pythagorean theorem, a fundamental mathematical concept that relates to right triangles. Learn how this theorem is applied in geometry, engineering, construction, art, and real-world scenarios. | 677.169 | 1 |
Page 295 - The sum of the angles of a spherical triangle is greater than two and less than six right angles ; that is, greater than 180° and less than 540°. (gr). If A'B'C' is the polar triangle of ABC...
Page 225 - Theorem. If each of two intersecting planes is perpendicular to a third plane, their line of intersection is also perpendicular to that plane. Given two planes, Q, R, intersecting in OP, and each perpendicular to plane M. To prove that OP _L M. | 677.169 | 1 |
Problem
Solution
(a) Note that, given the positions of and , can be in exactly two places. However, is on segment , and and are on the same side of line , so must be on the same side of as . This shows that, given the positions of , , and , we can determine the position of . Also note that must be in the circumcircle of , which means that must be outside triangle . Without loss of generality, assume that and are on opposite sides of . Now it suffices to show that for ever positioning of , , , and such that , there exists a on such that is cyclic. This is simple; merely extend so that it intersects the circumcircle of again at . An example of such an arrangement of points is shown below; , , , , and .
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Solution 2 (suli) Given the circle , construct equilateral inside with , on its interior such that . Then let the circle centered at with radius intersect circle at . Finally, extend to meet at and we are done!
(b) Let and . I shall now find all of the angles of . We know that , since is equilateral. Therefore . Triangle is isosceles, so we have that , and . Since and are inscribed angles that intercept minor arc , . Now note that . Since is isosceles, . We know that and , so . We are now able to use the Law of Sines on . However, we would only get an equation involving , so if we were to use it to find , we must find the length of . This can easily be done using the Law of Sines on :
And now we use the Law of Sines! On !
Therefore
Using the Law of Sines on gives that , so it suffices to show that
Canceling out 's and rearranging shows that this statement is equivalent to
Using the sine and cosine addition formulae gives that this statement is equivalent to
Cross-multiplying and using double-angle formulae gives that this statement is equivalent to
Canceling out like terms and dividing both sides by gives that this statement is equivalent to
This is a simple rearrangement of the Pythagorean Identity, which is true. We can work backwards to get that .
Soln. 2 (fmasroor)
Let <CBP=2x, so that <BPC=<BCP=90-x (no fractions). Then <ADC=180-<ABC=120-2x, <APD=180-<APB-<BPC=x+30, so then <PAD=x+30. Then PD=PA so se need to prove that ODA is equilateral where O is the center of ABCD. However, since <DAP=<DPA=x+30 DP=AP and so ABPD is a kite; <ABD=60/2=30 and then <DOA=2*30=60, so then ODA is equilateral. Done.
Solution 3 (suli)
Notice that the circumcenter of is , so . (Note that is equilateral.) Hence , so is equilateral and . Now if we let then it is easily seen via angle chasing that , so as desired. | 677.169 | 1 |
Right Triangle Trig Worksheet
Right Triangle Trig Worksheet. Pupils practise identifying the graphs for sine, cosine and tangent capabilities before using them to solve equations and to determine transformations on this glorious trigonometric worksheet. Displaying all worksheets associated to – Trigonometry Triangles. Videos, worksheets, 5-a-day and far more. Trigonometric identities involving the Pythagorean theorem are the most generally used ones.
Read Online Trigonometry Questions And Solutions Questions on Angles in Standard Position. Find quadrants of angles in Trigonometry Questions And Solutions The Corbettmaths Practice Questions on Trigonometry. Videos, worksheets, 5-a-day and rather more.
Proper Triangles And Trig Practice Geometry Choice Board Worksheets
This is a worksheet working angle of elevation and angle of melancholy word problems. The reply to the riddle is "Fo' Drizzle" which permits for simply checking scholar answers. Find the Indicated Side / Perimeter of a Right Triangle Using Trigonometric Ratios.
Corbettmaths – This is a component 3 of 3 of a trigonometry evaluate. It covers the inverse trigonometric functions and then discovering missing angles of right angled tr.
Fixing Proper Triangles With Trigonometric Capabilities Worksheet
This handout offers college students with an outline for notes that the trainer can guide them via. These notes cover the essential trig features and how to use them to resolve for lacking sides and angles of right triangles. Provided with this could be a model of the notes already crammed out, but there could be loads of room to add any additional content material.
In this worksheet, we'll follow finding and categorical the values of the three trigonometric ratios—sine, cosine, and tangent—for a given angle in a proper triangle.
Trigonometry- Vectors SHORT ANSWER. Write the word or phrase that best completes each assertion or answers the question.
Introduce the two methods to measure an angle, particularly degrees and radians with this set of worksheets.
Navigate by way of this regulation of sines worksheets that encompass an array of subjects like discovering the lacking aspect and the unknown angles, solving triangles, an ambiguous case in a triangle, discovering the realm of SAS triangle and extra.
From a given listing of equations, pupils establish and select the right reply to align to each graph. Engage your college students with this interactive drag and drop activity for Google Slides to strengthen understanding of right triangle trigonometry. Students will use inverse trigonometric functions to solve for sixteen lacking angles.
This worksheet stack consists of ample exercises to follow conversion between levels, minutes and seconds. We are a search engine for worksheets on the web – like google/bing.
Interactive resources you can assign in your digital classroom from TPT. Practice with Sin/Cos/Tan Practice using the Sin/Cos/Tan buttons on your calculator.
Displaying all worksheets associated to – Trigonometry Triangles. To specifically and precisely measure the scale of an angle in degrees, it is further damaged down into degrees, minutes and seconds.
Allied angle worksheets here enclose exercises like discovering the precise value of the trigonometric ratio providing angle measures in levels or radians, evaluating trig ratios of allied angles and proving the trigonometric statements to mention just some. Trigonometric identities help in simplifying trigonometric expressions. Trigonometric identities involving the Pythagorean theorem are essentially the most commonly used ones.
Adequate worksheets are offered to help in practicing immediate conversions of degrees to radians and vice-versa. Displaying all worksheets related to – Right Triangle Finding The Missing Sides.
Incorporate the regulation of cosines worksheets to raise your understanding of the concept and apply to find the lacking sides of a triangle, finding the unknown angles (SAS & SSS), fixing triangles and much more. With this set of evaluating trigonometric functions worksheets at your disposal, you have not any dearth of apply exercises. Begin with substituting the specified x-values in trigonometric functions and remedy for f.
This activity guides students through understanding the ideas behind trigonometric ratios of proper triangles. Students will discover ways to identify the opposite leg, adjacent leg and the hypotenuse of a right triangle with a given reference angle in addition to the method to set up trigonometric ratios for sine, cosine, and tangent. In this worksheet, we'll follow discovering and categorical the values of the three trigonometric ratios—sine, cosine, and tangent—for a given angle in a right triangle.
Students will follow making use of ideas of trigonometry to search out missing side measures in proper triangles. Access this huge collection of solving triangles worksheets to understand the topics like solving triangles, discovering the area of the triangle, solving the triangle using the given area and far more worksheets are included. This consists of the coed notes, teacher notes, and homework project with reply key for the lesson eleven.1 Pythagorean Theorem, Special Right Triangles, and Trig Functions.This is a part of my Unit 11 Trigonometry for Algebra 2 .
A fast activity to have college students practice labeling the edges of a right triangle because the hypotenuse, opposite side, or adjoining facet. Introduce the 2 ways to measure an angle, namely degrees and radians with this set of worksheets.
Reinforce the concept of principal solutions of trigonometric equations with this sufficient supply of worksheets like fixing linear trigonometric equations, solving trigonometric equations in quadratic kind and far more. Identify the quadrant encompassing the terminal facet of the angle with this set of quadrants and angles worksheets. Draw the indicated angle on the coordinate plane, measure the angles in the quadrant and symbolize as levels and radians and a lot more.
Included here are basic identities like quotient, reciprocal, cofunction and Pythagorean identities, sum and difference identities, sum-to-product, product-to-sum, double angle and half angle identities and ample trig expression to be simplified, proved and verified using the trigonometric formulation. Determine the reference angles in levels and radians, find the coterminal angles for the indicated angles, and optimistic and unfavorable coterminal angles with this assemblage of reference and coterminal angles worksheets.
Trigonometric graphs worksheet for GCSE Maths. Pupils practise figuring out the graphs for sine, cosine and tangent capabilities before using them to solve equations and to identify transformations on this excellent trigonometric worksheet.
Utilize this array of two-part worksheets, whose first includes workout routines to search out the labeled facet and the 2nd offers with discovering the perimeter. The printables are available in customary and metric items. This is a two-page, 6-question worksheet that involves angles of elevation and melancholy, SOH-CAH-TOA and right triangles in word issues.
We don't host any worksheets on our webservers unless acknowledged so or we have the permission of the original writer of the worksheet to host or it was created in-house. Search 50,000+ worksheets, curated by consultants, created by lecturers and aligning to mainstream curriculums. Nagwa is an academic expertise startup aiming to assist teachers train and college students learn.
Identify the legs, aspect and angles, introduce the six trigonometric ratios both main trig ratios and reciprocal trig ratios and rather more with these trigonometric ratio worksheets. Packed in these unit circle worksheets are workouts to search out the coordinates of some extent on the unit circle, decide the corresponding angle measure, use the unit circle to find the six trigonometric ratios and much more.
This is a great exercise for breakout rooms, an task examine, exit card or exit ticket, check for understanding, quiz and more! Assign straight by way of your Google Classroom or create a Google LTI project for Canvas LMS. Utilize this enough supply of inverse trigonometric ratio worksheets to search out the precise value of inverse trig ratios utilizing charts and calculators, discover the measure of angles, solve the equations, study to judge inverse and the composition of trigonometric capabilities and much more.
Questions on Amplitude, Period, range and Phase Shift of Trigonometric Functions with answers. Employ this assortment of general solutions of trigonometric equations worksheets that function ample of exercises to hone your expertise in solving several varieties of trigonometric equations to obtain the overall options.
Trigonometry – Formulas, Identities, Functions and Problems Here are some more worksheets about quantity and floor space . Navigate through this legislation of sines worksheets that embody an array of matters like discovering the missing side and the unknown angles, solving triangles, an ambiguous case in a triangle, discovering the world of SAS triangle and more. Kick begin your studying with these trig ratio worksheets.
A mixture of sin, cos & tan drilling exercise. A few isosceles triangles at the finish as extension.
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It's analysis time! Use this algebraic worksheet as a atom analysis for addition, subtraction, and asperity concepts. Catered to fifth-grade students, this worksheet appearance a array of accession and accession problems involving alloyed numbers and simple fractions, in accession to fill-in-the-blank comparisons. The final allocation of this worksheet additionally asks acceptance to compute the changed...
Equivalent Fractions Worksheet Pdf. Students should fill within the lacking numerator or denominator to make the . You can choose from supported sheets with diagrams for faculty kids who need further assist to harder worksheets for these extra assured. Then, in 5th grade, college students discover ways to add unlike fractions. Children work on quite... | 677.169 | 1 |
Rotating a regular pentagon about another regular pentagon
1. Draw a regular pentagon
2. Draw another regular pentagon of the same perimeter and connect the two via a pair of vertices
3. Rotate the second pentagon about the first
4. Draw the locus of points made by the vertices of the rotated pentagon | 677.169 | 1 |
Body (geometry)
A body is a three-dimensional figure in geometry that can be described by its surface . The surface of a body can be composed of flat or curved surface pieces. If the surface of a body consists only of flat pieces, it is a polyhedron . There are mathematical formulas for calculating the volume and surface area of many geometric bodies (see the Geometry formula collection ). More precisely, a geometric figure of the type just described is called a three-dimensional body , since this concept formation can also be generalized to higher dimensions.
contents
definition
Geometric bodies can be mathematically defined in various ways. If three-dimensional space is understood as a set of points , then a body is a subset of these points that fulfills certain properties.
In stereometry , a body is a restricted three-dimensional subset of three-dimensional space that is bounded on all sides by a finite number of flat or curved surface pieces, including these boundary surfaces. A set is called restricted if there is a correspondingly large sphere that completely encompasses the set. The union of the points of all the delimiting areas forms the surface of the body. The surface of a body divides space into two separate subsets, with the interior of the body being the subset that does not contain a straight line .
In geometric modeling , a body is a limited and regular subset of three-dimensional space. A quantity is called regular if it is the end of its interior. This condition ensures that a body also includes its edge and is completely three-dimensional, i.e. does not have any areas of lower dimensions. One speaks at this point of the homogeneity of a body. According to this definition, a body can also consist of several unconnected components.
The surface of a body can also consist of several unconnected parts. By assigning an orientation to these partial areas , a body can also be described using its surface. One then speaks of the surface representation ( boundary representation ) of the body.
Types of geometric bodies
polyhedron
A polyhedron is a geometric body whose boundaries are polygons . The most famous polyhedra include the regular polyhedra. These are the three-dimensional polygons delimited by regular polygons whose edges only point outwards and which are not infinitely large, such as the cube, the tetrahedron or the so-called soccer ball . There are only five types of these solids: the Platonic solids , which are dual with themselves or with one another, the Archimedean solids and the Catalan solids dual to them, and the Johnson solids . Then there are the prisms and the anti- prisms . There are only five regular polyhedra with which a complete alone chamber charge is possible: cubes, triangular and hexagonal prism twisted double wedge and truncated octahedron .
Body of revolution
Bodies, the surface of which is constructed by rotating a curve around a certain axis, are called bodies of revolution. Each cutting surface that is orthogonal to the axis of rotation has a circular or annular shape. These include spheres, cylinders, cones, truncated cones, torus and ellipsoid of revolution. The sphere occupies a special position because every straight line through its center is an axis of rotation | 677.169 | 1 |
Concept
Line
A line is a one-dimensional object of infinite length with no width or thickness that never bends or turns. Its graphical representation is a straight line with arrowheads on either end, indicating that it continues indefinitely in both directions.
Through any two different points there is exactly one line. A line can be named using any two points on it. The above line could be named AB,BA, or line ℓ. When two or more points lie on the same line, they are said to be collinear. | 677.169 | 1 |
How do you prove that consecutive angles of a parallelogram are supplementary?
How do you prove that consecutive angles of a parallelogram are supplementary?
Con angles on the same side of a transversal are supplementary. Therefore, ∠A + ∠D = 180°.
Are consecutive angles of a parallelogram are supplementary?
If a quadrilateral is a parallelogram, then consecutive angles are supplementary. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.
Is consecutive angles are supplementary?
Any pair of consecutive angles are supplementary . All angles are right angles. Opposite angles are congruent. Any pair of consecutive angles are supplementary.
Why consecutive angles are supplementary?
Note: Consecutive interior angles are supplementary angles, i.e., they add up to 180∘ . This can be proved by the consecutive interior angles theorem which states that "If a transversal intersects two parallel lines, each pair of consecutive interior angles are supplementary (their sum is 180∘ )."
What can you say about two consecutive angles in a parallelogram?
Theorem: Prove that any consecutive angles of a parallelogram are supplementary. Therefore, the sum of any two adjacent angles of a parallelogram is equal to 180°. Hence, it is proved that any two adjacent or consecutive angles of a parallelogram are supplementary.
What can you say about the consecutive angles of a parallelogram?
In a parallelogram, consecutive angles are supplementary and opposite angles are congruent.
What can you say about the two consecutive angles in a parallelogram?
What can you say about any two consecutive angles in parallelogram? A They are always congruent.
What type of angles are the consecutive angles of a parallelogram?
Explanation: In a parallelogram, consecutive angles are supplementary and opposite angles are congruent.
What are opposite angles in a parallelogram?
The opposite angles of a parallelogram are equal. The diagonals of a parallelogram bisect each other. Each diagonal of a parallelogram bisects it into two congruent triangles. If one pair of opposite sides of a quadrilateral is equal and parallel, then the quadrilateral is a parallelogram.
Can a quadrilateral be considered a parallelogram?
Prove theorem: if a quadrilateral is a parallelogram, then its consecutive angles are supplementary.
How are two parallel angles supplementary in Parallelograms get their names from having two pairs of parallel opposite sides.
What are the properties of
Can a two column proof prove a quadrilateral?
The two-column proof proved the quadrilateral is a parallelogram by proving opposite sides were parallel. You can also use the paragraph proof form for any of the six ways. Paragraph proofs are harder to write because you may skip a step or leave out an explanation for one of your statements. You may wish to work very slowly to avoid problems.
How do you prove that consecutive angles of a parallelogram are supplementary? Con… | 677.169 | 1 |
solved : PLEASE HELP Triangle ABC has been rotated 90 to create trian
PLEASE HELP Triangle ABC has been rotated 90 to create triangle…PLEASE HELPTriangle ABC has been rotated 90 to create triangle DEF. Write the equation, in slope-intercept form, of the side of triangle ABC that is perpendicular to EF You must show all work to receive credit.Math Geometry
How it works
Paste your instructions in the instructions box. You can also attach an instructions file | 677.169 | 1 |
You can use the Polygon and Point in the java.awt library: new Polygon(x_coordinates, y_coordinates, coordinates.length).contains(new Point(x, y)) where x_coordinates and y_coordinates are of type Array[Integer]
Yes, you are right. For a convex polygon you can compute the orientation of the point with each of the sides in anticlockwise order. It's inside if the orientation is on the left for all of the sides (the orientation is positive/clockwise). Orientation can be computed according to: (y2 - y1)*(x3 - x2) - (y3 - y2)*(x2 - x1). If 0 the points are colinear, if positive then points are oriented clockwise, if negative counterclockwise.
For those who would like to understand how the method written by Dean Povey above works, here is the explanation:
The method looks at a "ray" that starts at the tested spot and extends to infinity to the right side of the X axis. For each polygon segment, it checks if the ray crosses it. If the total number of segment crossing is odd then the tested point is considered inside the polygon, otherwise - it is outside.
To understand the way the crossing is calculated, consider the following figure:
For the intersection to occur, tested.y must be between the y values of the segment's vertices (v1 and v2). This is the first condition of the if statement in the method. If this is what happens, then the horizontal line must intersect the segment. It only remains to establish whether the intersection happens to the right of the
tested point or to the left of it. This requires finding the x coordinate of the intersection point, which is:
t.y - v1.y
c.x = v1.x + ----------- * (v2.x - v1.x)
v2.y - v1.y
All that remains to be done is examining the subtleties:
If v1.y == v2.y then the ray goes along the segment and
therefore the segment has no influence on the outcome. Indeed, the first part
of the if statement returns false in that case.
The code first multiplies and only then divides. This is done to support
very small differences between v1.x and v2.x, which
might lead to a zero after the division, due to rounding.
Finally, the problem of crossing exactly on a vertex should be
addressed. Consider the following two cases:
Now, to verify if it works, check for yourself what is returned for each
of the 4 segments by the if condition in the method body.
You should find that the segments above the ray (A1, C1, C2) receive
a positive result, while those below it (A2, B1, B2) receive a negative
one. This means that the A vertex contributes an odd number (1) to the crossing
count, while B and C contribute an even number (0 and 2, respectively), which
is exactly what is desired. A is indeed a real crossing of the polygon, while B
and C are just two cases of "fly-by".
Assuming that your point is at Y coordinate y, simply calculate the x positions where each
of the polygon's (non-horizontal) lines cross y. Count the number of x positions that are
less than the x position of your point. If the number of x positions is odd, your point is
inside the polygon. Note: this works for all polygons, not just convex. Think of it this way:
draw a line from infinitely far away straight in to your point. When that line crosses a
polygon line, it is now inside the polygon. Cross the line again, outside. Cross again,
inside (and so forth). Hope this helps!
Many answers are based on intersections. Here is a simple algorithm not based on intersections but only using vector products.
This method is less complex that methods based on intersections, but only works for convex polygons. Since the question is about convex polygons, this less complex method should be preferred.
Imagine that your convex polygon (P0, P1, ..., Pn), with (Pi-1, Pi) being its segments, is a needle clock and that the point C you want to check is inside the clock. The points (Pi) will either turn clockwise, or counterclockwise. Imagine a needle attached to C, that gives hours when turning clockwise. The vectors CP0→, CP1→, CP2→, CP...→, CPn→ all turn either clockwise, or counterclockwise. This means that the vector products CP0→⨯CP1→, CP1→⨯CP2→, CP2→⨯CP...→, CPn-1→⨯CPn→ and CPn→⨯CP0→ all have the same orientation. Since they are also colinear, their orientation is defined by the sign of their scalar product taken two by two.
This property only occurs when C is inside the polygon. Therefore, iif the signs of the scalar products of the vector products are constant, C is inside the polygon. | 677.169 | 1 |
Slopes of Perpendicular Lines 1
Perpendicular Lines
Two lines are perpendicular when they intersect at 90 degree angles.
Two lines are graphed in the plane. The slope of the first line is a/b. The slope of the second line is c/d. Change the values of a, b, c, and/or d and make note of the angle between the lines.
Slopes of Perpendicular Lines
Slopes of Perpendicular Lines
By examining the equations, how do you know when two lines will be perpendicular when graphed? | 677.169 | 1 |
Everything breaks down in my for loop when trying to calculate the third point. This is because the angle after 2 iterations is 0, and results in a bad spawn.
My knowledge of geometry is not really good enough for me to figure out what I should be doing here.
I have some work arounds which would involve adding static points on the circle, or redistributing points along the circle on a new spawn, but it would be pretty cool if I could get this function able to automatically find the correct mid point.
1 Answer
1
Firstly, atan2(p2.y - p1.y, p2.x - p1.x) isn't what you want, as that just gives the direction between the two points, not the distance, so comparing atan2(p2.y,p2.x) and atan2(p1.y,p1.x) is closer to what you want.
That still leaves an annoying edge case at the discontinuity when the atan2 jumps between -pi and pi, so I'd suggest calculating the biggest gap in a different way. You don't need the actual size of the gaps, you just need to compare relative sizes, so you could just use the direct square of the distance between the points instead of an angle: (p2.y-p1.y) * (p2.y-p1.y) + (p2.x-p1.x) * (p2.x-p1.x). This is faster to calculate, and completely avoids the discontinuity.
Even simpler would be a dot product: dot(p1,p2) = p1.x * p2.x + p1.y * p2.y, although in that case you need to reverse the comparison order: the dot product gets bigger when the points get closer together.
BTW, these solutions only work for points on the circumference - if you were wanting to compare general angles between points at varying distances from the center, you could use the dot product, but would have to normalize by dividing by distances from the center. | 677.169 | 1 |
1. Read the Problem Carefully
Start by carefully reading the problem statement to understand what is being asked. Identify any given information, such as measurements, angles, or geometric relationships.
Draw a Diagram
Visualize the problem by sketching a diagram that represents the geometric figures involved. Label the given information and any unknown quantities. A clear and accurate diagram can provide valuable insights and help you identify relevant geometric properties.
