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You may assume the polygon formed by given points is always a simple polygon. In other words, we ensure that exactly two edges intersect at each vertex and that edges otherwise don't intersect each other. | 677.169 | 1 |
Trigonometry (11th Edition) Clone
Chapter 3 - Test - Page 138: 7a
Answer
$\theta=\frac{4}{3}$ radians
Work Step by Step
Step 1: The formula to be used here is $s=r\theta$ where $s$ is the length of the arc intercepted on a circle of radius $r$ by a central angle of measure $\theta$ radians.
Step 2: Substituting the data provided by the question into the equation, $(200)=(150)(\theta)$
Step 3: Multiplying both sides by \frac{1}{150},
$200\times\frac{1}{150}=150\theta\times\frac{1}{150}$
Step 4: Simplifying, $\theta=\frac{4}{3}$ radians. | 677.169 | 1 |
For example, a flagpole is tilted back and has a slope of -5 (with the ground being the x-axis). What is it's angle to the ground in degrees?
1 Answer
we know that #m# is the slope of the line which it makes with #x#-axis and #c# is intercept on the #y#-axis.
Also that by definition #m-=tan theta#,
where #theta# is the angle line makes with the #x#-axis.
#:. theta=tan^-1m#
For the given problem #m=-5#
This implies that the angle #theta# is more than #90^@# and lies in the second quadrant as shown below.
# theta=tan^-1(-5)#
Now using tables or a calculator we get #theta=-78.7^@#, rounded to one decimal place.
This angle is as measured from #-x# axis in the opposite direction as indictaed by #-ve# sign. | 677.169 | 1 |
Concepts Covered - 1
Basic Terms and Definitions of Lines and Angles
Basic Terms and Definitions
The two simplest objects in geometry are points and lines.
Point: A point is a coordinate that marks a position in space (on a number line, on a plane or in three dimensions or even more) and is denoted by a dot. Points are usually labelled with a capital letter.
Line: A line is a continuous set of coordinates in space and can be thought of as being formed when many points are placed next to each other. Lines can be straight or curved, but are always continuous. This means that there are never any breaks in the lines. The endpoints of lines are labelled with capital letters.
Lines are labelled according to the start point and end point. We call the line that starts at a point A and ends at a point B, AB. Since the line from point BB to point A is the same as the line from point A to point B, we have that AB=BA.
Line Segment: A part (or portion) of a line with two end points is called a line-segment.
If three or more points lie on the same line, they are called collinear points; otherwise they are called non-collinear points.
Ray: A ray is a directed line segment. It consists of one point on a line and all points extending in one direction from that point. The first point is called the endpoint of the ray. We can refer to a specific ray by stating its endpoint and any other point on it.
Note that the line segment AB is denoted by and its length is denoted by AB. The ray AB is denoted by and a line is denoted by .
However, in this chapter, we will not use these symbols, and will denote the line segment AB, ray AB, length AB and line AB by the same symbol, AB. The meaning will be clear from the context.
Angle: An angle is the union of two rays having a common endpoint. The endpoint is called the vertex of the angle, and the two rays are the sides of the angle.
The angle in below figure is formed from and . Angles can be named using a point on each ray and the vertex, such as angle DEF, or in symbol form .
Angle creation is a dynamic process. We start with two rays lying on top of one another. We leave one fixed in place, and rotate the other. The fixed ray is the initial side, and the rotated ray is the terminal side. In order to identify the different sides, we indicate the rotation with a small arrow close to the vertex as in figure given below. | 677.169 | 1 |
Honors Geometry Companion Book, Volume 1
6.2.1 Properties of Special Parallelograms Key Objectives • Prove and apply properties of rectangles, rhombuses, and squares. • Use properties of rectangles, rhombuses, and squares to solve problems. Key Terms • A rectangle is a quadrilateral with four right angles. • A rhombus is a quadrilateral with four congruent sides. • A square is a quadrilateral with four right angles and four congruent sides.
Theorems, Postulates, Corollaries, and Properties • Theorem If a quadrilateral is a rectangle, then it is a parallelogram. • Theorem If a parallelogram is a rectangle, then its diagonals are congruent. • Theorem If a quadrilateral is a rhombus, then it is a parallelogram. • Theorem If a parallelogram is a rhombus, then its diagonals are perpendicular. • Theorem If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles. Example 1 Craft Application This is a property of a rectangle. If a quadrilateral is a rectangle, then it is a parallelogram.
This is a property of a rectangle. If a parallelogram is a rectangle, then its diagonals are congruent.
A property of rectangles is used to find the length of the sides of a quilt patch in this application example. It is given that the patch is a rectangle. The length of one side, EH , and the length of a diagonal, EG , are given. By the properties of rectangles, opposite sides are congruent, so FG ≅ EH . EH is given as 12 in., so by the definition of congruent line segments, FG is also 12 in. | 677.169 | 1 |
NareshPro
laura.reeshughes
This is a nice powerpoint which is simply one slide of questions on working out the area of triangles, parallelograms and trapeziums. You could use it as a consolidation activity in class, or you could copy and paste the questions into another resource, such as a game of bingo, or questions to use with mini-whiteboards. The answers are not provided so you will need to work these out | 677.169 | 1 |
The value of
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Class 8-9-10, JEE & NEET
If $\vec{A}, \vec{B}, \vec{C}$ are mutually perpendicular, show that $\vec{C} \times(\vec{A} \times \vec{B})=0$. Is the converse true?
Solution:
A, B and C are mutually perpendicular vectors. Now, if we take cross product between any two vectors, the resultant vector will be in parallel to the third vector, as there are only three axis perpendicular to each other.