3.Identify Knowns and Unknown
Determine what is known and what needs to be found. This may involve identifying side lengths, angles, areas, perimeters, or other geometric properties. List the known values and assign variables to the unknown quantities.
4. Apply Relevant Formulas and Theorems
Use your knowledge of geometric formulas and theorems to formulate equations or relationships based on the given information. This may involve formulas for area, perimeter, volume, angles, Pythagorean theorem, similar triangles, or circle properties, depending on the problem.
5. Solve for the Unknowns
Use algebraic techniques to solve the equations or relationships derived from the geometric properties. This may involve simplifying expressions, combining like terms, isolating variables, or using strategies such as substitution or elimination.
6. Check Your Solution
Once you have found a solution, verify its accuracy by checking whether it satisfies all the conditions stated in the problem. Ensure that your solution makes geometric sense and aligns with the properties of the given figures. If possible, re-read the problem to confirm that your solution addresses the original question.
7. Consider Alternative Approaches
If you encounter difficulties or uncertainties while solving the problem, consider alternative approaches or strategies. This may involve breaking the problem down into smaller steps, using geometric constructions, or applying different theorems or formulas to gain insight.
8. Practice and Review
Practice solving a variety of geometry problems to reinforce your understanding of geometric concepts and problem-solving techniques. Review theorems, formulas, and geometric properties regularly to ensure that you are familiar with them. Learning from mistakes and seeking feedback can also help improve your problem-solving skills over time.
By following these steps and practicing regularly, you can develop confidence and proficiency in solving geometry problems of varying complexity. Remember to approach each problem systematically, use your geometric knowledge effectively, and stay patient and persistent when faced with challenges. | 677.169 | 1 |
Navigation as a source of geometric knowledge: Young children's use of length, angle, distance, and direction in a reorientation task.
2012-01-01
Abstract | 677.169 | 1 |
The Neglected Importance of Geometry
When 'Geometry' was first introduced to me in my 5th grade, I really disliked it a lot. But, being a Fine Arts student as well, I learned how each and every piece of art begins and comprises of basis shapes only.
Today, in this fast-changing world of science and technology, geometry is playing an important role in a very hideous way that we even fail to notice it sometimes. It is surely no more confined only to mathematics. To be more precise, it has applications in our daily lives since centuries.
What is Geometry?
Geometry is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. A mathematician who works in the field of geometry is called a geometer.
While geometry has evolved significantly throughout the years, there are some general concepts that are more or less fundamental to geometry. These include the concepts of points, lines, planes, surfaces, angles, and curves, as well as the more advanced notions of manifolds and topology or metric.
Geometry is all around us-
Geometry briefly is used in various daily life applications such as surveying, astronomy, navigation and building and much more. Some of such applications of Geometry in daily life in different fields are described below-
Art
Image Courtesy: graphicine.com
Mathematics and art are related in a variety of ways. For instance, the theory of perspective (a graphical representation of a real 3D image on a flat surface of an image as seen by eyes) showed that there is more to geometry than just the metric properties of figures: and this perspective is the basis of the origin of projective geometry.
Technicals
The concept of geometry is also applied in the fields of robotics, computer, and video games. Geometry provides handy concepts both for computer and video game programmers. The way & the design of the characters that move through their virtual worlds requires geometric computations to create paths around the obstacles concentrating around the virtual world. Video game engines typically put to use raycasting, which is a technique that simulates a 3-D world using a 2-D map. Using this form of geometry helps speed up processing because calculations are only done for the vertical lines on the screen.
Architecture
Image Courtesy: toeyyboonchit.wordpress.com
Just like other forms of arts, architects use mathematics for several reasons. Apart from the mathematics needed when engineering buildings, architects use geometry: to define the spatial form of a building i.e. to create the design of the building its shape, height, structure basically the construction blueprint.
Also, one of the best examples of the application of geometry in daily life will be the stairs which are built in homes in consideration to angles of geometry constructed at 90 degrees.
Geometry concepts are also applied in CAD (Computer Aided Design) where it helps the software to render visual images on the screen.
Astronomy & Physics
Here, geometry is used in the field of astronomy, helping to map the positions of stars and planets on the celestial sphere and describing the relationship between movements of celestial bodies.
In the field of Physics, there is a deep link between pseudo-Riemannian geometry and general relativity
Geographic Information Systems
Geometry concepts are used in satellites in GPS systems, it calculates the position of the satellite and location of GPS gauged by the latitudes and longitudes.
Geometry in Video Game Graphics
Video game graphics is all about geometry. Geometric shapes and interactions between geometric shapes is the basic foundation of all videos games. Video games rely on the extensive use of circles, squares, ovals, rectangles, trapezoids, and many other geometric shapes to form shapes you see on your computer or TV screen as you play video games.
When video game programmers first begin developing figures and scenes, they start with a basic geometric shape of the objects. Then the video programmers start inserting additional lines, circles, squares, and other shapes into these original lines at places to given the video graphics depth – three dimensional look.
Another use of geometry used in video games is using geometric shapes which are congruent with other shapes. Also they use geometric shapes by rotating, reflecting, and the use of similarities in making the video graphics move around the screen during action sequences.
The process of building a human-like head in a video game starts with simple shapes, as shown in this image created by Ryan Bown. More shapes are added as the modeling process continues. This builds up the resolution of an image, adding greater detail to games.
With the development of adding three dimensional views to video games, the entire video game industry was revolutionized. Video game programmers can now integrate any geometric shape they want and get it to accomplish pretty much anything what it to do.
If you want to become a video game programmer one day, you need to develop a deep understanding of all geometry concepts. The best video game programmers will all tell you that one subject they have to be strong in is geometry | 677.169 | 1 |
7-3 Proving Triangles Similar Worksheet Answer Key
7-3 Proving Triangles Similar Worksheet Answer Key - Web 7.3 proving triangles similar 11 february 10, 2010 feb 55:38 pm example #5: Web determine whether the triangles are similar. The triangles in each pair are similar. Plan proving triangles similar objectives 1 to use aa, sas, and sss similarity statements 2 to apply aa,. If so, write a similarity statement. Rectangle abcd has a length of 2.6 cm and a width of 1.8 cm. Web edit proving triangles similar worksheet answer key pdf. If not, what would be sufficient to prove the triangles. Web proving triangles similar worksheet answer key geometry. Web about the similar triangles worksheet answer key.
Web proving triangles similar just as when we were proving triangles were congruent (using sss, sas, asa, or aas), we have similar. Web determine whether the triangles are similar. Web edit proving triangles similar worksheet answer key pdf. The triangles in each pair are similar. If not, what would be sufficient to prove the triangles. Rectangle abcd has a length of 2.6 cm and a width of 1.8 cm. Web proving triangles similar worksheet answer key geometry.
√ Proving Triangles Similar Worksheet Answer Key Nilus Appletouch
By triangle sum theorem, in δ. Web proving triangles similar abe = dce by the aa = postulate lmn = opn by the aa = postulate not similar; If so, write a similarity. Web explain why the triangles are similar, then find be and cd. ∆qrs8) not similar find the missing length.
√ Proving Triangles Similar Worksheet Answer Key Nilus Appletouch
B if two sides of one. In this blog of triangles triangle angles,. Web proving triangles similar just as when we were proving triangles were congruent (using sss, sas, asa, or aas), we have similar. Web goal show that two triangles are similar using the aa similarity postulate. Web 7.3 proving triangles similar 11 february 10, 2010 feb 55:38 pm.
Unit 6 Similar Triangles Homework 4 Similar Triangle Proofs / Unit 6
Web if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. In this blog of triangles triangle angles,. Key words similar polygons p. Web observe the triangle pairs and based on the proportionality of their sides and congruence of their angles, identify the similarity. Web determine whether the pair of triangles.
️Proving Triangle Similarity Worksheet Free Download Goodimg.co
Web explain why the triangles are similar, then find be and cd. By triangle sum theorem, in δ. Geometry 7.3 notes similar triangles. Web determine whether the two triangles shown below are similar. Web determine whether the triangles are similar.
Unit 6 Similar Triangles Homework 1 Ratio & Proportion Answer Key
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50 Proving Triangles Similar Worksheet Chessmuseum Template Library
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PPT 7.3 Proving Triangles Similar PowerPoint Presentation, free
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Triangle Similarity INB Pages Mrs. E Teaches Math
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Geometry 7.3 Notes Similar Triangles.
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Web Determine Whether The Pair Of Triangles Is Similar.
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The Triangles In Each Pair Are Similar.
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Key Words Similar Polygons P.
If so, write a similarity statement. Web proving triangles similar worksheet answer key geometry. The triangle similarity worksheet is designed to help kids master. Web proving triangles similar abe = dce by the aa = postulate lmn = opn by the aa = postulate not similar; | 677.169 | 1 |
Pythagorean Theorem
Geometry began to take shape thousands of years ago. Though any number of great minds of the past had a hand in building this branch of mathematics, you hear about some more than others. One of the most famous of these is Pythagoras. He was a philosopher, teacher and gifted mathematician, and he's responsible for one of the most important rules about right triangles you'll use in the world of geometry: the Pythagorean Theorem.
Pythagoras theorem goes like this: "The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides." When you're using it to find your hypotenuse, it'll be written:
a2 + b2 = c2
A right triangle always has one angle measuring 90 degrees. The hypotenuse, or the longest side, will be the one directly across from the right angle. When you're using the Pythagoras theorem, it'll always be your "c" side. Which one ends up being "a" or "b" really doesn't matter one way or the other, but the side across from the right angles must be "c"; otherwise, you're setting yourself up for a big fall.
All that being said, let's get started with a basic example of finding the length of the hypotenuse of a right triangle.
Say the length of side "a" in our example here is 12, and the length of side "b" is 8. Our job is to find the length of side "c", so our problem will start off looking like this:
122 + 82 = c2
We have to square sides "a" and "b", or multiply them by themselves, to get to our next step in the equation, bringing us to:
144 + 64 = c2
From there, we simply add the squares of "a" and "b" together to get closer to our mystery hypotenuse length.
208 = c2, or c2 = 208
Since 208 is actually the square of our hypotenuse, we're not quite finished just yet. We have to find the square root of 208, which means finding the number you'd multiply by itself to get 208. If you have a scientific calculator at your fingertips, you could type in 208, hit the button that looks like an upward arrow (^), and type in .5. Alternately, you could hit the "2nd" button followed by the "x2". Then, type in 208 and press the "enter/=" key.
If you're ciphering without technology to back you up, it's just a matter of finding two numbers that come close to 208 when squared, one slightly below and one just above it. As it happens, 142 is 195 and 152 is 225, so:
c ≈ 14.5 by mental math, or c ≈ 14.4 according to the calculator.
You can also use the Pythagoras theorem to find the length of side "a" or "b" if you already know the length of the hypotenuse. Things just need to be rearranged a bit. Here's another example:
Say we already know the length of the hypotenuse is 12, and the length of "b" is 9. Start off by writing the Pythagorean theorem equation just as you normally would, inserting the numbers we already have:
a2 + 92 = 122
This would then become:
a2 + 81 = 144
Since we have to balance out the equation to get "a" all by itself, we simply subtract 81 from both sides:
a2 + 81 – 81 = 144 – 81
a2 = 63
a ≈ 7.9
Pythagoras is said to have refused to allow his concepts to be written down, but his students passed along his teachings to others. Eventually, this theorem went public and gave us the power to find the length of any side of a right triangle as long as we have an idea of what the other two might be. | 677.169 | 1 |
Hope this helps! -Ron
Thanks Ron. I agree with your opinion. In fact while going thru these concepts, i realized that substantial part of these concepts is based on Pythagorean theorem on right triangle e.g. distance formulae, slope, intercepts of a line, and all that. so in order to perform well in coordinate geometry it is helpful to think from this perspective | 677.169 | 1 |
Properties of Four Sided Figures- Drawing trapezium
Introduction of four sided figures
Geometry is an important branch of mathematics that involves studying shapes, sizes, positions, and dimensions of objects. In this lesson, properties of four sided figures we will focus on trapezium, which is a four-sided figure, and learn how to draw it using a protractor, ruler, and set square.
Properties of four sided figures
Before we begin drawing the trapezium, let's review some of its properties. A trapezium is a quadrilateral, which means it has four sides. The two opposite sides of a trapezium are parallel, and the other two sides are non-parallel. Additionally, a trapezium has two angles on each side.
Drawing Trapezium- four sided figures
To draw a trapezium, we need to follow certain steps. First, we will sketch the figure and then draw the trapezium for the given dimensions.
Step 1: Sketch the Figure
We start by sketching the trapezium in our notebooks. We draw a horizontal line segment FG of length 9cm. Next, we use a protractor to draw an angle of 60° at point F. Then we label point E, which is on line FG, and EF is of length 5cm. Finally, we draw a vertical line from E to point H.
Step 2: Draw a Line Parallel to FG
Now we need to draw a line parallel to FG, passing through point E. For this, we use a set square and a ruler. We place the set square along FG and position the ruler at point E as shown in the picture. Then we slide the set square upwards along the ruler until it touches point E. Next, we draw a line parallel to EH from point E, which intersects the line FH at point G.
Step 3: Draw the Angle at G and Label Point H
Now we need to draw the angle at G, which is 50°. We use a protractor to draw this angle. The two lines, EF and GH, meet at point H. We label this point.
Step 4: Check the Drawing
Once we have completed drawing the trapezium, we need to check it. We can compare the trapezium we drew with the sketch we made earlier to see if it is similar.
Conclusion of four sided figures
Drawing a trapezium may seem challenging at first, but it is a simple process if we follow the steps carefully. By using a protractor, ruler, and set square, we can draw a trapezium for any given dimensions. Understanding the properties of trapezium is also important as it helps us to identify and differentiate it from other quadrilaterals. | 677.169 | 1 |
How Many Edges Does a Cone Have?
Table of Contents
Introduction
A cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a single point called the apex or vertex. It is a commonly encountered shape in various fields, including mathematics, engineering, and everyday objects. One question that often arises is: how many edges does a cone have? In this article, we will explore the answer to this question and provide valuable insights into the properties of cones.
The Definition of a Cone
Before delving into the number of edges a cone has, let's first understand the basic definition of a cone. A cone is a solid object that has a circular base and a curved surface that connects the base to a single point called the apex. The base of a cone is always a circle, and the apex is located directly above the center of the base.
The Faces of a Cone
A cone has two main types of faces: the circular base and the curved lateral surface. The circular base is a flat face that is formed by a closed curve, which is a circle in the case of a cone. The curved lateral surface is a conical surface that connects the base to the apex. It is formed by infinitely many lines that extend from the points on the base to the apex.
The Edges of a Cone
Now, let's focus on the edges of a cone. An edge is a line segment where two faces of a solid meet. In the case of a cone, there are two types of edges: the base edges and the lateral edges.
Base Edges
The base edges are the line segments that form the boundary of the circular base. Since a circle is a closed curve, it does not have any straight line segments. Therefore, a cone does not have any base edges.
Lateral Edges
The lateral edges of a cone are the line segments that connect the points on the base to the apex. These edges form the curved surface of the cone. The number of lateral edges a cone has depends on the number of points on the base. If the base of the cone is a regular polygon with n sides, then the cone will have n lateral edges. For example, if the base is a triangle, the cone will have three lateral edges. If the base is a square, the cone will have four lateral edges.
Real-World Examples
Cones can be found in various real-world examples, showcasing their practical applications and relevance. Here are a few examples:
An ice cream cone: The shape of an ice cream cone is a classic example of a cone. The ice cream sits on top of the conical surface, and the cone itself has a circular base.
A traffic cone: Traffic cones are used to redirect traffic or indicate hazards on the road. They have a conical shape with a circular base and are often made of bright orange plastic.
A volcano: The shape of a volcano resembles a cone, with the apex representing the volcanic vent and the base representing the wider area of the volcano.
Summary
In conclusion, a cone has two main types of faces: the circular base and the curved lateral surface. It does not have any base edges, but the number of lateral edges depends on the number of points on the base. Real-world examples such as ice cream cones, traffic cones, and volcanoes demonstrate the practical applications of cones. Understanding the properties of cones, including the number of edges, is essential in various fields of study and everyday life | 677.169 | 1 |
The great-circle distance or orthodromic distance is the shortest distance between two points on the surface of a sphere, measured along the surface of the sphere (as opposed to a straight line through the sphere's interior). The distance between two points in Euclidean space is the length of a straight line between them, but on the sphere there are no straight lines. In spaces with curvature, straight lines are replaced by geodesics. Geodesics on the sphere are circles on the sphere whose centers coincide with the center of the sphere, and are called great circles.
Note that the definition also mentions 'Geodesics'
Operationally what is the difference between them?
In ArcMap, the measurement tool allows measuring 'Geodesic' distances but not 'Great Circle' distances. Can these be considered equivalent?
2 Answers
2
Geodesics are the shortest path on a curved surface (e.g. sphere or ellipsoid) - like a straight line on a flat plane. A great circle is the shortest path on a sphere, but not on an ellipsoid (as long as both axes are not of equal length). So every great circle is a geodesic, but not every geodesic must be a great circle (on an ellipsoid for example)
A geodesic distance is one measured along a geodesic. On a sphere it is measured along a great circle.
On an ellipsoid with a small eccentricity the geodesic is close to a great circle. The length of this geodesic is usually approximated iteratively using Vicenty's formula.
Geodetic, geodesic: Synonymous
There is no reason for the adjective geodetic to be used to qualify a geodesic curve. The origin seems to be an imprecise translation from French. However the confusion is present and in practical both geodesic and geodetic are found. WGS 84 uses geodetic, so I guess the future is to geodetic.
Geodesic, shortest path: Not synonymous
On any surface, a person walking "straight ahead" (or a small vehicle with un-steerable wheels) follows a geodesic.
The shortest path on a given surface can be a geodesic. However this is misleading:
There are geodesics which are not the shortest paths: The shortest path on a sphere is an arc of great circle, but the complement of this arc, which generally is not the shortest path, is also a geodesic.
There are shortest paths which are not geodesics. The black geodesic is not the shortest path from A to B: | 677.169 | 1 |
Polygons!
Students can first look for the first word and search either vertically or horizontally.
After finding one word, they will go to the second.
Similarly, they will find all the words on the list.
Puzzle War!
It is time for a puzzle battle.
Find the words.
Words may be arranged vertically, horizontally, or diagonally.
We initially select a word and find its first letter; alternatively, we may search for the word's final letter.
Download Free Printable Worksheet
Download the following combined PDF and enjoy your practice session.
So today, we've discussed geometry vocabulary word search puzzle with answer key worksheets using the concepts of basic vocabulary, angle, triangle, circle, and some interactive puzzles. Download our free worksheets, and after practicing these worksheets, students will surely improve their mathematical skills and have a better understanding of geometry vocabulary word search | 677.169 | 1 |
Q. Make correct statements by filling in the symbols ⊂ or ⊄ in blank spaces: (i) {2, 3, 4}......{1, 2, 3, 4, 5} (ii) {a, b, c}...{b, c, d} (iii) {x:x is a student of Class XI of your school}. . .{x:x student of your school} (iv) {x:x is a circle in the plane} . . .{x:x is a circle in the same plane with radius 1 unit} (v) {x:x is a triangle in a plane} . . . {x:x is a rectangle in the plane} (vi) {x:x is an equilateral triangle in a plane} . . . {x:x is a triangle in the same plane} (vii) {x:x is an even natural number} . . . {x:x is an integer} | 677.169 | 1 |
Returns the straight-line shortest distance from the point to the plane.
The returned value is positive if the point is in front of the plane (on
the side with the normal), or negative in the point is behind the plane (on
the opposite side from the normal). It's zero if the point is exactly in
the plane.
Returns true if the plane intersects the infinite line passing through
points p1 and p2, false if the line is parallel. The points p1 and p2 are
used only to define the Euclidean line; they have no other bearing on the
intersection test. If true, sets intersection_point to the point of
intersection.
Returns true if the two planes intersect, false if they do not. If they do
intersect, then from and delta are filled in with the parametric
representation of the line of intersection: that is, from is a point on
that line, and delta is a vector showing the direction of the line. | 677.169 | 1 |
In any right angled triangle, the hypotenuse happens to be the longest side. In this question, the side with length 5 cm is the hypotenuse. Also this triangle satisfies Pythagoras property, so the triangle is right angled.
By exterior angle sum property, we know that an exterior angle of a triangle is equal to the sum of its interior opposite angles. Measure of exterior angle = 50 + 70 = 120 degree. | 677.169 | 1 |
Pythagorean Theorem
The right triangle is a very distinct triangle that has the special characteristic property of having all equal angles inside the triangle. This type of triangle is the most famous of the three types of triangles. One of the best ways to know the overall length and area of a right triangle is through the usage of the Pythagorean Theorem.
7. The Full Pythagorean Theorem
8. Right Triangles and Pythagorean Theorem
What Is the Pythagorean Theorem?
The Pythagorean theorem is a proven solution that allows the person to calculate the area of each square that a line will represent on each side of the triangle. The Pythagorean theorem states that the summation of the squares of both of the smaller sides is equal to the square of the hypotenuse. This theorem has many applications in different contexts, themes, and practical situations, such as construction and architecture. This theorem has a lot of place in history and has origins that trace back to the Greeks culture, ethnicity, ethnic group, and subculture of mathematicians.
How to Solve Using the Pythagorean Theorem
The Pythagorean theorem follows a very simple outline or outline format that people can easily use to obtain and generate missing information about a right triangle. If you want to learn more about how to use and solve the Pythagorean theorem, you can opt to look at any of the articles, examples, or math worksheets for students on the links above.
Step 1: Obtain or Familiarize Yourself with the Pythagorean Theorem
Begin by obtaining or familiarizing yourself with the Pythagorean theorem as this will help provide a suitable structure and starting point you can use to create a solution. Not only that but it will also provide you with the variables you need to identify in the problem or question. The Pythagorean theorem is A^2 + B^2 = C^2 where A and B are the shorter lines and C is the hypotenuse.
Step 2: Write Down or Separate the Variables
You must write down or separate the variables that you will use in the questions as this will help you easily substitute the correct values to the variables. You must also ensure that all variables have the same measurements. If they don't, you must convert them accordingly.
Step 3: Substitute the Variables to the Formula
When you have finished writing down the specific variables, you but now substitute them to the formula of the Pythagorean theorem. Ensure that the missing variable is situated on the left side of the equation, this means you will do the specific operations to transfer the variable if it is needed.
Step 4: Solve the Missing Variable
When you have finished substituting all the variables in the formula, you must solve for the missing variable. You must also ensure that the answer has the correct mode of measurement the question has asked of you. If it is not, then you must convert it accordingly.
FAQs
Can a person apply the Pythagorean theorem on obtuse triangles?
No, they cannot. The Pythagorean theorem is a specific formula that people can only apply to right triangles exclusively. This means people cannot use the formula to solve for specific areas or lengths of the sides of an obtuse triangle.
Can a person apply the Pythagorean theorem on acute triangles?
No, a person cannot use the Pythagorean theorem to solve for specific lengths and areas on acute triangles. The Pythagorean theorem can only solve for areas or lengths of each side of right triangles.
Who created the Pythagorean theorem?
The Pythagorean theorem owes its namesake to the Greek philosopher or mathematician who postulated the formula. Pythagoras is a Greek philosopher who lived from 570 – 495 BC, he did not only formulate the Pythagorean theory, but Pythagoras also posited the idea of the sphericity of the Earth.
The Pythagorean theorem is a specific and specialized formula people can use to identify the specific lengths and areas of the sides of a right triangle. It is important to know and understand the Pythagorean theorem and its modern-day applications. | 677.169 | 1 |
Centers 2023
25 August 2023
6 August 2023
curprev03:1603:16, 6 August 2023 Eric Lengyeltalkcontribs 3,359 bytes+3,359 Created page with "The ''center'' of a round object (a round point, dipole, circle, or sphere) is the round point having the same center and radius. The center of an object $$\mathbf x$$ is denoted by $$\operatorname{cen}(\mathbf x)$$, and it is given by the meet of $$\mathbf x$$ and its own anticarrier: :$$\operatorname{cen}(\mathbf x) = -\operatorname{car}(\mathbf x^*) \vee \mathbf x$$ . (The negative sign is not strictly necessary, but is included so the fu..." | 677.169 | 1 |
Introduction
Angle bisectors can be seen quite often in our daily life. They are defined as a ray or a line that splits an angle into two equal halves. It is important to know their use and methods to construct them.
What Are Angle Bisectors and Their Use
Before we learn what angle bisectors are, Let's look at what an angle is and what bisectors are. An angle is a combination of two rays having a common endpoint. This common endpoint is known as the vertex of the angle and the spacing between the rays is called the measure of an angle. A bisector is simply a line (a ray in our case) that divides the angle into two equal parts.
Scissor Showing an Angle
The scissor pivot is called the angle's vertex. The line passing through the vertex divides the scissor into two symmetrical portions and is called the bisector. If a bisector bisects an angle, it is called an angle bisector.
Real-world applications for angle bisectors include quilting, baking, and architecture. Numerous games, including rugby, football, and pool, use angle bisectors.
How to Draw an Angle Bisector?
Let's take a fun approach in trying to make an angle bisector. Take a paper and make an angle on it. Now fold the paper so that the first line completely overlaps the second line. Make a crease. Now unfold the paper and draw a line along the crease.
Angle Bisector Using Paper Fold
What are the Properties of Angle Bisectors?
It is essential to know the properties of angle bisectors. Since we cannot call all lines angle bisectors, we need to know a few distinct properties of angle bisectors. These are:
It must pass through the vertex of the angle.
It must be equally inclined[spaced] between the rays of the angle.
The angle bisector is also the line of symmetry of the angle.
In the case of a straight line, the angle bisector is perpendicular to the line.
For example, the first image shows a slice of pizza unevenly portioned by the line. Thus the line is not the angle bisector. However, the pizza is evenly portioned by the line in the second image. Hence it can be called an angle bisector.
Pizza Showing Angle Bisector
Construction of Angle Bisectors
Construction of Angle Bisectors using Protractor
We have to bisect ∠PQR. We measure the angle using a protractor. Let's say the measure is 110°. The half of 110° is 55°. So we make an angle of measure 55°, called ∠PQS. QS is the required angle bisector. And ∠SQR \[ = \] ∠PQS \[ = \] 55°.
Drawing Angle Bisector using a Protractor
Let us now move on to learn how to construct an angle bisector using a compass.
Construction of Angle Bisectors using a compass
We have to bisect ∠ABC. To do that,
We first place the needle end of the compass on vertex B.
Draw an arc, letting the arc intersect AB and BC at R and P, respectively.
Now we place the needle end of the compass on P and draw a big enough arc.
Do the same at point R and let these two arcs meet at point S.
Join BS.
We have our required angle bisector for ∠ABC.
Constructing Angle Bisector using Compass.
Great! You have now understood what angle bisectors are, their main properties, and their construction. Now, let us take a look at some solved examples below.
Solved Examples
1. Given an ∠ABC \[ = \] 40°. Find the measure of the angle after an angle bisector bisects it.
Using the property that the angle bisector divides the angle into two equal parts, angle bisectors, we can find the measure of the angle. Let this measure be x. We have \[{\rm{x + x}} = {40^ \circ }\]. This gives \[{\rm{x}} = {20^ \circ }\]. Hence, the measure of the angle after its bisection is 20°.
2. Consider an angle ∠AOB. Write the steps to construct its angle bisector using a compass.
Steps to construct an angle bisector using a compass. First, draw the angle using a protractor.
Steps to Construct an Angle Bisector using a Compass
1. Make an arc of any radius using the compass with the needle on point O. Let's name the points where the arc cuts the rays of angle AOB
Steps to Construct an Angle Bisector using a Compass
2. Now with the needle on Point C, draw an arc with a greater radius. Now put the needle on Point D and draw an arc of the same length. And let these arcs intersect at Point E.
Steps to Construct an Angle Bisector using a Compass
3. Join OE. This completes the construction of the angle bisector.
Steps to Construct an Angle Bisector using a Compass
Conclusion
The angle bisectors are very important in proving complicated proofs. Great! We now know how to identify angles, how to construct angle bisectors, and their construction and mathematical implementation of angle bisectors.
FAQs on Understanding Angle Bisectors
1. How many angle bisectors does an angle have?
An angle has a unique angle bisector. However if we also choose to include the external angle, then there are precisely two angle bisectors for any angle formed by a pair of lines ( an internal angle bisector, and an external angle bisector).
2. Give one example of an angle in real life.
Take a book. Now open the cover page but keep it inclined at a position. What do you see? An angle formed between the book's page and the book's cover page.
3. What is the importance of angle bisectors?
Angle bisectors are very important in identifying the corresponding similar or identical parts of a triangle and many such polygons.
4. What is the range of value of a bisected angle?
The value of an angle ranges from 0° to 180°. Hence the value of the bisected angle ranges from 0° to 90°. | 677.169 | 1 |
Equilateral Triangle Calculator
Calculations at an equilateral triangle or regular trigon. This is the most simple regular polygon (polygon with equal sides and angles). Enter one value and choose the number of decimal places. Then click Calculate.
Length, height, perimeter and radius have the same unit (e.g. meter), the area has this unit squared (e.g. square meter).
Anzeige
Heights, bisecting lines, median lines, perpendicular bisectors and symmetry axes coincide. To these, the equilateral triangle is axially symmetric. They meet with centroid, circumcircle and incircle center in one point. To this, the equilateral triangle is rotationally symmetric at a rotation of 120°or multiples of this, so it has the order 3.
perimeter p, area A
heights ha, hb, hc
incircle and circumcircle
angles and bisecting lines
median lines
perpendicular bisectors
The equilateral or regular triangle has three angles of 60 degrees each. The root of 3 appears in its height, incircle and circumference radius as well as the area, so all of these values are irrational. The equilateral triangle forms the sides of three of the five Platonic solids, these are tetrahedron, octahedron and icosahedron. With equilateral triangles, the plane can be tiled without any gaps, without creating a right angle at any point.
Regular polygons get closer and closer to a circle as the number of corners increases. The equilateral triangle has the fewest corners of all these polygons, so it can be understood as the regular shape that differs most from the circle. Since circular objects can roll and polygonal shapes roll better as the number of corners increases, shapes based on the equilateral triangle are the regular objects that resist rolling the most and therefore offer the best protection against unwanted rolling. | 677.169 | 1 |
Given an origin O and a point P on the curve, let B be the point where the extension of the line OP intersects the line $x=2a$ and C be the intersection of the circle of radius $a$ and center $(a,0)$ with the extension of OP. Then the Cissoid of Diocles is the curve which satisfies OP=CB.
But is this condition applicable to the points on the curve that lie outside the circle? Clearly, if point 'P' was outside the circle, then the extension of OP wouldn't have intersected the circle.
1 Answer
1
$OP$ is considered part of its own extension, so for $P$ outside the circle, $C$ is between $O$ and $P$ (and the required equality may be restated as $OC=PB$). There is thus no problem in the definition.
$\begingroup$I get your first point — that the point $C$ will be between $O$ and $P$, but are you sure that the required equality has to be restated to $OC =PB$? Wouldn't the original condition $OP = CB$ still work?$\endgroup$ | 677.169 | 1 |
Worksheet for constructing Plane Tessellation Exploration Activity
Constructing the above GeoGebra applet expose participants to various symmetric tessellation pattern constructions in the Cartesian plane. The basic regular tessellation also allows for exploration of many derivative tessellations with artistic qualities that are linked to 4-fold rotational symmetry. | 677.169 | 1 |
line $\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$ is parallel to the plane $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=7$. Also, find the distance between them. | 677.169 | 1 |
The "rotation" here refers to the angular displacement of a single rotation about a fixed axis. In 3D space, any rotation is always around a single axis of rotation, which may have any spatial orientation. This is called the "axis–angle representation" of a rotation.
In general, one may represent a rotation as three sequential rotations about a fixed set of Cartesian x-y-z axes (often called "Euler angles"). However, according to Euler's rotation theorem, any sequence of rotations in 3D space can always be equivalently represented by a single rotation about some fixed axis. | 677.169 | 1 |
A Supplement to the Elements of Euclid
common vertex equal to the half of one of the equal hypotenuses; i. e. they will all be at distances from that point, equal to the half of any one of those equal lines: It is manifest, ..., that they will be in the circumference of a circle, described from that point as the centre, at a distance equal to the half of one of the hypotenuses.
41. COR.2. A circle described from the bisection of the hypotenuse of a right-angled triangle as a centre, at the distance of half the hypotenuse, will pass through the summit of the right angle.
42. COR. 3. The vertical angle of a ▲ being a right angle, a point in the base, which is equidistant from the vertex and from either extremity of the base, bisects the base.
Let the point D, in the base BC of the ▲ ABC, having the B a right angle, be equidistant from either extremity, as B, of BC, and from the angular point A: The point D bisects BC.
For, if not, let G be the bisection of BC, and join D, A and E, A: Then, since (hyp.) DA= DB, .. (E. 5. 1.) the DAB DBA: also, since G is the bisection of BC, .. (S. 29. 1.) GA=GB; .. (E. 5. 1.) the GAB GBA; ..the GAB DAB, the greater to the less, which is absurd; .. no other point than D can be the bisection of BC.
PROP. XXX.
43. PROBLEM. Upon a given finite straight line, as a diameter, to describe a square.
Let AB be a given finite straight line: Upon
AB, as a diameter, it is required to describe a
square.
Bisect (E. 10. 1.) AB in E; through E draw (E. 11. 1.) DEC 1 to AB, and make (E. 3. 1.) ED and EC each of them equal to AE or EB: Join A, D, and D, B, and B, C, and C, A: The figure ADBC is a square, having AB for its di
ameter.
For since (constr.) DE=EC, and AE is common to the AED, AEC, and that the right < AED = right ▲ AEC, ... (E. 4. 1.) AD= AC; and in the same manner AD may be shewn to be equal to DB, and DB to BC; ... the figure ADBC is equilateral.
Again, since (constr.) AE= DE, the EAD =EDA (E. 5. 1.); but (constr.) ▲ AED is a right; each of the EAD, EDA, is half a right; and, in the same manner, may each of the EDB, DBE, CBE, BCE, ECA, EAC, be shewn to be half a right ; .. all the of the figure ADBC are right; and it has been proved that all its sides are equal; .. (E. 30. def. 1.) ADBC is a square.
PROP. XXXI.
44. THEOREM. If either of the acute angles of a given right-angled triangle be divided into any number of equal angles, then, of the segments of the base, subtending those equal angles, the nearest to the right angle is the least; and, of the rest, that which is nearer to the right angle is less than that which is more remote.
Let ACB be a right-angled A, right-angled at C,
and let the acute BAC be divided into any num
ber of equal, CAD, DAE, EAB, &c.; then is CD the least of the segments of the base subtending those equal, and of the rest DE< EB; and
Again, at the point E, in AE, make the ▲ AEK = AED; and it may, in like manner, be shewn that EKED: But (E. 16. 1.) 2 BKE > < AEK; .. < BKE > < AED; and ▲ AED > < ABE; much more then is BKE > < EBK; .. (E. 19. 1.) BE >EK or ED; i. e. ED < EB.
And in the same manner may EB be shewn to be less than the next segment that is more remote from C; and so on.
45. COR. It is manifest, from the demonstration, that if any three straight lines AB, AE, AD, be drawn to the given straight line XC from a given point A, without it, so that the BAE= <EAD, the segment BE, of XC, which is the further from the perpendicular AC, shall be greater than the segment ED, which is the nearer to AC.
PROP. XXXII.
46. THEOREM. If either angle at the base of a
triangle be a right angle, and if the base be divided into any number of equal parts, that which is adjacent to the right angle shall subtend the greatest angle at the vertex; and, of the rest, that which is nearer to the right angle shall subtend, at the vertex, a greater angle than that which is more remote.
Let ACB be a right-angled ▲, right-angled at
C, and let the base BC be divided into any number of equal parts CD, DH, HG, &c.: Of these segments DC shall subtend the greatest at the vertex A; and of the rest DH shall subtend, at A, a greater than HG; and so on. | 677.169 | 1 |
Explore 15 Isosceles Triangle Facts You Never Knew
Triangles, not just any triangles, but the isosceles triangles, have fascinated mathematicians since ancient times. From being integral in the design of the Egyptian pyramids built around 2600 BC to featuring prominently in Leonardo Da Vinci's studies in the 15th century, these triangles have an illustrious history.
Before we delve into some intriguing fun facts about isosceles triangles, did you know that the word "triangle" dates back to the late 14th century? It's derived from the Latin word "triangulus"!
1. The Ancient Definition: Tracing the Name "Isosceles"
The term "isosceles" wasn't just pulled out of thin air. It hails from ancient Greece, where the love for math and geometry was unprecedented. Breaking it down: "isos" means equal and "skelos" translates to leg. So, our beloved triangle was aptly named for its two equal legs around the 5th century BC.
By the way, while many ancient civilizations had their take on geometry, it was the Greeks, with legends like Pythagoras and Euclid, who laid down principles we reference even today.
2. Equal Length Drama: The Two Identical Sides Tale
Having two identical sides isn't just a cool feature, it's a defining characteristic of our isosceles triangles. These equal lengths were a topic of great discussion in Euclid's "Elements" penned around 300 BC. In fact, Book I Proposition 5 of "Elements" is dedicated entirely to proving the properties of isosceles triangles, underscoring its importance.
And, speaking of those equal lengths, the largest known isosceles triangle is part of the stars! In the Northern sky, the Summer Triangle's sides Vega, Deneb, and Altair form an almost perfect isosceles.
3. It's All About Angles: The Equal Angles Story
Angles, especially in isosceles triangles, tell their own tale. The base angles, always congruent, are one of the consistent features of this shape. This fact was so pivotal that by 2000 BC, ancient civilizations had drafts showcasing this property.
The consistency in these angles doesn't just give the isosceles its unique shape but has been integral in architectural designs. Structures like the Eiffel Tower use this principle, with congruent angles ensuring stability and symmetry.
Image by vecstock / Freepik
4. From Egypt to Greece: Historical Use in Architecture
Triangles, particularly the isosceles variety, are etched into the annals of human civilization. Let's zip back to around 2580 BC, the era of the Great Pyramid of Giza. This pyramid, with its triangular faces, leans on geometrical principles involving isosceles triangles. Now, a fast forward to Athens around 447 BC: the construction of the Parthenon began. Observing its façade, one can spot the influence of triangles in its Doric columns and pediments.
The wonders of isosceles triangles didn't stop with ancient marvels. Contemporary architects still nod to these early principles, weaving them into modern designs.
5. Nature's Isosceles: Discovering Triangles in the Wild
Nature, in its intrinsic wisdom, embraces geometry, and isosceles triangles are no exception. Ever observed the dorsal fin of certain fish like the sailfish? It's a classic example of an isosceles triangle! Even the wings of some birds, when fully spread, form a close resemblance.
Such designs aren't mere coincidences. They serve evolutionary purposes. For instance, the sailfish, with its fin resembling an isosceles triangle, cuts through water with greater agility. Mother Nature sure knows her geometry!
6. Right Angle Rendezvous: When Isosceles Meets Right Triangle
Here's a fun fact: not all isosceles triangles are the same! Enter the right isosceles triangle—a delightful union of the isosceles and right triangle characteristics. With one right angle and two equal angles, this unique shape has properties from both worlds. Pythagoras would be giddy knowing that his theorem applies perfectly here!
From kites to flags, and even in the realm of computer graphics, the right isosceles triangle has made a significant mark. It balances both symmetry and function seamlessly.
Image: artfile.me
7. The Golden Triangle Connection: Beauty in Geometry
The Golden Ratio, symbolized by the Greek letter φ (phi), approximately equals 1.6180339887… This number has been an object of fascination for artists, architects, and mathematicians alike. It's found in various aspects of nature, art, and architecture. Now, here's the golden connection: the Golden Triangle, a unique isosceles triangle, fits right into this narrative. This triangle, unlike the usual isosceles, has angles measuring 36°, 72°, and 72°, ensuring its sides are in that magical golden proportion.
While the Parthenon or the pyramids might not be direct manifestations of the Golden Triangle, the principle of the Golden Ratio is evident in their proportions. It's a testimony to how geometry, particularly triangles, plays a role in things we deem beautiful.
8. The Lopsided Sibling: Isosceles vs. Scalene Showdown
In the realm of triangles, isosceles and scalene triangles serve as a study in contrasts. Let's dive into the nitty-gritty: The isosceles triangle is defined by two equal sides and two equal angles, lending it a sense of symmetry. Scalene triangles, on the other hand, don't have any sides or angles of the same length or measure. This makes them incredibly versatile in geometry.
In applications, the isosceles triangle's symmetry offers stability, often seen in truss bridges and roof structures. The unpredictable nature of scalene triangles, conversely, makes them indispensable in triangulation methods, crucial for satellite navigation systems and 3D computer graphics. So, while they may appear vastly different, both have carved out essential niches in the world of practical applications.
Image: franklinmslibrary.pbworks.com
9. Equilateral's Close Cousin: Almost Three Equal Sides!
The equilateral triangle, with its three equal sides, might seem like the perfect sibling, but the isosceles isn't far behind. While it might have just two sides of equal length, it shares more similarities with the equilateral than with any other triangle. Both are champions of symmetry in their own right!
This familial connection between the isosceles and equilateral triangles means they often find themselves used interchangeably in various fields, from graphic design to structural engineering.
10. Divide and Conquer: Bisecting the Vertex Angle
When you bisect the vertex angle of an isosceles triangle, you're in for a treat. Not only do you get two equal angles, but you also split the triangle into two congruent right triangles. That's right! Each side of the bisection becomes a height for the original triangle, offering a neat lesson in geometry.
This bisection property plays a pivotal role in the world of optics, particularly in lens designs where light needs to be split or refracted equally.
11. Hitting the Slopes: Inclination and the Isosceles
Ski slopes, ramps, and slides often boast an isosceles profile. Why? Well, the triangle's equal sides ensure that the inclination or
slope is the same on both sides, providing a smooth, predictable descent. This symmetry doesn't just serve thrill-seekers on snowy mountains; it also plays a pivotal role in engineering and design, ensuring stability and safety.
In skate parks, for instance, ramps often use the isosceles design to give skaters a consistent experience, allowing for more controlled tricks and jumps.
Image: offerup.com
12. Trigonometry's Favorite Child: Sine, Cosine, Tangent Galore
Dive deep into the waters of trigonometry, and you'll see the isosceles triangle swimming alongside sine, cosine, and tangent. It's a match made in geometry heaven! The equal angles opposite the equal sides simplify many trigonometric relationships. Remember SOHCAHTOA from math class? Yep, the isosceles makes those ratios come to life.
In applications like signal processing and physics, these trigonometric relations are paramount. They help in predicting wave behaviors or analyzing forces.
13. Area Amazement: Unveiling the Space Within
The area of an isosceles triangle can be a magical revelation. Using the formula ½ x base x height, one can uncover the space it encapsulates. This isn't just a mathematical exercise; architects and designers use this to optimize space in structural designs.
Take the Louvre Pyramid, for instance. Its isosceles triangular faces are a perfect embodiment of area optimization while ensuring an aesthetic appeal.
14. A Staple in Space: NASA's Isosceles Affair
You might not associate triangles with space exploration, but isosceles triangles have been crucial in the design of certain spacecrafts and satellites. Their symmetrical properties offer balanced aerodynamics, ensuring stability during launch and orbit.
15. Bridging the Gap: Isosceles in Modern Infrastructure
Drive through any modern city, and you'll likely cross a bridge or two. Many of these bridges employ isosceles triangles in their trusses and supports. The equal length of the two sides ensures an even distribution of weight and force, making for more stable and resilient structures.
The iconic Golden Gate Bridge in San Francisco, with its repeating triangular patterns, is a testament to this design principle.
Image by wirestock / Freepik
FAQ
What is unique about an isosceles triangle?
An isosceles triangle stands out in the triangular family because of its symmetry. Specifically, it has two sides of equal length, known as the legs, and the angles opposite these equal sides are also congruent. This symmetry not only gives it a distinct shape but also means it has certain consistent properties that other triangles might not have.
What are the 5 properties of an isosceles triangle?
Equal Sides: The most defining property of an isosceles triangle is its two sides of equal length.
Equal Angles: The angles opposite the equal sides (known as the base angles) are also congruent.
Vertex Angle: The angle that is not a base angle, situated between the two equal sides, is referred to as the vertex angle.
Perpendicular Bisector: The altitude drawn from the vertex angle to the base bisects the base into two equal halves and is perpendicular to the base.
Angle Bisector: The angle bisector drawn from the vertex angle bisects the base into two equal segments and divides the triangle into two congruent right triangles.
Why are isosceles triangles important?
Isosceles triangles are paramount in both theoretical and practical aspects. Their symmetrical properties make them easier to analyze and understand in geometry, simplifying many mathematical proofs. In real-world applications, their predictable proportions offer stability, making them a preferred choice in architectural and engineering designs, such as truss bridges and certain roof structures.
What proof does isosceles triangles have?
The most classic proof concerning isosceles triangles is the "Base Angles Theorem" or the "Isosceles Triangle Theorem." It states that if two sides of a triangle are congruent (thus making it isosceles), then the angles opposite those sides are congruent. The converse is also true: if two angles of a triangle are equal, then the sides opposite those angles are congruent. This theorem can be proven using various methods, including transformations, congruent triangles, or algebraic proofs.
What is a real life example of an isosceles triangle?
Real-life examples of isosceles triangles abound. A classic example can be seen in the "A-frame" houses, where the two sides of the roof slope down at the same angle, forming an isosceles triangle. Another instance is the yield traffic sign, which is an upside-down isosceles triangle. The structural stability and aesthetic appeal of the isosceles triangle make it a popular choice in various designs and symbols.
Are all isosceles triangles similar?
No, all isosceles triangles are not similar. While they do share the characteristic of having two congruent sides and angles, the actual measurements of those sides and angles can vary. Similar triangles must have corresponding angles that are congruent and corresponding sides that are proportional. While two isosceles triangles might have congruent base angles, their vertex angles (and thus their overall shape) might differ, preventing them from being similar. | 677.169 | 1 |
SSC JE Mechanical Engineering 2015
In the given figure, the circle stands for intelligent, square for hardworking, triangle for Post graduate and the rectangle for loyal employees. Study the figure and answer the following questions. Employees who are intelligent, hardworking and loyal but not Post graduate are represented by | 677.169 | 1 |
Trigonometric Ratios
The 6 trigonometric ratios can be said to be the sine (sin), the cosine (cos), the tangent (tan), the cotangent (cot), the cosecant (cosec), and the secant (sec). In the geometry, the trigonometry can be said to be a branch of the mathematics that actually deals with the sides as well as the angles of any right-angled triangle. Hence, the trig ratios are actually evaluated in relation to the sides as well as the angles.
The trigonometry ratios in relation to a specific angle 'θ' has been given below:
Trigonometric Ratios
Sin θ
Opposite Side to the θ/Hypotenuse
Cos θ
Adjacent Side to the θ/Hypotenuse
Tan θ
Opposite Side/Adjacent Side and Sin θ/Cos θ
Cot θ
Adjacent Side/Opposite Side and 1/tan θ
Sec θ
Hypotenuse/Adjacent Side and 1/cos θ
Cosec θ
Hypotenuse/Opposite Side and 1/sin θ
Trigonometric Ratios can be said to be defined as the values of every trigonometric function is actually based upon the value in relation to the ratio of the sides in any right-angled triangle.The ratios of the sides of any right-angled triangle in relation to any of its critical or acute angles are known to be the trigonometric ratios of that specific angle.
How to Find Trigonometric Ratios?
Trigonometric ratios are possible to be calculated by taking the specific ratio of any 2 sides of the right-angled triangle. One shall be able to evaluate the 3rd side using the Pythagoras theorem, provided the measure of the other 2 sides. One shall be able to utilize the abbreviated version of the trigonometric ratios in order to compare the length of any 2 sides with the specific angle in the particular base. The angle θ can be said to be an acute angle (θ < 90º) and in general has been measured with reference to positive x-axis, in the anti-clockwise direction. The rudimentary trigonometric ratios formulas have been given below: -
sin θ = the Perpendicular/the Hypotenuse
cos θ = the Base/the Hypotenuse
tan θ = the Perpendicular/the Base
sec θ = the Hypotenuse/the Base
cosec θ = the Hypotenuse/the Perpendicular
cot θ = the Base/the Perpendicular
Therefore, one should first observe the reciprocal trigonometric ratio formulas in relation to the above-mentioned trigonometric ratios. As one actually observes, he or she notices that sin θ is actually a reciprocal of the cosec θ, the cos θ would be a reciprocal of the sec θ, the tan θ can be said to be a reciprocal of the cot θ, as well as vice-versa. Hence, the new set or series of the formulas for the trigonometric ratios is: -
the sin θ = 1/cosec θ
the cos θ = 1/sec θ
the tan θ = 1/cot θ
the cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
Trigonometric Ratios Table
In the particular trigonometric ratios table, one actually utilizes the values of the trigonometric ratios for the purposes of the standard angles 0°, 30°, 45°, 60°, as well as 90º. It can be said to be very easy to actually predict the values in relation to the table as well as to utilize the table as a locus in order to calculate the values in relation to the trigonometric ratios for the numerous other angles, through the utilization of the trigonometric ratio formulas for the existing patterns in the trigonometric ratios as well as amidst the angles. After such process, one shall be able to summarize the value in relation to the trigonometric ratios for the particular angles in the specific table that has been mentioned below: -
Trigonometry Applications
Apart from the astronomy as well as the geography, the trigonometry can be said to be applicable in numerous fields such as the satellite navigation, developing the computer music, the chemistry number theory, the medical imaging, the electronics, the electrical engineering, the civil engineering, the architecture, the mechanical engineering, the oceanography, the seismology, the phonetics, the image compression as well as the game development. It should be noted that there are Trigonometry Applications in the real life. It might not have the direct applications in solving the practical issues but has been utilized in numerous fields. For instance, the trigonometry is actually utilized in developing the computer music: as one would be familiar that sound actually travels in the version of the waves and such wave pattern, with the help of a sine or a cosine function for the development of the computer music. Following can be said to be few applications where the trigonometry as well as its functions can be said to be applicable: -
Solved Example
Example: If the distance from where the building is observed is 90 ft from its base and the angle of elevation to the top of the building is 35°, then find the height of the building.
Solution: Given:
Distance from the building is 90 feet from its base.
The angle of elevation from to the top of the building is 35°.
Now, one should search for the height of the building by recalling the trigonometric formulas. Here, the angle as well as the adjacent side length are actually provided. Therefore, through the utilization of the formula of the tan.
tan35∘=Opposite Side/Adjacent Side
tan 35°= h/90
h = 90 × tan 35°
h = 90 × 0.7002
h = 63.018 ft
Therefore, the height of the specific building is actually 63.018 ft.
Most Popular Questions Searched By Students
What are the three primary trigonometric ratios?
The 3 primary trig ratios can be said to be the sine (sin), the cosine (cos) and the tangent (tan).
What is SOH CAH TOA?
SOH CAH TOA can be said to be a mnemonic device that is actually utilized to remember the particular ratios of the sine, the cosine, and the tangent in the trigonometry | 677.169 | 1 |
Finsler Geometry Intricacies: An Overview
F
The Core of Finsler Geometry
The crux of Finsler geometry is the Finsler metric, a function that attributes a positive real figure to every tangent vector in the tangent space at a point in a differentiable manifold. This metric is instrumental in defining lengths, angles, and areas in the manifold, establishing the foundation of Finslerian concepts and explorations.
Finsler Geometry: A Unique Perspective
The uniqueness of Finsler geometry is its non-restrictive nature compared to others like Riemannian geometry. While Riemannian metrics are required to be quadratic in tangent vectors, Finsler metrics can be any smooth function that is positive-definite and homogeneous of degree one in vectors. This flexibility broadens the modeling of diverse spaces, making Finsler geometry an influential tool in multiple scientific domains.
Exploring Further into Finsler Geometry Intricacies
To fully grasp the depth and intricacy of Finsler geometry, it's crucial to understand its key principles and methodologies. They include:
Geodesics and Distance Functions: Geodesics, which are the shortest paths between two points in a Finsler manifold, are vital in understanding the structure and properties of the space. The distance function, defined using these geodesics, is an integral part of Finsler studies.
Finsler Structures: The Finsler structure, defined by a Finsler function on the tangent bundle of a manifold, offers an intrinsic approach to study the manifold's geometric and topological characteristics.
Curvature in Finsler Geometry: Curvature, a measure of how much a geometric object deviates from being flat, is an essential concept in Finsler geometry. The different types of curvature (such as scalar, sectional, and Ricci curvature) provide crucial insights into the manifold's properties.
Utilizing Finsler Geometry
The usage of Finsler geometry extends to numerous areas. In physics, it models spacetime in general relativity and studies quantum mechanics. In computer graphics, it facilitates the modeling of intricate shapes and textures. In biology, it aids in comprehending the shape and structure of proteins.
Epilogue: Finsler Geometry Intricacies and Their Impact
To wrap up, Finsler geometry provides a comprehensive and flexible framework for investigating diverse geometric phenomena. Its unique blend of adaptability and precision makes it a valuable asset in several scientific and mathematical areas. As research progresses, we anticipate discovering more intriguing applications and insights from this dynamic field of study. | 677.169 | 1 |
Anyone can download RD Sharma's Class 7 Maths, Chapter - Constructions which are available here for free. Solutions of RD Sharma Class 7 Maths, Chapter 17, are developed to help students enhance their proficiency and subject awareness. With the help of specialized tools, the figures with different measurements are drawn and constructed, which is the easiest method of construction. RD Sharma Solutions/Answers of Class 7 Maths, Chapter 17, therefore, provides a comprehensive set of examples from which students may learn and practice. RD Sharma's Class 7 Maths, has provided the Solutions of Chapter 17 - Construction, free of charge so that anyone who has any sort of doubt/query regarding the subject matter and content of the chapter can easily clear it out.
Introduction to RD Sharma's Class 7 Maths Chapter - "Construction" :
RD Sharma Maths Solutions Class 7 Chapter 17 - Construction, provided byVedantumakes students comprehend the basics of geometry, such as - various definitions of certain topics and creation of angles, distances, and other geometric figures with the assistance of tools like a compass and a ruler. By class 7, students are familiar with the basic shapes/figures and it's construction, which makes it much easier for them to score well in their examinations. Maths Solutions offered by RD Sharma are high in demand amongst students of class 7, as sequenced and detailed methods are provided for the construction of geometric figures, which now further has extended in as to how to draw a parallel line and triangles with different measurements.
SAS triangle constructions
ASA triangle constructions and
and RHS triangle constructions.
Preparation Tips to Score Well in the Exams:
Forming study groups in each of your classes might be a beneficial and enjoyable method to prepare for exams, in addition to making friends.
Working together as a team would result in building healthy study habits which would strengthen team spirit and increase trust. Exchange of ideas, approaches, methods of studying and interpreting the concepts would also help anyone in general.
Another technique or tip which is fruitful for students involves rewriting important topics like essential facts, ideas, and meanings on flashcards. Flashcards enhance your memorization skills and comprehension.
Conclusion: Solutions of RD Sharma Class 7 Maths allows students to aim and earn a successful academic score in their exams. These methods are planned and produced by experts to improve students' confidence in learning and understanding the concepts covered in this chapter and also bring forth the strategies to quickly solve the problems. These study materials are composed on the basis of the Class 7 CBSE syllabus, taking into account the various kinds of questions presented in RD Sharma Class 7 Maths Chapter Constructions. The chapter broadly discusses the construction of triangles which have four parts/exercises. Likewise, the expert faculty at Vedantu have provided answers for each chapter to make students grasp the concept well in a simple and easy manner so that they can score well. These solutions provided by Vedantu can be obtained by anyone at any time from the downloaded PDF so that one can get the most out of it and make their learning process more effective.
FAQs on RD Sharma Class 7 Maths Solutions Chapter 17 - Constructions
1. In Architectural Construction, Which Branch of Mathematics is Used?
In architectural construction, geometry, arithmetic, and trigonometry all play key roles. To plan their blueprints or original draught plans, architects apply these mathematical forms. They also measure the possibility of challenges that could be encountered by the development team when they put the concept vision to reality in three dimensions.
2. What is a Congruent Angle?
Congruent angles are at the same angle (in degrees and radians, both are units of measure for angles). As long as the angles are the same, angles are congruent, they do not have to point in the same direction, they do not have to be generated with lines of the same length.
3. What is Meant by Construction?
Construction involves constructing things in the general sense. Yet it has a peculiar significance in geometry. Here, construction is the process of drawing geometric forms using only a compass and a straight line. Precisely draw a shape, line, or angle using a compass and a straight line (ruler). You can also have a protractor and triangle occasionally.
4. How can I prepare well from RD Sharma Solutions Class 7 Maths?
Solutions provided by Vedantu of RD Sharma Class 7 Maths are extremely useful for all the students who wish to score well in their examinations. These solutions are 100% accurate with proper guidelines and steps with which you can answer and attempt any question paper of Maths. Vedantu experts have always been cautious while solving questions and preparing solutions. So, if you follow these notes provided by Vedantu, which are made as per the guidelines of CBSE and even in accordance with NCERT.
5. Is this Chapter Important as per the examination and would these solutions be enough?
Yes, the Construction chapter of RD Sharma is quite important and scoring as well. It is not at all difficult and can guarantee you brilliant marks in examinations. Students of class 7 should always prefer this chapter because as compared to the other chapters, it is easy and scoring, which is a bonus for all of you. Moreover, the solutions provided by Vedantu can boost your answering skills, just follow the instructions step-by-step and practice each and every exercise in the book.
6. What is known as Construction in Simple Words?
Construction generally involves constructing things/objects in general. However, it has a different connotation in geometry. Here, Construction is the process of drawing geometric figures using only tools like a compass and a ruler, according to the measurements. One can also use a protractor and triangle if required.
7. Is the Construction Chapter's Solutions of RD Sharma Scoring or Hard?
Yes, Construction is one of the most scoring chapters in Maths and anyone who possesses a good understanding of the chapter with great clarity can score full marks from this chapter itself. Simple rules and techniques are enough to earn good and satisfactory marks. Hence, this chapter is not at all hard.
8. How should we construct figures which are appropriate and perfect?
A clean paper itself can make your geometrical figures look perfect. Always sharpen your pencils and don't use them harshly as well. Keep checking the measurements, always mark the points, and don't erase the figure again and again because it might tear the page. Students can download free study material from Vedantu mobile app and site for free. | 677.169 | 1 |
KCSE MATHEMATICS QUESTIONS AND SOLUTIONS ~ Topically Analyzed
QUESTION 22 | KCSE 2023 | LOCI | PAPER 2 | FORM 4 LEVEL
In this question, use a ruler and a pair of compasses only. The following figure is drawn to scale. It shows sides AB and AD of a trapezium ABCD in which AB and DC are parallel. Angle DAB = 40° and vertex C is not shown.
KCSE 2020 MATHEMATICS ALT A PAPER 2 QUESTION 9
a) Using a ruler and pair of compasses only construct triangle ABC in which AB = 6.5cm, BC= 5.0cm and angle ABC = 600. Measure AC (3mks)
b) On same side of AB as C (3mks)
i) Determine the locus of a point P such that angle APB = 600 (3mks)
ii) Construct the locus of R such that AR = 3cm. (1mk)
iii) Identify the region T such that AR 3 and APB 600 by shading the unwanted part. (3mks)
Form 4 Mathematics
The figure KLMN below is a scale drawing of a rectangular piece of land of length KL = 80m
(a) On the figure, construct
(i) The locus of a point P which is both equidistant from points L and M It and from lines KL and LM.
(ii) the locus of a point Q such that ∠KQL = 90°.
(b) (i) Shade the region R bounded by the locus of Q and the Locus of points equidistant from KL and LM.
(ii) Find the area of the region R in m². (Take ℼ= 3.142).
Form 4 Mathematics
(a) BCD is a rectangle in which AB = 7.6 cm and AD = 5.2 cm. draw the rectangle and construct the lucus of a point P within the rectangle such that P is equidistant from CB and CD ( 3 marks)
(b) Q is a variable point within the rectangle ABCD drawn in (a) above such that 600 ≤ AQB≤ 900 On the same diagram, construct and show the locus of point Q, by leaving unshaded, the region in which point Q lies
Form 4 Mathematics
Figure ABCD below is a scale drawing representing a square plot of side 80 metres.
On the drawing, construct:
(i) the locus of a point P, such that it is equidistant from AD and BC.
(ii) the locus of a point Q such that <AQB = 60°.
(i) Mark on the drawing the point Q , the intersection of the locus of Q and line AD.
Determine the length of BQ1 in metres.
(ii) Calculate, correct to the nearest m2, the area of the region bounded by the locus of P, the locus of Q and the line BQ1
Using ruler and compasses only, construct a parallelogram ABCD such that AB = 10cm, BC = 7cm and ∠ABC = 1050. Also construct the loci of P and Q within the parallel such that AP ≤ 4 cm, and BC ≤ 6 cm. Calculate the area within the parallelogram and outside the regions bounded by the loci. | 677.169 | 1 |
Geometry Spot: Exploring the Beauty of Shapes
Geometry spot, the branch of mathematics dealing with shapes, sizes, and properties of space, has fascinated humanity for centuries. From the ancient Egyptians' use of geometry in construction to modern architects' reliance on geometric principles, this field remains vital in various domains. In this article, we delve into the intriguing world of geometry, exploring its historical significance, practical applications, and enduring beauty.
Historical Roots:
The origins of geometry can be traced back to ancient civilizations, where practical applications drove its development. The Egyptians used geometry to survey and construct the pyramids with astonishing precision, showcasing early mastery of geometric principles. Similarly, the Greeks, particularly Euclid, formalized geometry into a system of axioms and theorems in his renowned work, "Elements," laying the foundation for modern geometry.
Key Concepts:
Geometry encompasses a vast array of concepts, but some fundamental principles underpin its study:
1. Points, Lines, and Planes: The basic building blocks of geometry, points have no size, lines extend infinitely in both directions, and planes are flat surfaces that extend indefinitely.
2. Angles: Formed by two rays with a common endpoint, angles are fundamental to understanding shapes and their relationships.
3. Polygons: Closed figures formed by connecting line segments, polygons range from triangles to polygons with numerous sides, each possessing unique properties.
4. Circles: Defined as the set of all points equidistant from a central point, circles play a crucial role in geometry, with properties distinct from polygons.
Practical Applications:
Geometry finds applications in various fields, underscoring its practical significance:
1. Architecture: Architects employ geometric principles to design structures that are aesthetically pleasing, structurally sound, and efficient in space utilization. From the symmetry of classical buildings to the innovative forms of modern skyscrapers, geometry shapes architectural marvels.
2. Engineering: Engineers rely on geometry to design infrastructure, machinery, and technological systems. Whether calculating the angles of support beams in a bridge or optimizing the aerodynamics of a vehicle, geometric principles inform engineering decisions.
3. Art and Design: Artists and designers leverage geometry to create visually captivating compositions. From the intricate patterns of Islamic art to the geometric abstraction of modern art movements, geometry serves as a source of inspiration and a means of expression.
Beauty and Symmetry:
Beyond its practical applications, geometry possesses inherent beauty and symmetry that captivate the imagination. The elegant simplicity of geometric forms, from the perfect symmetry of a snowflake to the harmonious proportions of a nautilus shell, inspires awe and wonder. Artists and mathematicians alike have long been drawn to the aesthetic allure of geometric patterns, exploring themes of symmetry, repetition, and proportion in their work.
Challenges and Innovations:
While geometry offers profound insights into the structure of the physical world, it also presents challenges and unanswered questions. Mathematicians continue to explore unsolved problems, such as the elusive proof of the Riemann hypothesis or the classification of finite simple groups, pushing the boundaries of human knowledge and understanding.
Furthermore, advances in technology have opened new frontiers in geometric research, enabling simulations, visualizations, and computational techniques that were once unimaginable. From fractals and chaos theory to the emerging field of computational geometry, contemporary mathematics continues to evolve, fueled by the enduring fascination with shapes and patterns.
Conclusion:
Geometry stands as a testament to the enduring power of human curiosity and ingenuity. From its humble beginnings in ancient civilizations to its pivotal role in modern science and technology, geometry spot continues to shape our understanding of the world around us. Whether unlocking the secrets of the universe or simply appreciating the beauty of a geometric design, the study of shapes and space remains a timeless pursuit, enriching our lives and expanding the boundaries of human knowledge.
So, let us continue to explore the geometry spot, where beauty, utility, and wonder converge in a symphony of shapes and patternsplaying the | 677.169 | 1 |
Sequence Of Transformations Worksheet
Sequence Of Transformations Worksheet - Web sequences of transformations problem 1. Web this scaffolded worksheets helps students solidify the skill of describing a sequence of rigid transformations. Web given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper,. Sequence of transformations problem #2. Web this scaffolded worksheets helps students solidify the skill of describing a sequence of rigid transformations. Web we have classifying and naming transformations, reading protractors and measuring transformations, finding. Web sequence of transformations worksheets. Web sequence of transformations (perform & describe) this 2 page assignment includes a total of 8 practice problems: Web this geometric sequence of transformations maze is just what you need to help students practice their math skills. In earlier grades, college students describe a sequence of inflexible.
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Describe in words a sequence of transformations that maps ∆ABC to ∆A'B
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Sequence Of Transformations Worksheet Worksheet for Education
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31 Sequence Of Transformations Worksheet support worksheet
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In Earlier Grades, College Students Describe A Sequence Of Inflexible.
Web Sequence Of Transformations Examples, Solutions, Videos, And Lessons To Help High School Students When Given A Geometric Figure And A.
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1 Answer
The three basic trigonometric functions are sine, cosine, and tangent. The other three trigonometric functions are cosecant, secant, and cotangent, these are the reciprocals of the three basic trigonometric functions.
The easiest way to evaluate the six trigonometric functions is using a right triangle. Given a right triangle with an interior angle of the triangle, , and the legs opposite and adjacent the angle , we have , , | 677.169 | 1 |
How do you rotate a matrix about the Z-axis?
Description. R = rotz( ang ) creates a 3-by-3 matrix used to rotate a 3-by-1 vector or 3-by-N matrix of vectors around the z-axis by ang degrees. When acting on a matrix, each column of the matrix represents a different vector. For the rotation matrix R and vector v , the rotated vector is given by R*v .
The most general three-dimensional rotation matrix represents a counterclockwise rotation by an angle θ about a fixed axis that lies along the unit vector n. The rotation matrix operates on vectors to produce rotated vectors, while the coordinate axes are held fixed. This is called an active transformation.
What axis rotates around the Z axis?
3) Rotation about the z-axis: In this kind of rotation, the object is rotated parallel to the z-axis (principal axis), where the z coordinate remains unchanged and the rest of the two coordinates x and y only change.
What is the Z axis?
z-axis. noun. a reference axis of a three-dimensional Cartesian coordinate system along which the z- coordinate is measured.
How do you rotate 45 degrees?
If we represent the point (x,y) by the complex number x+iy, then we can rotate it 45 degrees clockwise simply by multiplying by the complex number (1−i)/√2 and then reading off their x and y coordinates.
What will be the equation for z coordinates if an object undergoes 3D rotation around Y axis?
Which of the following equation is correct for the new Y co-ordinate if an object undergoes 3D rotation around z axis? When the matrix form for 3D rotation around z axis is expanded then we get the following equations – Xnew = Xold x cosθ – Yold x sinθ; Ynew = Xold x sinθ + Yold x cosθ; Znew = Zold.
How do you find the Z axis?
The direction of the z-axis is determined by the right-hand rule as illustrated in Figure 2: If you curl the fingers of your right hand around the z-axis in the direction of a 90° counterclockwise rotation from the positive x-axis to the positive y-axis, then your thumb points in the positive direction of the z-axis | 677.169 | 1 |
Check that your answer is reasonable. The hypotenuse is the longest side in a right triangle.
The following figure shows how to use SOHCAHTOA to decide whether to use sine, cosine or tangent in a given problem. Scroll down the page for examples and solutions.
Example:
Calculate the value of cos * θ* in the following triangle.
Solution:
Use Pythagoras' theorem to evaluate the length of PR.
Find Missing Sides
Geometry - Math worksheet - Find the missing length of a triangle using the cosine function
This video shows you how to find the missing length of a triangle using the cosine function. We are given a right triangle and two side lengths and a missing length. We show a right triangle and label the hypotenuse, and two legs the opposite side and adjacent side according to their relation to theta.
Find Missing Angles
Cosine Word Problems
Trigonometric Functions To Find Unknown Sides of Right Triangles,
This video uses information about the length of the hypotenuse of a right triangle as well as a trig function to find the length of a missing side.
Example:
A 60 ft ladder is leaning against a wall so that the top of the ladder makes a 15-degree angle with the wall. How high up the wall does the ladder reach?
How to Use Cosine to Solve a Word Problem?
Example:
A ramp is pulled out of the back of truck. There is a 38 degrees angle between the ramp and the pavement.If the distance from the end of the ramp to to the back of the truck is 10 feet. How long is the ramp?
Step 1: Find the values of the givens
Step 2: Substitute the values into the cosine ratio
Step 3: Solve for the missing side
Step 4: Write the units
An airplane over the Pacific sights an atoll at a 20° angle of depression. If the plane is 435 m above water, how many kilometers is it from a point 435 m directly above the center of the atoll?
To find the height of a pole, a surveyor moves 70 feet away from the base of the pole and then, with a transit 2 feet tall, measures the angle of elevation to the top of the poleto be 30°. What is the height of the pole to the nearest foot?
A ladder 16 feet long makes an angle of 35° with the ground as it leans against a store. To the nearest hundredth, how far up the store does the ladder reach? | 677.169 | 1 |
math4finance
If lmn xyz which congruences are true cpctc check all that apply i.will give you brainiest
3 months ago
Q:
if lmn xyz which congruences are true cpctc check all that apply i.will give you brainiest
Accepted Solution
A:
Answer:A; B; EExplanation:CPCTC stands for "Congruent Parts of Congruent Triangles are Congruent."Using the congruence statement, we match up corresponding pieces based on where they are in the statement:LM corresponds with XY; MN corresponds with YZ; LN corresponds with XZ; L corresponds with X; M corresponds with Y; N corresponds with Z.Using these pieces of information, we have that A, B and E are the only true options. | 677.169 | 1 |
-b) formula, which allows us to express the sine of the difference of two angles in terms of the sines and cosines of those angles. In this article, we will explore the sin(a-b) formula in detail, understand its derivation, and examine its practical applications.
Understanding-b) formula is one such identity that helps us express the sine of the difference of two angles in terms of the sines and cosines of those angles.
The Sin(a-b) Formula
The sin(a-b) formula states that:
sin(a – b) = sin(a)cos(b) – cos(a)sin(b)
This formula allows us to find the sine of the difference of two angles by using the sines and cosines of those angles. It is derived from the sum-to-product identities, which express the sum or difference of two trigonometric functions in terms of their products.
Derivation of the Sin(a-b) Formula
To derive the sin(a-b) formula, we start with the sum-to-product identity for sine:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
Let's substitute x with a and y with -b:
sin(a + (-b)) = sin(a)cos(-b) + cos(a)sin(-b)
Since cosine is an even function (cos(-x) = cos(x)) and sine is an odd function (sin(-x) = -sin(x)), we can simplify the equation:
sin(a – b) = sin(a)cos(b) – cos(a)sin(b)
Thus, we have derived the sin(a-b) formula.
Applications of the Sin(a-b) Formula
The sin(a-b) formula finds applications in various fields, including physics, engineering, and geometry. Let's explore some practical examples where this formula is useful:
1. Vectors and Forces
In physics and engineering, vectors are quantities that have both magnitude and direction. The sin(a-b) formula can be used to find the angle between two vectors or the resultant force of two forces acting at an angle. By applying the sin(a-b) formula, we can calculate the sine of the angle between the vectors or forces and use it to determine their relationship.
2. Trigonometric Equations
Trigonometric equations involve trigonometric functions and are often used to model periodic phenomena. The sin(a-b) formula can be used to simplify and solve such equations by expressing them in terms of sines and cosines. This simplification allows us to find the solutions more easily and efficiently.
3. Geometry and Trigonometry
The sin(a-b) formula is also useful in solving geometric and trigonometric problems. For example, it can be used to find the angles of a triangle when the lengths of its sides are known. By applying the sin(a-b) formula, we can relate the angles to the sides and solve for the unknown angles.
Summary
The sin(a-b) formula is a powerful trigonometric identity that allows us to express the sine of the difference of two angles in terms of the sines and cosines of those angles. It is derived from the sum-to-product identities and finds applications in various fields, including physics, engineering, and geometry. By understanding and applying the sin(a-b) formula, we can solve trigonometric equations, analyze vectors and forces, and solve geometric problems more efficiently.
Q&A
1. What is the sin(a-b) formula?
The sin(a-b) formula states that sin(a – b) = sin(a)cos(b) – cos(a)sin(b). It allows us to find the sine of the difference of two angles by using the sines and cosines of those angles.
2. How is the sin(a-b) formula derived?
The sin(a-b) formula is derived from the sum-to-product identities, which express the sum or difference of two trigonometric functions in terms of their products. By substituting the appropriate values and simplifying, we can derive the sin(a-b) formula.
3. What are some practical applications of the sin(a-b) formula?
The sin(a-b) formula finds applications in physics, engineering, and geometry. It can be used to analyze vectors and forces, solve trigonometric equations, and solve geometric problems involving triangles.
4. How does the sin(a-b) formula simplify trigonometric equations?
The sin(a-b) formula allows us to express trigonometric equations in terms of sines and cosines, which simplifies the equations and makes it easier to find their solutions. By applying the sin(a-b) formula, we can eliminate complex trigonometric functions and reduce the equations to simpler forms.
5. Can the sin(a-b) formula be used to find the angles of a triangle?
Yes, the sin(a-b) formula can be used to find the angles of a triangle when the lengths of its sides are known. By relating the angles to the sides using the sin(a-b) formula, we can solve for the unknown angles | 677.169 | 1 |
??? - 2014 JUNE Q6
6.
A circle with centre O has radius 12.4 cm. A segment of the circle is shown shaded in Fig. 6. The segment is bounded by the arc AB and the chord AB, where the angle AOB is 2.1 radians. Calculate the area of the segment. [4] | 677.169 | 1 |
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Law of reflection:angle of incidence= angle of reflection This formula is only used when the angles are measured from the normal to the ray and not from the ray to the mirror. In this diagram we see that the angle is measured from the ray to the mirrot .The normal measures 90° so we need to subtract 90° from 35° to get the angle of incidence which is 55° . We now that angle of incidence= Angle of reflection hence the answer will be 55°
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if a person moves closer to the plane miror , ,would not size of the imagge become larger ,??????? WHy the statement saying that if a eprson moves near to a plane mirror , the image of the person would emain same in size??? SOF , please correct the answer of the question.. I am finding too many wrong quesions nowadays | 677.169 | 1 |
Types Of Angles
What are the different types of angles? To mathematics, angles are a very important concept, especially when it comes to geometry. They are composed of a vertex and two endpoints. There are many different types of angles. They depend on the place and the direction of the endpoints that are connected to the vertex.
The most common of the types of angles is the right angle. This is the types of angles wherein one of the endpoints is pointing directly upwards from the vertex and the other one is horizontal from it. This angle has a measurement of 90 degrees. When compared to a clock, the right angle forms at 3 o'clock, at 6:15, at 6:45, and many others.
An acute angle is among the types of angles that have a measurement of below 90 degrees. If you look at a clock, you can see an acute angle when the time is 3:10, 2:55, 10:40, and many others.
Obtuse angles, on the other hand, are types of angles that have measurements of more than 90 degrees but less than 180 degrees. When you look at the clock, you can see an obtuse angle when it is 3:55, 6:10, 8:30, and many others.
Straight angles are, well, angles that form a straight line. They measure exactly 180 degrees. When you look at a clock, you can see a straight angle when it is 6:00, 3:45, 10:15, 4:50, and many others.
Reflex angles are types of angles that form a measurement of more than 180 degrees but are not as big as 360 degrees. In clocks, reflex angles can be seen when it is 3:35, 1:25, 6:50, and many others.
Next, we have adjacent angles. These are angles that are composed of two angles that have a common vertex. This common vertex is called the adjacent angle. Unlike the many other types of angles, the adjacent angle cannot easily be seen on the clock. You can see it, however, when a clock has an arm for seconds. In this case, an example of an adjacent angle would be 3:50 where the arm for seconds can be found somewhere in the 10 seconds mark. The common arm would be the seconds arm and the vertex will be the point where the three arms meet.
Complementary angles are types of angles that are composed of two angles that have the measurement sum of 90 degrees. They can be adjacent angles sometimes and sometimes they are not. Much like adjacent angles you can only see complementary angles when there is an arm for seconds on the clock. In these cases, a good example would be 6:15 where the seconds arm is somewhere around the 20 to 25 seconds marks.
Lastly, we have the supplementary angles. These are types of angles, two angles that have the sum measurement of 180 degrees. They can also be adjacent angles. Supplementary angles can be made up of a straight angle and an acute angle. For instance, when you look at a clock, you can see this at 3:45 with the seconds arm somewhere between the 45 second and the 15 second mark. | 677.169 | 1 |
There are 12 problems total. For each problem, the identify how the triangles are congruent and the valid congruency statements. This will match the numbers with the colors in order to color the dinosaur an excellent resource. I used it as a review activity to help the students consolidate their understanding on congruent triangles, particularly on how to name triangles correctly. My class were engaged with the task and enjoyed the challenge. Thank you.
—SUZANNE P.
My students always love the activities from All Things Algebra, as it breaks up the monotony of just working problem after problem. Great way to review for quizzes or tests, or offer extra credit assignments.
—MAE MARGARET D.
This was a great review for congruent triangles. I like how it required determining the congruency rule and also the congruence statement for the triangles. I always enjoy resources by All Things Algebra | 677.169 | 1 |
Math Formula Sheet on Co-Ordinate Geometry
All grade math formula sheet on co-ordinate geometry. These math formula charts can be used by 10th grade, 11th grade, 12th grade and college grade students to solve co-ordinate geometry.
Co-Ordinate Geometry
● Rectangular Cartesian Co-ordinates:
(i) If the pole and initial line of the polar system coincides respectively with the origin and positive x-axis of the Cartesian system and (x, y), (r, θ) be the Cartesian and polar co-ordinates respectively of a point P on the plane then,
x = r cos θ, y = r sin θ
and r = √(x2 + y2), θ = tan-1(y/x).
(ii) The distance between two given points P (x1, y1) and Q (x2, y2) is
PQ = √{(x2 - x1)2 + (y2 - y1)2}.
(iii) Let P (x1, y1) and Q (x2, y2) be two given points.
(a) If the point R divides the line-segment PQ internally in the ratio m : n, then the co-ordinates of R
are {(mx2 + nx1)/(m + n) , (my2 + ny1)/(m + n)}.
(b) If the point R divides the line-segment PQ externally in the ratio m : n, then the co-ordinates of R are
{(mx2 - nx1)/(m - n), (my2 - ny1)/(m - n)}.
(c) If R is the mid-point of the line-segment PQ, then the co-ordinates of R are {(x1 + x2)/2, (y1 + y2)/2}.
(iv) The co-ordinates of the centroid of the triangle formed by joining the points (x1, y1) , (x2, y2) and (x3, y3) are
({x1 + x2 + x3}/3 , {y1 + y2 + y3}/3
(v) The area of a triangle formed by joining the points (x1, y1), (x2, y2) and (x3, y3) is
½ | y1 (x2 - x3) + y2 (x3 - x1) + y3 (x1 - x2) | sq. units
or, ½ | x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2) | sq. units.
● Straight Line:
(i) The slope or gradient of a straight line is the trigonometric tangent of the angle θ which the line makes with the positive directive of x-axis.
(ii) The slope of x-axis or of a line parallel to x-axis is zero.
(iii) The slope of y-axis or of a line parallel to y-axis is undefined.
(iv) The slope of the line joining the points (x1, y1) and (x2, y2) is
m = (y2 - y1)/(x2 - x1).
(v) The equation of x-axis is y = 0 and the equation of a line parallel to x-axis is y = b.
(vi) The equation of y-axis is x = 0 and the equation of a line parallel to y-axis is x = a.
(vii) The equation of a straight line in
(a) slope-intercept form: y = mx + c where m is the slope of the line and c is its y-intercept;
(b) point-slope form: y - y1 = m (x - x1) where m is the slope of the line and (x1 , y1) is a given point on the line;
(c) symmetrical form: (x - x1)/cos θ = (y - y1)/sin θ = r, where θ is the inclination of the line, (x1, y1) is a given point on the line and r is the distance between the points (x, y) and (x1, y1);
(f) normal form: x cos α + y sin α = p where p is the perpendicular distance of the line from the origin and α is the angle which the perpendicular line makes with the positive direction of the x-axis.
(g) general form: ax + by + c = 0 where a, b, c are constants and a, b are not both zero.
(vi) The point (x1, y1) lies outside, on or inside the parabola y2 = 4ax according as y12 = 4ax1 >, = or,<0
● Ellipse:
(i) Standard equation of ellipse is
x2/a2 + y2/b2 = 1 ……….(1)
(a) Its centre is the origin and major and minor axes are along x and y-axes respectively ; length of major axis = 2a and that of minor axis = 2b and eccentricity = e = √[1 – (b2/a2)]
(b) If S and S' be the two foci and P (x, y) any point on it then SP = a - ex, S'P = a + ex and SP + S'P = 2a.
(c) The point (x1, y1) lies outside, on or inside the ellipse (1) according as x12/a2 + y12/b2 - 1 > , = or < 0.
(d) The parametric equations of the ellipse (1) are x = a cos θ, y = b sin θ where θ is the eccentric angle of the point P (x, y) on the ellipse (1) ; (a cos θ, b sin θ) are called the parametric co-ordinates of P.
(e) The equation of auxiliary circle of the ellipse (1) is x2 + y2 = a2.
(ii) Other forms of the equations of ellipse:
(a) x2/a2 + y2/b2 = 1. Its centre is at the origin and the major and minor axes are along y and x-axes respectively.
(b) [(x - α)2]/a2 + [(y - β)2]/b2 = 1.
The centre of this ellipse is at (α, β) and the major and minor ones are parallel to x-axis and y-axis respectively.
● Hyperbola:
(i) Standard equation of hyperbola is x2/a2 - y2/b2 = 1 . . . (1)
(a) Its centre is the origin and transverse and conjugate axes are along x and y-axes respectively ; its length of transverse axis = 2a and that of conjugate axis = 2b and eccentricity = e = √[1 + (b2/a2)].
(b) If S and S' be the two foci and P (x, y) any point on it then SP = ex - a, S'P = ex + a and S'P - SP = 2a | 677.169 | 1 |
In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $\frac{BC}{3},$ then find $\frac{AC}{BC}.$ | 677.169 | 1 |
Advanced algebra w trig name trig identities review date period simplify each expression. A find an expression for the perimeter of the triangle in terms of r giving your answer in its simplest form. 4 5 4 35 | 677.169 | 1 |
How Do You Graph a Linear Equation by Making a Table?
Note:
Graphing a function? It would be really helpful if you had a table of values that fit your equation. You could plot those values on a coordinate plane and connect the point to make your graph. See it all in this tutorial!
Ordered pairs are a fundamental part of graphing. Ordered pairs make up functions on a graph, and very often, you need to plot ordered pairs in order to see what the graph of a function looks like. This tutorial will introduce you to ordered pairsObtuse angles are those which are greater than 90° but less than 180°, that is, 90° < ϴ < 180°.
Acute Angles
Acute angles are angles which are greater than 0° but less than 90°, that is, 0° < ϴ < 90°.
Reflex Angles
Reflex angles are angles which are greater than 180° but less than 360°, that is, 180° < ϴ < 360°.
Adjacent Angles
Two angles which share the same vertex (centre, usually represented by 0) and have a common side (line) are called adjacent angles.
Complementary Angles
Complementary angles are two angles which when summed equals 90°.
Note: <A and <B, are 'angle A' and 'angle B' respectively.
Supplementary Angles
Supplementary angles are two angles which when summed equals 180°.
Vertically Opposite Angles
Vertically opposite angles are the angles opposite to each other when two straight lines intersect. Their defining property is that, vertically opposite angles are equal in magnitude.
Corresponding Angles
When two parallel lines are crossed by a line called the transversal, the angles formed which are in corresponding positions, are called corresponding angles. Corresponding angles are equal in magnitude.
EXAMPLES: Calculate the size of the marked angle in the diagram below
560 950
470 x 950 780 x
POLYGON: A part from triangle and quadrilateral, we have other polygons which are also named according to the number of sides they have.
Examples are pentagon = 5, Hexagon = 6. It can be regular or irregular. A polygon is said to be regular when all the sides and angles are equal. An irregular polygon has neither of the sides or angles equal.
TYPES OF POLYGON (Regular Polygons)
Regular polygons have all sides, and all angles equal.
Size of Internal Angles
To find the size of the internal angles of a regular polygon with 'n' sides, use the formula:
For example, the size of the interior angles of the pentagon (five sides) above is:
The sum of all the interior angles of a polygon with 'n' sides is found using the formula:
(n – 2) x 180°
Therefore, the sum of all the interior angles of the pentagon above is:
(5 – 2) x 180° = 3 x 180° = 540°
Size of Exterior Angles
Interior and Exterior angles are measured on the same line, that is, they add up to 180°.
Therefore, the size of an exterior angle = 180° – Interior angle.
For example, the size of the external angle of the pentagon above is:
Since, interior angle = 108°
Then, exterior angle = 180° – Interior angle
180° – 108° = 72°
Below is a list of the names and the number of sides, of some of the most popular polygons.
What is angle of elevation?
The angle of elevation is the angle between a horizontal line from the observer and the line of sight to an object that is above the horizontal line.
In the diagram below, AB is the horizontal line. q is the angle of elevation from the observer at A to the object at C.
A B
What is angle of depression?
The angle of depression is the angle between a horizontal line from the observer and the line of sight to an object that is below the horizontal line.
In the diagram below, PQ is the horizontal line. q is the angle of depression from the observer at P to the object at R.
How to solve word problems that involves angle of elevation or depression?
Step 1: Draw a sketch of the situation.
Step 2: Mark in the given angle of elevation or depression.
Step 3: Use trigonometry to find the required missing length
Example:
Two poles on horizontal ground are 60 m apart. The shorter pole is 3 m high. The angle of depression of the top of the shorter pole from the top of the longer pole is 20˚. Sketch a diagram to represent the situation.
Solution:
Step 1: Draw two vertical lines to represent the shorter pole and the longer pole.
Step 2: Draw a line from the top of the longer pole to the top of the shorter pole. (This is the line of sight).
Step 3: Draw a horizontal line to the top of the pole and mark in the angle of depression.
Example:
A man who is 2 m tall stands on horizontal ground 30 m from a tree. The angle of elevation of the top of the tree from his eyes is 28˚. Estimate the height of the tree.
Solution:
Let the height of the tree be h. Sketch a diagram to represent the situation.
Average is a single value used to represent a set of numbers (i.e all value in as et data)
The most common ly used statistics is average.
MEDIAN = THE NUMBER AT THE MIDDLE AFTER THE ARRANGEMENT OF THE DATA.
MODE IS THE VALUE THAT OCCURS MOST FREQUENTLY.
EXAMPLES: Calculate the mean, media and mode of the following data
45, 50, 55, 54, 48, 53, 50, 55
38, 35, 36, 30.8, 34.7, 37.9, 33.1
3, 0,4,7, 0, 5, 3, 4, 0, 3, 6, 5, 5 ,4, 6, 5
Solution:
Mean = 45+50+ 55+ 54+ 48+ 53+ 50+ 55
8
= 410/8
= 51.25
Median = 45, 48, 50, 50, 53, 54, 55,55
50 + 53
2
= 52
Mode = 50 and 55
Average
The average or mean of a set of numbers is defined by the formula:
Example
Bar Charts
A Bar chart is a series of rectangular bars of the same width, drawn vertically or horizontally, with an equal space between them, with the height of each bar being a depiction of the data it is representing.
Example
The table below lists several models of Blackberry cellular phones and the amount of each that an electronic store has in stock. Draw a vertical and horizontal bar chart to represent the data.
Blackberry Phones
Stock Amount
Blackberry Curve 8310
75
Blackberry Curve 8320
100
Blackberry Pearl 8100
50
Blackberry Bold 9650
200
Blackberry Bold 9000
150
Blackberry Curve 8520
125
Pie Charts
A Pie chart is a circular diagram divided into sectors, with the size of each sector representing the magnitude of data it is depicting. Each sector of a pie chart can either be displayed in percentages (note all sectors must add up to 100%) or as an angle (note all sectors must add up to 360o). [mediator_tech]
Example
The table below lists some of the most popular football clubs and the number of students at a given institution that supports each. Use a Pie chart to represent the information given in the table.
Football Clubs
Number of Students
Chelsea
50
Manchester United
200
Barcelona
350
Real Madrid
150
Inter Milan
25
Arsenal
100
Liverpool
40
AC Milan
75
[mediator_tech]
The Pie Chart above depicts each sector as percentages. To calculate the percentages for each sector use the formula below:
% of a sector = Number of students x 100
Total number of students
So, to calculate the percentage of Chelsea fans:
% of Chelsea fans = 50 x 100
990
% of Chelsea fans = 5%
For Pie charts which depicts each sector as angles, the angles for each sector is found using the formula below:
Angle of a sector = Number of students x 360
Total number of students
So, to calculate the angle of the Chelsea sector:
Angle of Chelsea sector = 50 x 360
990
Angle of Chelsea sector = 18o
Note: In most cases the questions set on Pie charts require those drawn depicting sectors in percentages.
Line Graphs
Line graphs are mostly used in depicting trends, and as such, values are in most cases plotted against time. A line graph is drawn by connecting a line to consecutive values, with a circle/point made at each value being depicted.
Example
The table below lists the amount of Toyota motor vehicles produced in the month of April over the period 2000- 2010.
Year
Number of Toyota Motor Vehicles Produced
2000
220,382
2001
260,879
2002
213,546
2003
238,890
2004
227,678
2005
245,376
2006
240,224
2007
224,100
2008
258,100
2009
248,024
2010
249,123
WEEK 7 REVIEW OF FIRST HALF TERM'S
WEEK 8AND 9 PROBABILITY
PROBABILITY
Probability is the measure of the likelihood that an event will occur. Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty). The higher the probability of an event, the more certain we are that the event will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is unbiased, the two outcomes ("head" and "tail") are equally probable; the probability of "head" equals the probability of "tail." Since no other outcome is possible, the probability is 1/2 (or 50%) of either "head" or "tail". In other words, the probability of "head" is 1 out of 2 outcomes and the probability of "tail" is also, 1 out of 2 outcomes.
The probability of an eventA is written as , , or .[24] This mathematical definition of probability can extend to infinite sample spaces, and even uncountable sample spaces, using the concept of a measure.
The opposite or complement of an event A is the event [not A] (that is, the event of A not occurring), often denoted as , or ; its probability is given by P(not A) = 1 − P(A). As an example, the chance of not rolling a six on a six-sided die is 1 – (chance of rolling a six) . If two events A and B occur on a single performance of an experiment, this is called the intersection or of A and B, denoted as .
Independent events
for example, if two coins are flipped the chance of both being heads is .[26]
Mutually exclusive events
If either event A or event B occurs on a single performance of an experiment this is called the union of the events A and B denoted as . If two events are mutually exclusive then the probability of either occurring is
Not mutually exclusive events
If the events are not mutually exclusive then
For example, when drawing a single card at random from a regular deck of cards, the chance of getting a heart or a face card (J,Q,K) (or one that is both) is , because of the 52 cards of a deck 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards" but should only be counted once.
Event
Probability
A
not A
A or B
A and B
A given B
[mediator_tech]
PROBABILITY SCALE AND TERMS
EVENT: An event is something that happens. Example tossing a coin or throwing a dice is an event.
OUTCOME: An outcome is the result of an event. Example if you toss a coin, you will either get a Head or Tail. This means there are 2 possible outcomes.
IMPOOSSIBLE: An event that is impossible will definitely happen is given a probability of 0.
UNLIKELY: When the probability tends towards 0, then there is less chances that an event will happen.
LIKELY: When the probability tends towards 1, then there is a likely chance that is 50-50
Example 2: There are 7 red balls, 8 white balls and 5 blue balls in a box. Find the probability thst the ball is
White
Red
Blue or red
Neither red nor white
green
Solution:
Total number of balls = 7+ 8+ 5
= 20
White ball = 8, pro. Of selecting a white ball is = 8/20 = 2/5
Number of red balls = 7
Pro. Of selecting a red balls = 7
7/20
Number of blue and red balls = 5+ 7 = 12
Pro. Of selecting a blue or red ball = 12/ 20
3/5
If the ball is neither red nor white, then it must be blue. Pro. Of selecting a blue ball = 5/20 = ¼
There are no green balls therefore the pro. Of green is 0
Example 3: A card is selected from a well shuffled standard pack of 52 cards. What is the probability of getting,
A diamond
A queen
An ace
A red card
The ace of spades
Any card other than an ace
NOTE: A PACK OF CARDS ARE IN 4 SUITS. EACH SUIT HAS 13 DIAMONDS, 13 HEARTS, 13 SPADES, 13 CLUBS. THE DIAMOND AND THE HEART ARE BOTH RED WHILE THE CLUB AND THE SPADE ARE BLUE. THE SIZE OF NUMBERS ON THE CARD ARE: A 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. | 677.169 | 1 |
2
§4.2: Reference Angles and Unit Circle Reference angle is a positive acute angle formed by the terminal side of ө and the 𝒙-axis. They are viewed as linear pairs. (Think: REFER's back to the 𝒙-axis) The main angle should be within 0 and 360 degrees. No reference trigonometric values of measure are greater than 𝟗𝟎° or less than to 𝟎° Quadrant I: 𝜽 Quadrant II: 𝟏𝟖𝟎°−𝜽 Quadrant III: 𝜽−𝟏𝟖𝟎° Quadrant IV: 𝟑𝟔𝟎°−𝛉 11/10/ :27 AM §4.2: Reference Angles and Unit Circle | 677.169 | 1 |
...The radius of a circle is a straight line drawn from the centre to the circumference, as OB. 4. The diameter of a circle is a straight line drawn through the centre, and terminated both ways at the circumference, as C 0 A. 5. A chord is a straight line joining any two points of the circumference,...-cen<re, diameter meets the curve are called the vertices...
...circumference, are equal to one another : xvi. And this point is called the centre of the circle. xvn. A diameter of a circle is a straight line drawn through...centre, and terminated both ways by the circumference. xvin. A semicircle is the figure contained by a diameter and the part of the circumference it cut off....
...circumference, are equal to one another.circumference, are equal to one another. XVI. And this point is called the centre of the circle. XVII. A diameter of a circle is a straight line drawn through the centre, and terminated both wa\s XVIII. A semicircle is the figure contained by a, dia-meter and the part of the circumference...
...circumference, are equal to one another. XVI. And this point is called the centre of the circle. XVII. A diameter of a circle is a straight line drawn through the centre, and terminated both way s XVIII. A semicircle is the figure contained by a diameter and the part of the circumference cut...
...circumference of a circle : therefore all radii of the same or equal circles are equal to one another. 16. A diameter of a circle is a straight line drawn through the centre to the circumference on either side ; — of a parallelogram is that straight line which joins opposite...
...circumference, are equal to one another : —, and the part of the circumference cut off by... | 677.169 | 1 |
The 22 worksheets in this booklet provide practice in solving geometry problems or riders simple They focus on Grade 8 geometry content and include solutions for each question The pack is called TRY ngles because we know that geometry is difficult to learn and to teach 3 filtered results 8th grade Angles Show interactive only Sort by Parallel Lines Cut By A Transversal Worksheet Transversals of Parallel Lines Angle Relationships Worksheet Transversals of Parallel Lines Worksheet 1 Browse Printable 8th Grade Angle Worksheets Award winning educational materials designed to help kids succeed
An isosceles triangle has two equal sides and the angles opposite the equal sides are equal A scalene triangle has no sides equal A right angled triangle has a right angle 90 An obtuse triangle has one obtuse angle between 90 and 180 An acute triangle has three acute angles 90 MEMORANDUM DAY 1 Activity 1 2 Grade 8 Angles Find the measure of one angle of the following polygon 135 1080 1440 180 Grade 8 Angles CCSS 8 G A 5 Lines m and k are parallel with angle 7 measuring 100 degrees Find the measure of angle 3 in degrees
Grade 8 Math Angles Worksheets Pdf - The 22 worksheets in this booklet provide practice in solving geometry problems or riders simple They focus on Grade 8 geometry content and include solutions for each question The pack is called TRY ngles because we know that geometry is difficult to learn and to teach | 677.169 | 1 |
Square ABCD has side length 1 unit. Points E and F are on sides AB and CB, respectively, with AE = CF. When the square is folded along the lines DE and DF, sides AD and CD coincide and lie on diagonal BD. The length of segment AE can be expressed in the form \(\sqrt k-m\) units. What is the integer value of k + m? | 677.169 | 1 |
Identify and plot points in the four quadrants of a Cartesian plane using integral ordered pairs. Students who have achieved this outcome should be able to: A. Label the axes of a four quadrant Cartesian plane and identify the origin. B. Identify the location of a given point in any quadrant of a Cartesian plane using an integral ordered pair. C. Plot the point corresponding to a given integral ordered pair on a Cartesian plane with units of 1, 2, 5 or 10 on its axes. D. Draw shapes and designs, using given integral ordered pairs, in a Cartesian plane. E. Create shapes and designs, and identify the points used to produce the shapes and designs in any quadrant of a Cartesian plane. | 677.169 | 1 |
Tamara was worried about her six-year-old dog, Suze, so she
took her to three different veterinarians. Each vet told her the
approximate age of her dog in dog years based on Suze's
health. Tamara graphed her dog's age in dog years and
included the three vets' approximations at age 6.
Do
you
think this graph is a function?
Related Questions
Answers
What does it mean by "interpret its meaning"? I underlined it for you guys.
Answers
Like what is it's meaning in that problem.
Answer:
1: 300 miles/hour; 2: 5$/hour; 3: 10$ sales price/20$ retail price.
Step-by-step explanation:
1: You have to find the amount of distance traveled per hour.
Answer: 300 miles per hour.
2. You have to find the amount of money earned per hour.
Answer: 5$ per hour.
3: You have to find the sales price per retail price.
Answer: 10$ of sales price per 20$ of retail price.
Find the perimeter of $\triangle CDE$ . Round your answer to the nearest hundredth. A trapezoid A B C D is plotted on a coordinate plane. A D represents the longer base and B C represents the shorter base. Vertex A lies at ordered pair negative 5 comma 4. Vertex B lies at ordered pair 0 comma 3. Vertex C lies at ordered pair 4 comma negative 1. Vertex D lies at ordered pair 4 comma negative 5. A line is drawn from vertex B and intersects line A D at point F plotted at ordered pair negative 2 comma 1. A line is drawn from vertex C and intersects the line A D at point E plotted at ordered pair 2 comma negative 3. The perimeter is about units.
Answers
Answer:
The answer is below
Step-by-step explanation:
We are asked to find the perimeter of triangle CDE. The perimeter of a shape is simply the sum of all its sides, hence:
Perimeter of tiangle CDE = |CD| + |DE| + |CE|
Given that C(4, -1), D(4, -5), E(2, -3).
The distance between two points [tex]X(x_1,y_1)\ and\ Y(x_2,y_2)[/tex] is given as:
A conveyor belt moves at a constant rate of 12 feet in 3 seconds. A second conveyor belt moves at a constant rate of 21 in 7 seconds. Does this situation represent a proportional relationship?
Answers
Answer:
No.
Step-by-step explanation: If we look at the first conveyor belt, the rate is 12ft in 3 seconds, or 4 feet a second. The second conveyor belt moves at 21 feet in 7 seconds, or 3 feet a second. Since the two values are unequal, we can conclude it is not a proportional relationship.
the value of y varies directly with x when y=24, x=2 . what is the value of y when x is 7?
Answers
Answer:
84
Step-by-step explanation:
y∞x
y=kx (where k is the constant)
24=2k
24/2=k
12=k
therefore the relationship between k,y and x is y=12x
y=12×7
y=84
Can you please help me? I will give brainliest, I appreciate it :) Thanks!
Answers
Answer:
I going to say this should be easy because there must be a f(x) in a function so I think the answer is A.
Step-by-step explanation:
I think that functions must have f(x) in the equation.
a game consists of placing seeds in 24 cups. a player puts two seeds in the first cup, four in the second, six in the third and so on. how many seeds will be required altogether if the player continues to the 24th cup in this way?
Answers
Step-by-step explanation:
The next cup will always have 2 more than the previous one. Continue the pattern of adding two seeds to each cup until you reach the 24th cup like so:
#s on left side = # of seed , #s on right side = # cup
6 | 3 8 | 4 10 | 5 12 | 6 14 | 7 16 | 8 18 | 9 20 | 10 22 | 11
24 | 12 26 | 13 28 | 14 30 | 15 32 | 16 34 | 17 36 | 18 38 | 19
40 | 20 42 | 21 44 | 22 46 | 23 48 | 24
What you want to know is the number of TOTAL SEEDS NEEDED TO PLAY THIS GAME. NOT how many seeds will be in the 24th cup. To find the TOTAL you need to add 2 + 4 + 6 and so on like so:
notice there is a patter of 7 being used? there are probably a lot of other patterns that would also work.. but. if your class is working on multiplying by 7 , then this is probably the right pattern.
from the above pattern, then,
x=4x7 when y=56
x=28
Min read his book for 1/3 hour before dinner. He read his book for 1 1/4 hours after dinner. Altogether, how many hours did min read his book?
Answers
Answer:
1 7/12
Step-by-step explanation:
The answer is 1/3+1 1/4 and if you make them common fractions it is 4/12+1 3/12=1 7/12
Answer:37/12
Step-by-step explanation:
1/3+11/4=37/12
HELPPP 16 POINTS. It was estimated that 325 people would attend game night, but 315 people actually attended.
What is the percent error, to the nearest percent, of the estimate? Enter the answer in the box.
Answers
Answer:
3%
Step-by-step explanation: 315/325 as a percent would be 96.92%. After finding that, we would do 100 - 96.92, yielding 3.08, 3.08 to the nearest percent would result in 3%
Answer:
the answer to the nearest percent is 3%
Find the vertex and the axis of symmetry of the graph of y=-6(x+4)^2 – 3.
Answers
Answer:
vertex = (- 4, - 3 ), equation is x = - 4
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
y = 6(x + 4)² - 3 ← is in vertex form
with (h, k ) = (- 4, 3 ) ← vertex
The equation of the axis of symmetry is a vertical line passing through the vertex, with equation
Given :A line passes through a point ( 3 , -2 ) and the slope of the line is ⅓ .
ToFind:The equation of that line .
Solution:Here we are provided with a point ( 3 , -2 ) and slope of the line which is ⅓ . So clearly here to represent the line we will use point - slope form , which is ;
[tex]\large\underline{\boxed{\red{\bf y - b = m ( x - a ) }}}[/tex]
Where ,
a is x - coordinate.b is y - coordinate .m is the slope of the line .
Here ,
a = 3 b = -2m = ⅓ .
Now , put the respective values ;
[tex]\boxed{\purple{\sf y - 2 = \dfrac{1}{3}(x-3)}}[/tex]
Find the value of x.
Answers
in this problem x equals 10
Step-by-step explanation:
jd
Answer:
13
Step-by-step explanation:
Pls help dont lie
Or else I get wrecked
Answers
Answer:
19.20
Step-by-step explanation:
P.S im not a genius
40 POINTS FOR AWNSER
Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2, and the profit on every wrap is $3. Sal made a profit of $1,470 from lunch specials last month. The equation 2x + 3y = 1,470 represents Sal's profits last month, where x is the number of sandwich lunch specials sold and y is the number of wrap lunch specials sold.
Change the equation to slope-intercept form. Identify the slope and y-intercept of the equation.
Describe how you would graph this line using the slope-intercept method. Be sure to write using complete sentences.
Write the equation in function notation. Explain what the graph of the function represents. Be sure to use complete sentences.
Graph the function. On the graph, make sure to label the intercepts. You may graph your equation by hand on a piece of paper and scan your work or you may use graphing technology.
Suppose Sal's total profit on lunch specials for the next month is $1,593. The profit amounts are the same: $2 for each sandwich and $3 for each wrap. In a paragraph of at least three complete sentences, explain how the graphs of the functions for the two months are similar and how they are different.
Answers
The answer is 247=x-73y
For a given recipe, 12 cups of flour are mixed with 24 cups of sugar. How many cups of sugar should be used if 21 cups of flour are used?
Assuming a constant ratio, fill out the table of equivalent ratios until you have found the value of x. PLEASE ANSWER THIS QUICK
Answers
Answer:
33 cups of sugar should be used
Step-by-step explanation:
subtract 12 from 21 then add that answer to 24 and you get 33 so 33 cups of sugar should be used
42 cups of sugar should be used if 21 cups of flour are used
What is Ratio?
A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0.
Given,
For a given recipe, 12 cups of flour are mixed with 24 cups of sugar
We need to find cups of sugar should be used if 21 cups of flour are used
Let us consider x as cups of sugar should be used if 21 cups of flour are used
Let us form an equation to find x
12/24=21/x
Apply cross multiplication
12x=21×24
12x=504
Divide both sides by 12
x=504/12
x=42
Hence, 42 cups of sugar should be used if 21 cups of flour are used
To learn more on Ratios click:
#SPJ2
Use the given line and the point not on the line to
answer the question,
What is the point on the line perpendicular to the given
line, passing through the given point that is also on the
y-axis?
O (-3.6,0)
O (-2, 0)
O (0, -3,6)
O (0, -2)
Answers
Answer:
It's the third one, (0,-3.6)
Step-by-step explanation:
who does gymnastics?
answer if you do only
Answers
ME!!! I have to because of volleyball and track practicing
PLEASE HELP ASAP!! I'LL GIVE YOU BRAINLIEST!!
Answers
Answer:
-5
Step-by-step explanation:
This is the correct answer, I messed up at first so I edited it. (my bad)
i need help can yall help
Answers
Answer:
360"
Step-by-step explanation:
Firstly, you need to be able to convert your feet to inches. Remember, one foot is equal to twelve inches. You multiply six inches by five feet and later multiply by twelve to convert to inches.
i believe she can cut 360 6 inch strips from all of the rolls
Because:
the 5 rolls of ribbon each have six feet of ribbon on each roll, times the amount of inches in a foot
5(rolls of ribbon)*6(feet on each roll)*12(inches in a foot) = 360 (6 inch strips she can cut)
PLEASE give brainliest if correct! have a nice day :-)
Please help me with this problem
Answers
Answer: D) [tex]m \angle 2 = m\angle 5[/tex]
================================================
Explanation:
Let's go through the answer choices to see what's true and what's false.
A) This is true because they are corresponding angles. Corresponding angles are congruent when we have parallel lines.B) This is true as well. The reasoning is the same as choice A.C) These angles are vertical, so they are congruent angles. This is true.D) This is false. The angles are supplementary. They would only be congruent if both angles were 90 degrees; otherwise, they aren't congruent.
bob claims that he can map rectangle 1 to rectangle 2 select all of the transformations or series of transformation that support his claim
Answers
Answer:
The first thing you need to notice is that each mark in the coordinate axis represents 2 units.
Now, let's analyze each option that maps rectangle 1 to rectangle 2.
Reflection over the line x = 1, and a translation of 2 units down.
You can see that the right end of rectangle 1 us located at x = -2
The distance between x = -2 and the line of reflection is:
1 - (-2) = 3
Then the left end of the reflected rectangle will e 3 units at the right of x = 1.
1 + 3 = 4
Will be at x = 4, same as rectangle 2.
Now we move it 2 units down, and we will have rectangle 1 mapped into rectangle 2.
Translation of 12 units right, and then 2 units down.
The right end of rectangle 1 is at x = -2
The right side of rectangle 2 is at x = 10
The difference is 10 -(-2) = 12
Then we must move rectangle 1 12 units to the right.
Now, same as before, we can move rectangle 1 2 units down and we will have rectangle 1 mapped into rectangle 2.
Rotation of 180° around the point (1, 4)
This means a reflection over the line x = 1, followed of a reflection over the line y = 4.
We already see that a reflection over the line x = 1 leaves rectangle 1 right above rectangle 2. And the line that separates them is the line y = 4, then when we do a reflection over that line, we will have rectangle 1 mapped into rectangle 2.
Answer:
D, E, and F
Step-by-step explanation:
Frwinds? XD
Question uh....
65x9
UwU
Answers
Answer:
585
Step-by-step explanation:
65 times 9= 585
60 times 9=540
5 times 9=45
540 plus 45= 585
~~~Hope this helps ~~~ :)
Answer:
yws can we be frwinds? `(*>﹏<*)′
Step-by-step explanation:
65x9 = 585
ε=ε=ε=┏(゜ロ゜)┛
Can someone please help me.
Answers
Answer:
linear
Step-by-step explanation:
to be a linear equation it needs to have consistent rate of change for x and y. its constant for x at +4 but not for y at +16 +8 and +16
Answers
Answer:
A. △GHD ≅ △EDH by SAS
Step-by-step explanation:
This is right on edge. 2020
In triangles GHD and EDH angle GHD and angle EDH is congruent, side GH is congruent to the side ED, and side HD is congruent to side HD. So, from side angle side postulate triangle GHD and triangle EDH are congruent. | 677.169 | 1 |
NCERT Solutions for Class 9 Maths Chapter 10 Circles
Last Updated : 09 Nov, 2023
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NCERT Solutions for Class 9 Maths Chapter 10 Circles is a curated article by professionals at GFG, to help students solve problems related to circles with ease. All the solutions provided here are factually correct and
The NCERT Class 9 Maths Chapter 10 Circles covered a variety of issues, including how to determine the separation between equal chords from the centre and the angles that a chord at a location subtended. There are also studies of cyclic quadrilaterals and their properties, as well as the angles that a circle's arc subtends.
Question 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the Centre makes equal angles with the chords.
Solution:
Given: Equal chords AB and CD intersect at P.
To prove: angle1=angle=2
Construction: Draw OM perpendicular AB & ON perpendicular CD.
Solution: In ∆OMP & ∆ONP
Angle M= Angel N [90 ° each]
OP=OP [common]
OM=ON —————[ Equal chords are equal distant from center]
∴∆OMP≅∆ONP ———-[R.H.S]
∴∠1=∠2 ———–[C.P.C.T]
Question 4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Figure).
Solution:
Given : two concentric circle with O. A line intersect them at A, B, C , and D
To prove: AB=CD
construction: Draw OM ⊥ AD ,In bigger circle AD is chord OM ⊥ AD.
∴AM=MD —————-[⊥ from center of circle of a circle bisects the chord] __________ 1
The smaller circle :
BC is chord OM ⊥ BC
BM=MC ——————-[⊥ from center of circle of a circle bisects the chord] __________ 2
subtracting 1-2
AM-BM=MD-MC
AB=CD
Question 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Solution:
To find RM=?
Let Reshma, Salma and Mandip be R,S,M
Construction: Draw OP ⊥ RS join OR and OS.
RP=½RS ___________[⊥ from center bisects the chord]
RP=½*6=3m
In right ΔORP
OP²=OR²- PR²
OP= √ 5² -3²
=√259 =√16 =4
Area of ΔORS=½*RS*OP
=½*6*4=12m² —————–1
Now, ∠N=90°
Area of ΔORS=½*SO*RN
=½*SO*RN ——————-2
Above ,1=2
12=½*5*RN
12/5*2=RN
RN=4.8
RM=2*RN _________________[⊥ from center bisects the chord]
=2*4.8
9.6m
Question 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary. Each boy has toy telephone in his hand to talk with each other. Find the length of the string of each phone.
Solution:
Draw AM⊥SD
AS=SD=AD
∴ASD is the equilateral Δ
Let each side of Δ-2xm
SM=2x/2=x
Now in Δ DMS, by the Pythagoras theorem
AM²+SM²=AS²
AM²= AS²- SM²
AM=√(2x²+x² )
==√(3x² )
AM =√3x
OM=AM-AO
OM=√3x-20
Now in right ΔOMS
OM²+SM²=SO²
(√3x-20)²+2x²+x²=20²
20²+400-40√3x+x^2=400
4x²=40√3x
4xx=40√3x
X=(40√3)/4
X=10√3x
Length of each string =2x
=2*10√3xm
NCERT solutions for Class 9 Maths Chapter 10 Circles: Exercise 10.5
Question 1.In fig. 10.36, A, B, and C are three points on a circle with Centre O such that ∠BOC=30° and ∠AOB=60°. If D isa point on the circle other than the arc ABC, find ∠ADC.
Solution:
Given: ∠BOC=30° and ∠AOB=60° To find: ∠ADC Solution: ∠AOC=2∠ADC ———[The angle subtended by an arc at the centre is double the angle the angle subtended by it any point on the remaining part of the circle.] ∠AOB+∠BOC=2∠ADC 60°+30°=2∠ADC 90+30=2∠ADC 90/2=∠ADC 45=∠ADC
Question 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord ata point on the minor arc and also at a point on the major arc.
Solution:
Given: PQ=OP To find: Angle on major arc is ∠A=? Angle on the minor arc is ∠B=? Since, =PO=OQ ∴∠POQ=60° ∠POQ=2∠PAQ [The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining point of the circle] Reflex ∠POQ=360°-60° Reflex ∠POQ=300° Reflex ∠POQ=2∠POQ 300°=2∠PBQ 300°/2=∠PBQ 150°=∠PBQ
Question 3. In fig. 10.37, ∠PQR=100°,where P, Q and R are the points on a circle with centre O. Find ∠OPR.
Solution:
Given: ∠PQR=100° To find: ∠OPR=? Reflex ∠POR=2∠PQR ——–[ The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining point of the circle] Reflex ∠PQR=2*100 =200° ∠POR=360°-200° Now in ∆POR,OP=QR [ Radii of same circle] ∠P=∠R and let each =x. ∴∠P+∠O+∠R=180° [angle sum property of ∆] x+160°+x=180°-160° 2x+160°=180° x=20°/2=10° ∴∠OPR=10°
Question 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Given: ABCD is a cyclic quadrilateral. Diagonals of ABCD are also diameters of circle. To prove: ABCD is a rectangle AC=BD ———-[diameters of same circle] OA=OA ———[radii of the same circle] OA=OC=1/2AC ———2 OB=OD=1/2BD ———-2 From I and 2 diagonals are equal and bisect each other ∴ABCD is a rectangle
Question 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Question 9. Two circles intersect at two points B and C. Through B, two-line segments ABD and PBQ are drawn to intersect the circles at A, D, and P, Q respectively (see fig. 10.40). Prove that ∠ACP=∠QCD.
Solution:
To prove: ∠ACP=∠QCD or ∠1=∠2 ∠1=∠2 —— [angles in the same segment are equal] 1 ∠ 3=∠ 4 ——- [angles in the same segment are equal] 2 ∠2=∠4 ——- [vertically opposite angles] 3 From 1 2 and 3 ∠1=∠3 ∴∠ACP=∠QCB
Question 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Given: ABC is ∆ and AB and AC are diameters of two circles To prove: Point of intersection is D, lies on the BC. Construction: Join AD ∠ADB=90° ——-[angles in semicircle] 1 ∠ADC=90 ° ——[angles in semicircle] 2 Adding 1 and 2 ∠ADB+∠ADC=90°+90° ∠BDC=180° BDC is a straight line therefore D lies on BC.
Question 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD=∠CBD.
Solution:
Given: ABC and ADC are two right angle triangles with common hypotenuse AC. To prove: ∠ADB=∠CBD Solution: ∠ABC=∠ADC=90° Circle drawn by taking AC as diameter passes through B and D. For chord CD ∠CAD=∠CBD ——-[angle in the same segment]
Question 12. Prove that a cyclic parallelogram is rectangle.
Solution:
Given: ABC is a cyclic ||gm To prove: ABCD is a rectangle. Because ABCD is a cyclic ||gm ∴∠A+∠C=180° ∠A=∠C [opposite angle of ||gm] ∴∠A=∠C=(180°)/2=90° ∠A=90° ∠C=90° Similarly, ∠B+∠D=180° ∴∠B=∠D =(180°)/2=90° ———-[opposite of a ||gm] Each angle of ABCD is 90° ∠B=90° ∠D=90° Thus, ABCD is a rectangle.
NCERT solutions for Class 9 Maths Chapter 10 Circles: Exercise 10.6
Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two circles with Centre A and B circle intersects at C and D.
To prove: ∠ACB=∠ADB
Construction: Join AD,BC and BD
Proof: In ∆ACB and ∆ADB
AC=AD ——–[radii of the same circle]
BC=BD ———[radii of the same circle]
AB=AB ——–[common]
∴∆ACB≅∆ADB ——— [by S.S.S]
∠ACB=∠ADB ——–[c.p.c.t.]
Question 2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution:
Let O be the centre of circle and be r cm
Given: AB=5cm, CD =11cm
Construction: Draw OM perpendicular AB and OL perpendicular CD.
Because OM perpendicular AB and OL perpendicular CD and AB||CD.
∴Points O,L, and M are collinear, than ∠M=6cm
Let OL=x
Then OM=6=x
Join AO and CO
OA=OC =r
OL=1/2CD=1/2*11=5.5cm —–[perpendicular from bisects the chord]
AM=1/2AB=1/2*5=2.3cm —–[perpendicular from bisects the chord]
Now, In right ∆DLC
r2=(OL)2+(CL)2
r2=x2+(5.5)2
r2=x2+30.25 ———–1
Now in right ∆OMA
r2=(OM)2+(MA)2
r2=(6-x)2+(2.5)2
r2=36+x2=12x+6.25
r2=x2-12x+42.25 ———–2
Now equating equation 1 and 2
X2+30.25=x2-12x+30.25
12x=42.25-30.25
X=12/12=1
Putting value of x in equation 1
r2=x2+30.25
r2=(1)2+30.25
r2=31.25
r=√31.25=5.6 (approx.)
Radius of circle is 5.6cm.
Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the centre?
Solution:
Let AB and CD are|| chord of circle with centre O which AB=6cm and CD=8cm and radius of circle =r cm.
Construction: Draw OP perpendicular AB and OM perpendicular CD.
Because AB||CD and OP perpendicular AB and OM perpendicular CD therefore. Point O, M and P are collinear.
Clearly, OP=4cm ———-[According to question]
OM=to find?
P is midpoint of AB.
∴AP=1/2 AB=1/2*6=3cm
M is midpoint of AB.
CM=1/2 CD=1/2*8cm=4cm
Join AO and CO
Now in Right ∆OPA,
r2=AP2+PO2
r2=32+42
r2=9+16=25
Now in ∆OMC
r2=CM2+MO2
25=42+MO2
25-16=MO2
9=MO2
√9=MO
3=MO
∴Therefore distance of the other chord from the centre is 3cm
Question 4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Give: Vertex B of ∆ABC lie outside the circle,chord AD=CE
To prove: ∠ABC=1/2(∠DOE-∠AOC)
Construction: Join AE
Solution: Chord DE subtends ∠DOE at the center and ∠DAE at point A on the circle.
∴∠DAE=1/2∠DOE ———-1
chords AC subtends ∠AOC at the centre and ∠AEC at point
∴∠AEC=1/2∠AOC ———2
In ∆ ABE,∠DAE is exterior angle
∠DAE=∠ABC +∠AEC
1/2∠DOE=∠ABC+1/2∠AOC
½(∠DOE-∠AOC)= ∠ABC
Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
Given: A rhombus ABCD in which O is intersecting point of diagonals AC and BD.
A circle is drawn taking CD as diameter.
To prove: circle points through O or Lies on the circles.
Proof: In rhombus ABCD,
∠DOA=90° ——–[diagonals of rhombus intersect at 90°] 1
In circle:
∠COD=90° ——–[angle made in segment O is right angle] 2
From 1 and 2
O lies on the circle.
Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
ABCD is a ||gm. The circle through A,B and C intersect at E.
To prove: AE=AD
Proof: Here ABCE is a cyclic quadrilateral
∠2+∠4=180° —–[sum of opposite is of a cyclic quadrilateral is 180°]
∠4=180°-∠1 ——-1
Now ∠4+∠6=180°-∠6 ———2
From 1 and 2
180°-∠2=∠180°-∠6
∠2=∠6 ———–3
Also ∠2=∠5 ———[opposite angles of ||gm are equal] —–4
From 3 and 4
∠5=∠6
Now, In ∆ADE,
∠5=∠6
∴AE=AD ——[sides apposite to equal angles in a∆ are equal]
Question 7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Given:∆ABC and it circum-circle AD,BE and CF are bisectors of ∠A,∠B and ∠C
Respectively.
To proof:∠ D=90°-1/2∠A , ∠E=90°-1/2∠B , ∠F=90°-1/2∠C
Construction: Join AE and AF.
Solution: ∠ADE=∠ABE ———-1 [angle in the same segment are equal]
∠ADF=∠ACF ———–2 [angle in the same segment are equal]
Adding 1 and 2
∠ADE+∠ABF=∠ABE+∠ACF
∠D=1/2∠B+1/2∠C ——[BC and CF are bisector of ∠B & ∠c]
∠D=1/2(∠B+∠C)
∠D=1/2(180°-∠A)
∠D=1/2(180°-∠A)
∠D=90°-1/2∠AC
Question 9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
Given: two congruent circles which intersect at A and B.
PAB is a line segment
To prove: BA=BQ
Construction: join AB
Proof: AB is a common chord of both the congruent circle.
Segment of both circles will be equal
∠P=∠Q
Now, in ∆ BPQ,
∠P=∠Q
BP=BQ ——[sides opposite to equal angles are equal]
Question 10. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Solution:
Given: A ∆ABC, in which AD is angle bisector of ∠A and OD is ⊥ bisector of BC.
To prove: D lies on circumcircle.
Construction: Join OB and OC
Proof: Since BC subtends ∠BAC at A on the remaining of the circle.
∠BOC=2∠BAC ——-1
Now, In ∆BOE and ∆ COE
BO=OE ——–(radii of the same circle)
BE=CE —–(give)
∴∆BOE≅COE ——-(S.S.S)
∠1=∠2 ——-(c.p.c.t)
Now,
∠1+∠2=∠BOC
2∠1=∠BOC
2∠1=2∠BAC ———- (from 1)
∠1=∠BAC
∠BOE=∠BAF
∠BOD=∠BAC
∠BOD=2∠BAD [AD is bisector of ∠BAC]
This is possible only if BD is chord of the circle.
D lies on the circle.
Important Points to Remember
NCERT Solutions for Class 9 will help students to learn the solution for all the NCERT Problems.
These solutions are entirely accurate and can be used by students to prepare for their board exams.
All the solutions provided are in a step-by-step format for better understanding.
Key Takeaways of NCERT Solutions Class 9 Maths Chapter 10
By utilizing the class 9 questions with solutions on circles, students can reap various advantages that would aid their academic growth. Firstly, these comprehensive resources enable students to improve their scores significantly while simultaneously enhancing their understanding of the subject matter. Moreover, what further adds to their accessibility is the fact that these helpful PDF files are conveniently available free of charge. Therefore, students have the flexibility to choose between a soft copy or a hard copy, based on their personal preference and convenience.
FAQs – NCERT Solutions for Class 9 Maths Chapter 10 Circles
Q1: Why is it important to learn about Circles?
It is crucial to understand circles since they are fundamental geometric forms having numerous uses in a variety of industries, including math, physics, engineering, and architecture. Understanding circles enables you to apply ideas like tangents, chords, and sectors to solve issues involving their properties, such as calculating circumference, area, and arc length. In general, understanding circles is crucial for building a solid geometry foundation and for practical applications in a variety of fields.
Q2: What topics are covered in NCERT Solutions for Chapter 10- Circles?
NCERT Solutions for Class 9 Maths Chapter 10- Circles can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.
Q4: How many exercises are there in Class 9 Maths Chapter 10-Circles?
There are 6 exercises in the Class 9 Maths Chapter 10- Circles which covers all the important topics and sub-topics.
Q5: Where can I find NCERT Solutions for Class 9 Maths Chapter 10-Circles?
You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.
To excel in CBSE board exams, students often rely on NCERT textbooks as their primary study material. These textbooks are renowned for their comprehensive content, making them an invaluable resource. In order to enhance their understanding of the textbook problems, students can conveniently refer to the NCERT solutions for Class 9 Maths Chapter 10 offered by GeeksforGeeks. These solutions are meticulously prepared by experts, providing precise and accurate methods to solve the problems effectively. By utilizing these solutions, students can significantly boost their proficiency and master the subject in no time. | 677.169 | 1 |
Chords are an essential element in the study of geometry, and their properties have fascinated mathematicians for centuries. In this article, we will delve into the topic of chord PQ, specifically focusing on its length of 8 cm. Through a combination of research, examples, and case studies, we will explore the significance of this measurement and its implications in various mathematical contexts.
Understanding Chords and Their Properties
Before we dive into the specifics of chord PQ, let's establish a foundational understanding of chords in geometry. A chord is a line segment that connects two points on a curve, typically within a circle. In the case of chord PQ, it connects points P and Q on a given curve.
Chords possess several interesting properties that make them worthy of study. One such property is that the perpendicular bisector of a chord passes through the center of the circle. This property holds true for chord PQ as well, indicating that the perpendicular bisector of PQ intersects the center of the circle.
Another important property of chords is that they divide the circle into two segments: the minor segment and the major segment. The minor segment is the region enclosed by the chord and the arc opposite to it, while the major segment is the remaining portion of the circle. Understanding these segments is crucial when analyzing the length of a chord.
The Length of Chord PQ: Exploring Its Significance
Now that we have established a basic understanding of chords, let's focus on chord PQ and its length of 8 cm. This specific measurement holds significance in various mathematical contexts, and we will explore some of them in this section.
1. Circle Geometry
In the realm of circle geometry, the length of chord PQ can provide valuable insights into the properties of the circle. For instance, if we know the length of a chord and the radius of the circle, we can determine the distance between the chord and the center of the circle using the Pythagorean theorem.
Let's consider an example to illustrate this concept. Suppose we have a circle with a radius of 5 cm, and chord PQ has a length of 8 cm. By applying the Pythagorean theorem, we can calculate the distance between the chord and the center of the circle as follows:
Distance = √(r^2 – (c/2)^2)
Distance = √(5^2 – (8/2)^2)
Distance = √(25 – 16)
Distance = √9
Distance = 3 cm
In this example, we find that the distance between chord PQ and the center of the circle is 3 cm. This calculation showcases how the length of a chord can provide valuable information about the geometry of a circle.
2. Trigonometry
Chord PQ's length of 8 cm also has implications in the field of trigonometry. By considering the angle subtended by the chord at the center of the circle, we can apply trigonometric functions to analyze the relationship between the length of the chord and other elements of the circle.
For instance, let's assume that the angle subtended by chord PQ at the center of the circle is 60 degrees. By utilizing trigonometric functions, we can determine the radius of the circle using the following formula:
Radius = (c/2) / sin(θ/2)
Radius = (8/2) / sin(60/2)
Radius = 4 / sin(30)
Radius = 4 / 0.5
Radius = 8 cm
In this example, we find that the radius of the circle is 8 cm. This calculation demonstrates how the length of chord PQ, combined with the angle subtended by the chord, can help us determine the radius of the circle using trigonometric functions.
Case Studies: Real-World Applications
While the study of chord PQ's length of 8 cm has primarily focused on its mathematical implications, it is worth exploring some real-world applications where this measurement finds relevance. Let's examine two case studies that highlight the practical significance of this mathematical concept.
1. Architecture and Construction
In the field of architecture and construction, precise measurements are crucial for ensuring structural integrity and aesthetic appeal. The length of chord PQ can be utilized in the design and construction of circular structures, such as domes or arches.
By understanding the properties of chords and their relationship with the circle's geometry, architects and engineers can accurately determine the dimensions and proportions of circular elements in their designs. The length of chord PQ, along with other relevant measurements, aids in creating visually pleasing and structurally sound architectural marvels.
2. Navigation and GPS Systems
Navigation systems, including GPS (Global Positioning System), rely on mathematical principles to determine precise locations on Earth's surface. The concept of chord length plays a significant role in these systems, particularly when calculating distances between two points.
By considering the Earth as a sphere, GPS systems utilize chords to calculate the shortest distance between two locations. The length of chord PQ, in this context, represents the distance between two points on the Earth's surface, allowing for accurate navigation and positioning.
Q&A: Exploring Common Questions
Now, let's address some common questions that often arise when discussing the length of chord PQ and its mathematical implications.
1. Can the length of chord PQ be greater than the diameter of the circle?
No, the length of chord PQ cannot be greater than the diameter of the circle. The longest possible chord in a circle is the diameter itself, which passes through the center of the circle. Therefore, the length of chord PQ must always be less than or equal to the diameter of the circle.
2. How can the length of chord PQ be calculated if the radius is known?
If the radius of the circle is known, the length of chord PQ can be calculated using the following formula:
Length of Chord PQ = 2 * √(r^2 – d^2)
In this formula, "r" represents the radius of the circle, and "d" represents the perpendicular distance between chord PQ and the center of the circle.
3. What other properties of chord PQ can be explored?
Aside from its length, chord PQ possesses several other properties that can be explored. These include the angle subtended by the chord at | 677.169 | 1 |
Table of Contents
Introduction Law of Cosines:
Have you ever wondered how to solve triangles when you know the lengths of all three sides but none of the angles? This is where the Law of Cosines comes into play. In this comprehensive guide, we'll unravel the mysteries of the Cosines of law, exploring its significance, applications, and formulas.
Understanding the Basics:
The Law of Cosines is a fundamental principle in trigonometry, used to find the lengths of the sides or angles of a triangle when other measurements are known. Unlike the more commonly known Pythagorean theorem, which applies only to right triangles, the Cosines of law is applicable to all triangles, whether they are acute, obtuse, or right-angled.
Formulas and Applications:
The Law of Cosines can be expressed in various forms, depending on what information is known about the triangle. The most common form relates the lengths of the sides of a triangle to the cosine of one of its angles:
c2 =a2 + b2 −2ab⋅cos(C)
Where a, b, and c represents the lengths of the sides of the triangle, and C is the angle opposite side c.
This formula allows us to find the length of one side of a triangle when the lengths of the other two sides and the included angle are known. Similarly, it can be rearranged to solve for angles when the lengths of all three sides are known.
Real-World Applications:
The Law of Cosines has numerous applications in various fields, including engineering, physics, navigation, and even computer graphics. For example, it's used in surveying to measure distances between points on the Earth's surface, in physics to analyze forces acting on objects at angles, and in navigation to determine the position of ships and aircraft.
FAQs:
What is the difference between the Law of Cosines and the Pythagorean theorem?
The Pythagorean theorem applies only to right triangles, while the Cosines of law applies to all triangles, regardless of their angles.
When should I use the Law of Cosines?
You should use the Cosines of law when you need to find the length of a side or angle in a triangle where the lengths of all three sides are known, or when you have two sides and the included angle.
Is it possible to utilize the Law of Cosines for determining the area of a triangle?
No, the Cosines of law is used to find side lengths and angles in a triangle. To find the area, you would typically use formulas such as 12 sin() 2 1 ab sin(C).
How do I remember the formula for the Law of Cosines?
One helpful mnemonic is "The Cosines of law deals with 'Sides and Angles,' so the formula involves side lengths and the cosine of an angle."
What if I only have the lengths of two sides and an angle not between them?
In this case, you can use the Cosines of law to find the length of the third side, and then use trigonometric ratios to find the other angles.
Conclusion:
In conclusion, the Law of Cosines is a powerful tool in trigonometry, allowing us to solve triangles in various scenarios where the Pythagorean theorem falls short. By understanding its applications and formulas.
You can tackle a wide range of problems in mathematics and beyond. So, the next time you encounter a triangle with known side lengths, remember the Law of Cosines and unlock its potential to solve the unknown. | 677.169 | 1 |
Money
Geometry and Mensuration
Concept and understanding of different shapes Line Segment, Ray, Angles, Types of Angles, Types of Triangles, Types of Quadrilaterals, circles, Length, Perimeter and Area of different Geometrical figures, Weight, (Time, Capacity and Volume. | 677.169 | 1 |
This free course looks at various aspects of shape and space. It uses a lot of mathematical vocabulary, so you should make sure that you are clear about the precise meaning of words such as circumference, parallel, similar and cross-section. You may find it helpful to note down the meaning of each new word, perhaps illustrating it with a diagram. | 677.169 | 1 |
visualsubjectmatter
The isosceles trapezoid ABDE is part of an isosceles triangle ACE. Find the measure of the vertex an...
4 months ago
Q:
The isosceles trapezoid ABDE is part of an isosceles triangle ACE. Find the measure of the vertex angle of ACE. (See attachment)A. 130 degreesB. 60 degreesC. 65 degreesD. 50 degreesI really need an explanation along with the answer, thank you!! | 677.169 | 1 |
The A Square + B Square Formula: Understanding its Significance and Applications
significance of this formula, its applications in various fields, and explore some real-life examples to better understand its practicality.
The Pythagorean Theorem: Unveiling the Formula
The Pythagorean theorem, named after the ancient Greek mathematician Pythagorasa2 + b2 = c2
Here, 'a' and 'b' represent the lengths of the two shorter sides of the triangle, while 'c' represents the length of the hypotenuse.
The Significance of the A Square + B Square Formula
The A square + B square formula holds immense significance in mathematics and various other fields. Let's explore some of its key applications:
1. Geometry and Trigonometry
The Pythagorean theorem is a fundamental concept in geometry and trigonometry. It provides a basis for understanding the relationships between the sides and angles of right-angled triangles. By using this formula, mathematicians and engineers can solve complex geometric problems, calculate distances, and determine unknown angles.
2. Architecture and Construction
In the field of architecture and construction, the A square + B square formula is crucial for ensuring structural stability and accuracy. Architects and engineers use this formula to calculate the lengths of diagonal beams, determine the dimensions of rooms, and ensure that buildings are built on a solid foundation.
3. Navigation and GPS
The Pythagorean theorem plays a vital role in navigation and GPS systems. By using this formula, GPS devices can accurately calculate distances between two points on a map, determine the shortest route, and provide real-time directions. This application has revolutionized the way we navigate and travel.
4. Computer Graphics and Animation
In the world of computer graphics and animation, the A square + B square formula is used extensively. It helps in creating realistic 3D models, calculating distances between objects, and determining the angles of rotation. Without this formula, the creation of visually stunning graphics and animations would be significantly more challenging.
Real-Life Examples of the A Square + B Square Formula
To better understand the practicality of the A square + B square formula, let's explore some real-life examples:
Example 1: Building a Fence
Imagine you are building a fence in your backyard. You want to ensure that the fence is perfectly square, so you measure the diagonals to confirm. By using the Pythagorean theorem, you can calculate the length of the diagonal and compare it to the sum of the squares of the two sides. If they are equal, your fence is square.
Example 2: Calculating Distance
Suppose you are planning a road trip and want to calculate the distance between two cities. By using the A square + B square formula, you can determine the straight-line distance between the two cities, even if the roads are not direct. This calculation can help you estimate travel time and plan your journey accordingly.
Example 3: Roofing a House
When roofing a house, it is essential to ensure that the roof is properly aligned and the angles are correct. By using the Pythagorean theorem, roofers can measure the diagonals of the roof to ensure they are equal. This ensures that the roof is square and prevents any potential structural issues in the future.
Q&A
Q1: Who discovered the Pythagorean theorem?
A1: The Pythagorean theorem was discovered by the ancient Greek mathematician Pythagoras.
Q2: What is the formula for the Pythagorean theorem?
A2: The formula for the Pythagorean theorem is a2 + b2 = c2, where 'a' and 'b' represent the lengths of the two shorter sides of a right-angled triangle, and 'c' represents the length of the hypotenuse.
Q3: How is the Pythagorean theorem used in navigation?
A3: The Pythagorean theorem is used in navigation to calculate distances between two points on a map, determine the shortest route, and provide real-time directions.
Q4: Can the Pythagorean theorem be applied to non-right-angled triangles?
A4: No, the Pythagorean theorem is only applicable to right-angled triangles.
Q5: What are some other applications of the Pythagorean theorem?
A5: Apart from the fields mentioned earlier, the Pythagorean theorem is also used in physics, engineering, surveying, and even in everyday life for various measurements and calculations.
Summary
The A square + B square formula, also known as the Pythagorean theorem, is a fundamental concept in mathematics with wide-ranging applications. It is used in geometry, trigonometry, architecture, navigation, computer graphics, and many other fields. By understanding and applying this formula, we can solve complex problems, calculate distances, and ensure structural stability. Real-life examples such as building a fence, calculating distances, and roofing a house demonstrate the practicality of this formula. The Pythagorean theorem continues to be a cornerstone of mathematics, enabling us to explore and understand | 677.169 | 1 |
The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to :
A
$\frac{\sqrt{1691}}{5}$
B
$\frac{\sqrt{2009}}{5}$
C
$\frac{\sqrt{1541}}{5}$
D
$\frac{\sqrt{1741}}{5}$
2
JEE Main 2023 (Online) 13th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Let the tangent and normal at the point $$(3 \sqrt{3}, 1)$$ on the ellipse $$\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$$ meet the $$y$$-axis at the points $$A$$ and $$B$$ respectively. Let the circle $$C$$ be drawn taking $$A B$$ as a diameter and the line $$x=2 \sqrt{5}$$ intersect $$C$$ at the points $$P$$ and $$Q$$. If the tangents at the points $$P$$ and $$Q$$ on the circle intersect at the point $$(\alpha, \beta)$$, then $$\alpha^{2}-\beta^{2}$$ is equal to :
A
61
B
$$\frac{304}{5}
$$
C
60
D
$$\frac{314}{5}
$$
3
JEE Main 2023 (Online) 12th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let $$\mathrm{P}\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), \mathrm{Q}, \mathrm{R}$$ and $$\mathrm{S}$$ be four points on the ellipse $$9 x^{2}+4 y^{2}=36$$. Let $$\mathrm{PQ}$$ and $$\mathrm{RS}$$ be mutually perpendicular and pass through the origin. If $$\frac{1}{(P Q)^{2}}+\frac{1}{(R S)^{2}}=\frac{p}{q}$$, where $$p$$ and $$q$$ are coprime, then $$p+q$$ is equal to :
A
143
B
147
C
137
D
157
4
JEE Main 2023 (Online) 11th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
If the radius of the largest circle with centre (2,0) inscribed in the ellipse $$x^2+4y^2=36$$ is r, then 12r$$^2$$ is equal to : | 677.169 | 1 |
NCERT solutions class 10 coordinate geometry exercise 7.2
Q1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Solution 1
Q2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Solution 2.
Q3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1 4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1 5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution 3.
Q4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
Solution 4
Q5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Solution 5
Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution 6
Q7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution 7
Q8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.
Solution 8
Q9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Solution 9
Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint : Area of a rhombus = 1/2 (product of its diagonals)]. | 677.169 | 1 |
The "shape" of a planar curve and/or landmark configuration is considered its equivalence class under translation, rotation and scaling, its "form" its equivalence class under translation and rotation while scale is preserved. | 677.169 | 1 |
Pages
30 January 2016
New GCSE: Pythagoras Questions
This week my department took a detailed look at GCSE sample questions for topics that are coming up on our Year 10 Scheme of Work. This was the most useful thing we've ever done in a department meeting! We used Mel's (@Just_Maths) excellent Questions by Topic resources.
I worked through the Higher Pythagoras questions. Have you looked at these in detail? It's well worth doing. The questions are excellent. They will be challenging for even our highest attainers. I usually use a nice set Pythagoras problems when I teach Pythagoras at GCSE but I think I'm going to have to up my game.
Let's take a look at some of the new GCSE questions.
Here's a three mark question from OCR. Have a go. I think my students would find this pretty challenging. They need to know Pythagoras and how to work with surds 'backwards' (I don't do enough of this when I teach surds...).
Here's another question - this is a four marker from AQA. It's a great question... Again, a different level of challenge to what we're used to at GCSE level.
This question is from Edexcel - here we have bounds and Pythagoras:
And this question is from OCR. It's not quite as challenging as the others though there are a lot of steps.
Finally, here's another question from AQA. I particularly like this one.
All of these examples show that new GCSE questions will often combine two or more topics. Rightly so. Previous GCSE questions often treated topics in isolation.
When you're planning your Year 10 Pythagoras lessons I recommend that you try the rest of the sample questions. Even the 3D Pythagoras question has a spin on it. There's some really great problems here. These questions also remind us of the notable step up in challenge level that we're dealing with. I'm trying not to panic! We have our work cut out. | 677.169 | 1 |
This trigonometry textbook this book is not just about mathematical content but is also about the process of learning and doing mathematics. That is, this book is designed not to be just casually read but rather to be engaged. Since this can be a difficult task, there are several features of the book designed to assist students in this endeavor. In particular, most sections of the book start with a beginning activity that review prior mathematical work that is necessary for the new section or introduce new concepts and definitions that will be used later in that section. Each section also contains several progress checks that are short exercises or activities designed to help readers determine if they are understanding the material. In addition, the text contains links to several interactive Geogebra applets or worksheets. These applets are usually part of a beginning activity or a progress check and are intended to be used as part of the textbook.
Thumbnail: For some problems it may help to remember that when a right triangle has a hypotenuse of length \(r\) and an acute angle \(θ\), as in the picture below, the adjacent side will have length \(r\cos θ\) and the opposite side will have length \( r\ sin θ\). You can think of those lengths as the horizontal and vertical "components'' of the hypotenuse. (GNU FDL; Michael Corral | 677.169 | 1 |
Jun 8, 2016 · Volodymyr Zelenskyy shapes 'political battlefield' by confronting developing powers on whatsapp (opens in a new window) Save. Shapes define the boundary of an object and can be differentiated in many ways based on their properties. . Level 2. . Cone. 00:00. a "big, yellow, square", "a small, blue square", etc. With the blade shape different on each side of the midrib. Hexagon Shapes are a type of geometric shape with six faces, six edges, and six vertices. Lemon. Describe the shapes by the number of angles and sides. the form characteristic of a particular person or thing, or class of things. Jun 8, 2016 · com/user/EnglishSingsing9Kids vocabulary - Shape - Name of the Shape - Learn English for kids - English educational videoThis "Kids Vocabu. Shapes in English triangle circle square rectangle diamond oval start heart octagon. 7ESL Learning English. Shapes in Spanish tri. A cube's basic shapes. They have also been used in the design of jewelry, pottery and other art forms. . Sekadar informasi, sangat jarang vocabulary " triangle " dan " rectangle " disajikan dalam bentuk jamaknya). . Ring = Annulus. Wiltshire Council said the "critical" £30m scheme should help protect hundreds of homes from. . Shapes. 28M subscribers. May 24, 2018 · Describing simple shapes in English (especially when they're 3D) isn't as easy as it seems. Shapes are everywhere we look. Triangle, square, rectangle and circle are the most well-known geometric shapes. . ) – more common. . Hexagon. Dec 3, 2013 · The following list covers the most common words for shapes in the English language, with explanations (sorry, no pictures!). 7ESL Learning English Courses: Explore the English vocabulary of Shapes in this sound integrated guide. It has 3 faces, 2 edges and 0 vertex. Liquid-mediated patterning, which is a fabrication method exploiting the nature of liquids, has achieved great advances in unconventional micro- and nano. It is one of the three regular polygon shapes. Think of an ice cream cone. Hexagons have been used in architecture and engineering since ancient times. Kite. asymmetrical. . Shapes in Spanish tri. X. The noun and adjective forms for each are shown below. . . Hexagons have been used in architecture and engineering since ancient times. Triangle, square, rectangle and circle are the most well-known geometric shapes. . The last two adverbs are very uncommon. . Types of triangles. In this video we'll learn. . . 2M views 4 years ago English Vocabulary. Shapes in English triangle circle square rectangle diamond oval start heart octagon.
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. ) / donut (n. 2. It is one of the three regular polygon shapes. com/ Learn Shapes Names: Below you'll find an extensive list of shapes and shape names in English with pictures. You can also elicit colors and sizes (e. In this post, you're going to: Learn words for common 2D and 3D shapes in English. . . . Introduce Shapes. . . ) / donut (n. g. 7. Learn some common English idioms and phrases with shapes. . Prove your vocabulary mastery. Drawings on the board or flashcards will be the easiest way to introduce shapes. com/shapes-vocabulary/List of Shapes in English with American English Pr. May 24, 2018 · Describing simple shapes in English (especially when they're 3D) isn't as easy as it seems. You may choose to only teach square, rectangle, circle, and triangle but feel free to include other vocabulary such as star and diamond if appropriate. Feb 18, 2019 · November 10, 2020 February 18, 2019 by English Tutor Shapes Names ! This lesson provides common shapes names with pictures in English to help you expand your English vocabulary. . . There is a line on the top and. Hyperboloid. ) – more common. Basic Shape Names. Dec 3, 2013 · The following list covers the most common words for shapes in the English language, with explanations (sorry, no pictures!). Peter Karol and Bhamati Viswanathan of New England Law Boston join us to discuss. . Inside bomb-proof frozen vaults underneath the English countryside hides a treasure trove of 40,000 species of wild plant seeds from around the world, many of which are in danger of disappearing. It is one of the three regular polygon shapes. . . .
Our Best Stories in Your 00:00. Prove your vocabulary mastery by completing challenges. Level 1. Explore the English vocabulary of Shapes in this sound integrated guide. May 21, 2023 · NHK Trophy 2023: Hashimoto Daiki's pain threshold put to test in winning third successive title as Japan's gymnastics worlds team takes shape. The world's largest seed bank, located in the sleepy countryside south of London, is in a race against time because two out of five plant. Drawings on the board or flashcards will be the easiest way to introduce shapes. 7. Dec 3, 2013 · The following list covers the most common words for shapes in the English language, with explanations (sorry, no pictures!). Listen. Hexagon. . Hexagon. Sekadar informasi, sangat jarang vocabulary " triangle " dan " rectangle " disajikan dalam bentuk jamaknya). Average: 3. They have also been used in the design of jewelry, pottery and other art forms. . . May 21, 2023 · NHK Trophy 2023: Hashimoto Daiki's pain threshold put to test in winning third successive title as Japan's gymnastics worlds team takes shape. 3D Shapes. . . com/ Learn Shapes Names: These solids are. Circle. . A shape or figure is a graphical representation of an object or its external boundary, outline, or external surface, as opposed to other properties such as color, texture, or material type. . . ) – more common. How a Supreme Court ruling on an Andy Warhol piece could shape the future of art 13:18. . List Of Shapes In English. Ron DeSantis officially announced his run Wednesday. [ C ] the physical form or. . A few examples of shapes are circles, squares, rectangle, triangles, and so on. the form characteristic of a particular person or thing, or class of things. . We give an example of each shape. The triangle is a closed geometric shape that is formed by joining three straight lines. Prove your vocabulary mastery by completing challenges. They have also been used in the design of jewelry, pottery, and other art forms.
youtube. ) – more common. Inside bomb-proof frozen vaults underneath the English countryside hides a treasure trove of 40,000 species of wild plant seeds from around the world, many of which are in danger of disappearing. | 677.169 | 1 |
Dentro del libro
Resultados 1-5 de 51
Pįgina 19 ... parallel to one another , when , being in the same plane , they are incapable of meeting in a single point , however they be produced . 11. A FIGURE is that which is enclosed by one or more boundaries . [ The space enclosed within a ...
Pįgina 20 ... parallel . 27. Of parallelograms , that is a SQUARE which has all its sides and angles equal . 28. An OBLONG is that which has equal angles , but not equal sides . 29. A RHOMB is that which has equal sides , but not equal angles . 30. A ...
Pįgina 21 ... parallel , is called a TRAPEZium . [ The straight line joining the opposite angles of a quadrilateral figure is called a DIAGONAL . ] POSTULATES . 1. Let it be granted , that a straight line may be drawn from any one point to any other ...
Pįgina 22 ... Two straight lines which intersect each other cannot be both parallel to the same straight line . * * For Comments on the Definitions , & c . see APPENDIX A. PROPOSITION I. PROBLEM . On a given finite straight line 22 EUCLID'S ELEMENTS .
Pįgina 37 ... parallel . A F E / B G For if the lines AB , CD be not parallel , they shall meet when produced . Suppose that they meet in the point G , then EGF is a triangle , and its side GE being produced , the exterior angle AEF is greater than | 677.169 | 1 |
What does the term Trigonometry mean?
The mathematical branch that deals with the angles and the sides of a right-angled triangle is called trigonometry. Two Greek words: "trigonon" and "metron" form the word trigonometry, where the meaning of the word trigonon is a triangle and the meaning of the word metron is to measure. One of the complex parts of mathematics is trigonometry, and its functions are known as trigonometric functions.
Measurements of Angles
The intersection of two lines or rays results in the formation of an angle, and the two rays are then called the sides or arms of the angle. The arm where the angle starts is known as the initial side, and the other arm is known as the terminal side. As soon as the initial ray goes along the positive x-axis and the origin contains the vertex, this condition states that the initial side is in standard position.
The rotation between the terminal and the initial side is used to determine the angle. When the rotation is anti-clockwise, the angle is positive, and when it is clockwise, the angle is negative.
The measurement of the angle can be done in two different ways:
1. Degrees
2. Radians
1. Degrees: - A degree is considered as 1360th of the angle which represents one revolution.Initially, the angle is measured in degree.
2. Radians: - One revolution measured in radians is 2π, where π is a constant, whose value is approximately 3.14.
Unit Circle
A circle whose radius is always one and the centre is at the origin is known as the unit circle. The trigonometry ratios are defined using the unit circle.
What is Trigonometric Function (Trig Function)?
There are six ratios (trigonometric functions) that justify the angles of a right-angled triangle (right triangle) and the side lengths. Remember that a right-angled triangle contains an angle of 90°. Consider a right-angled triangle that consists of an acute angle 𝜃 with the initial side of length 'x' and the terminal side of length 'r'. Consider that the length of the remaining side is 'y'. Here 'r' denotes the hypotenuse.
The Pythagorean theorem states that the square of the hypotenuse of a right triangle is the same as the sum of the squares of the lengths of the other two sides.
According to the Pythagorean Theorem: x2+y2=r2 .
Therefore, r=x2+y2.
𝜃 (theta) is used as an angle for the following six trigonometric functions:
sine (sin)
cosine (cos)
tangent (tan)
cosecant (csc)
secant (sec)
cotangent (cot)
The ratios for the three basic trigonometric functions are as follows:
1. sinθ=opposite sidehypotenuse⇒sinθ=yr
2. cosθ=adjacent sidehypotenuse⇒cosθ=xr
3. tanθ=opposite sideadjacent side⇒tanθ=yx
These three trigonometric functions have three reciprocal functions.
The reciprocal function of sine is called the cosecant function.
cscθ=1sinθcscθ=ry
5.The reciprocal function of cosine is called the secant function.
secθ=1cosθsecθ=rx
6. The reciprocal function of tangent is called the cotangent function.
cotθ=1tanθcotθ=xy
Example:
Determine the values of the six trigonometric functions having angle theta.
Before we can get the six trigonometric ratios, the missing side length is to be found.
Quotient Identities
Pythagorean Identities
Negative Angle Identities
1. sin(−𝜃) = − sin 𝜃.
2. cos(−𝜃) = cos 𝜃.
3. tan(−𝜃) = − tan 𝜃.
4. csc(−𝜃) = − csc 𝜃.
5. sec(−𝜃) = sec 𝜃.
6. cot(−𝜃) = − cot 𝜃.
Co-Function Identities
When the summation of two angles is 90°, they are said to be complementary to each other. The functions f and g are said to be co-functions of each other if f(a)=g(b) whenever a and b are complementary angles. Below mentioned are the trigonometric co-function identities:
Product to Sum Formulas
Example: - Change the term sin 75° sin 15° to sum using the product to sum rule.
sin 75° sin 15° = 1/2 [cos(75° − 15°) − cos(75° + 15°)].
sin 75° sin 15° = 1/2 [cos 60° − cos 90°].
sin 75° sin 15° = 1/2 [ 1/2 − 0].
sin 75° sin 15° = 1/4.
Sine and Cosine Rule/Law
The rule is helpful for oblique triangles.
Sine Law
Let the side length opposite to vertex A be a. Similarly, the side length opposite to vertex B is b,and the side length opposite to vertex C is c. Then, as per the sine law, we have:
sinAa=sinBb=sinCc
The sine law can be used when some side lengths and angles are given in the problem.
Law of Cosine
c²=a²+b²−2abcosC.b²=a²+c²−2accosB.a²=b²+c²−2bccosA.
Cosine law is used when only one angle and the other three sides are given in the problem.
Graphing Trigonometry Functions
This section will be used to brief the graphing of three trigonometric functions, sine (sine function), cosine (cosine function), and tangent (tangent function). It will state to identify the graph of the trigonometric function and its key features. The sine and the cosine are referred to as periodic functions, which means if the graph is made, it will repeat after intervals for that trig function.
A periodic trig function relation can be defined as:
(𝑥) = 𝑓(𝑥 + 𝑛𝑝), where n denotes an integer.
We know that one revolution of the unit circle is measured as 2π radians, which means that the circumference of the unit circle is 2π. It is the period for sine and cosine.
Graph of the Sine Function (𝒚 = 𝒔𝒊𝒏𝒙)
. The range and domain of the sine function is [-1, 1] and (-∞, ∞), respectively.
2. It is defined as an odd function because of the symmetry with the origin.
3. The repetition of the sine function occurs at intervals of 2𝜋.
Graph of the Cosine Function (𝒚 = 𝒄𝒐𝒔 𝒙)
The scope (range) of the cosine function is [-1, 1], and the domain for the cosine function is (-∞, ∞).
2. This is an even function because it is symmetric with respect to the y-axis.
Graph of the Tangent Function (𝒚 = tan 𝒙)
3. When the numerator representing the sine function is zero, the tangent stands out to be zero.
4. When the value of the denominator representing the cosine function is zero, the tangent stands out to be undefined.
5. Vertical asymptotes are present in the tangent function graph, and the asymptotes' equations are given as 𝑥 = 𝑛𝜋 + 𝜋/2, where n is an integer.
6. The function is an odd function as the graph is symmetrical about the origin.
Inverse Trigonometric Function
The inverse trigonometric function is not the same as the reciprocal of the trigonometric function. Instead, the inverse trigonometric functions are found using the inverse relation of the given trigonometric functions. Some of the inverse trigonometric functions are as follows.
Inverse Sine:
y=sin−1x⇔siny=xfor−π2≤y≤π2
Inverse Cos:
y=cos−1x⇔cosy=xfor0≤y≤π
Inverse Tan:
y=tan−1x⇔tany=xfor−π2<y<π2
Common Mistakes
All the trigonometric functions (sine, cosine, tangent, cosec, secant, and cot) are positive in the first quadrant.
The trigonometry functions sine and cosecant are positive in the second quadrant.
The trigonometry functions tangent and cotangent are positive in the third quadrant.
The trigonometry functions cosine and secant are positive in the fourth quadrant.
Formulas
Law of Cosine: -
c²=a²+b²−2abcosC.b²=a²+c²−2accosB.a²=b²+c²−2bccosA.
Product to Sum Formulas
cos 𝐴 cos𝐵 = 1/2 [cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵)].
sin 𝐴 sin𝐵 = 1/2 [cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)].
sin 𝐴 cos𝐵 = 1/2 [sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)].
cos 𝐴 sin 𝐵 = 1/2 [sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵)].
For sine and cosine ratios formula
sin(x + t) = sin x cos t + cos x sin t.
sin(x − t) = sin x cos t − cos x sin t.
cos(x + t) = cos x cos t − sin x sin t.
cos(x − t) = cos x cos t + sin x sin t.
Context and Applications
This topic is significant in the professional exams for undergraduate and postgraduate courses, especially for:
B.Sc. in Mathematics
M.Sc. in Mathematics
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We've got you covered with step-by-step solutions to millions of textbook problems, subject matter experts on standby 24/7 when you're stumped, and more. | 677.169 | 1 |
2024 Geometry unit 7 polygons and quadrilaterals quiz 7 2 answer key - | 677.169 | 1 |
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Página 10 ... polygons , by more than four straight lines . 19. Of three sided figures , an equilateral triangle is that which has three equal sides . 20. An isosceles triangle is that which has only two sides equal . ΔΔΔ 21. A scalene triangle is ...
Página 31 ... right angle . SCHOLIUM . When ( Cor . 1. ) is applied to polygons , which have re - entrant angles , as ABC each re - entrant angle must be regarded as greater than two right angles . And , by joining BD , BE , BF , OF GEOMETRY . BOOK I.
Página 32 ... polygons with salient angles , which might otherwise be named convex polygons . Every convex polygon is such that a straight line , drawn at pleasure , cannot meet the contour of the polygon in more than two points . A PROP . XXXIII ...
Página 45 ... polygon ABCDE will be equivalent to the polygon ABCF , which has one side less than the original polygon . For the triangles CDE , CFE , have the base ČE common , and they are between the same paral- lels ; since their vertices D , F ...
Página 46 ... polygon ; for , by suc- cessively diminishing the number of its sides , one being retrenched at each step of the process , the equivalent triangle will at length be found . COR . Since a triangle may be converted into an equivalent | 677.169 | 1 |
First Part of an Elementary Treatise on Spherical Trigonometry
From inside the book
Page 3 ... perpendicular to each other . Moreover , the angle BOA , measured by BA , is equal to BA or h ; BOC is equal to its ... perpendicular to OA . Its intersections A'C ' and A'B ' , with the planes COA and BOA must ( 430 ) be perpendicular ...
Page 27 ... perpendicular let fall from one of its verticles upon the opposite side ; and if , in the two right triangles , the middle parts are so taken that the perpendicular is an adjacent part in both of them ; then ( 603 ) The sines of the ...
Page 28 ... perpendicular BP . First . To find PC , we know , in the right tri- angle BPC , the hypothenuse a and the angle C. Hence , by means of ( 474 ) , tang . PC cos . C tang . a . Secondly . AP is the difference between AC and PC , that is | 677.169 | 1 |
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