So if we consider $(A \times B)$, then it is parallel to $C$, and so angle between the resultant vector and $C$ is $0^{\circ}$, and $\sin \left(0^{\circ}\right)=0$. So, $C \times(A \times B)=0$ | 677.169 | 1 |
Elementary Trigonometry
From inside the book
Results 6-10 of 27
Page 17 ... centre of any circle , and for its denominator the radius of that circle . Let EOD be any angle . About O as centre and with any radius , describe a circle cutting OE in A , and OD in R. LA Make angle AOP equal to the unit of circular ...
Page i ... . The Circular Measure . 32. In this method , which is chiefly used in the higher branches of Mathematics , the unit of angular measurement may be described as ( 1 ) The angle subtended at the centre of ON THE MEASUREMENT OF ANGLES . 15.
Page ii James Hamblin Smith. ( 1 ) The angle subtended at the centre of a circle by an arc equal to the radius of the circle , or , which is the same thing , as we proved in Art . 21 , as ( 2 ) The angle whose magnitude is the 7th part of two ...
Page iii ... centre of any circle , and for its denominator the radius of that circle . Let EOD be any angle . About O as centre and with any radius , describe a circle cutting OE in A , and OD in R. R A Make angle AOP equal to the unit of circular ...
Page 8 ... centre of the circle . 13. If the angles of a triangle are in Arithmetical Progression , shew that one of them is 60o . 14. Determine in grades the magnitude of the angle sub- tended by an arc two feet long at the centre of a circle ...
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Page 178 202 - An Analysis of the Exposition of the Creed, written by the Right Rev. Father in God, JOHN PEARSON, DD, late Lord Bishop of Chester. Compiled, with some additional matter occasionally interspersed, for the use of the Students of Bishop's College, Calcutta, by WH MILL, DD late Principal of Bishop's College, and Regius Professor of Hebrew in the University of Cambridge.
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Draw 6 Circles In A 5 Group
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Web explore math with our beautiful, free online graphing calculator. Make groups of 5 circles. Then write a fraction to name the shaded part of the groups Web a venn diagram represents each set by a circle, usually drawn inside of a containing box representing the universal set. Decompositions of 6, 7, and 8 into.
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6 circles in a circle
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Web Explore Math With Our Beautiful, Free Online Graphing Calculator.
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How Many Circles Are There?
Fill in the number bond. Then draw a circle tangent to this circle and two other sides. Web explore math with our beautiful, free online graphing calculator. Starting with a triangle, draw a circle touching two sides. | 677.169 | 1 |
A circle has its Centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.
Answers (1)
Answer. [True] Solution. The centre of the circle is O (0, 0).
If point P(5, 0) lies on the circle then the distance between O(0, 0) and P(5, 0) is the radius of the circle
OP = 5
Radius of circle = 5
If point Q (6, 8) is outside the circle then the distance between O(0, 0) and Q(6, 8) is grater then the radius of the circle
(x1, y1) = (0, 0) (x2, y2) = (6, 8)
Here point OQ is greater than the radius of circle
Hence, point Q(6, 8) lies outside the circle | 677.169 | 1 |
Answer to Question #137433 in Civil and Environmental Engineering for Zero
A field is in the form of a regular pentagon. It is required to determine the directions of bounding sides which are referenced from an assumed meridian 05°30' to the right (easterly) of the true meridian. If the assumed bearing of side AB is N 33°20'W, determine the true azimuths (from south) of the following sides of the field: AB, BC, CD, DE and EA. Assume that the corners are labeled in a counterclockwise direction.
Tabulate answers in a convenient format.
1
Expert's answer
2020-10-15T10:37:37-0400
Dear Zero | 677.169 | 1 |
The area of a triangle ABC is 24. D,
E, and F are the midpoints of BC, AC, and AB, respectively.
Perpendiculars from E to AB and F to AC meet at G,
perpendiculars from F to BC and D to AB meet at H,
perpendiculars from D to AC and E to BC meet at M. Find the area
of the hexagon DMEGFH. | 677.169 | 1 |
Lesson Plan: Visualizing Slope on a Graph
Lesson Objectives
Standards
CCSS.MATH.CONTENT.8.EE.B.6: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane.
CCSS.MATH.CONTENT.8.F.B.4:Explore (10 minutes)
Distribute graph paper and pencils to students. Provide them with a set of coordinates and ask them to plot the points on the coordinate plane. Then, have them connect the points to form a line. Encourage students to observe the steepness or flatness of the line they have drawn.
Use this Desmos activity to have students explore slope as the ratio of rise over run. Students click and drag on the two points and then use their understanding of right triangles to find the ratio of the rise over the run. | 677.169 | 1 |
I'm working on a programming project, in this project I'm receiving an angle as a quaternion value, I partially understand how they work but I don't find any math to get the values I need.
What I would need is the angle between a fictional line/vector going to the the quaternion point from the origin (yes I know what you are thinking, but I couldn't think of a better explanation) and the "earth" a plane that is perpendicular to the gravitational vector, in this case one of your planes of reference.
Also I would need to get the rotation of the line/vector, this time the rotation should be according to the plane perpendicular to itself.
If possible all angles should be described as an angle between -180° and 180° (that's were my troubles are from.
In this picture γ complementary angle of the first questing and R is the secondary angle. Angles
\$\begingroup\$When you say "the quaternion point", do you mean a vector representing the axis of rotation specified by the quaternion, or the image of a particular input vector (say, a standard "forward" vector) after rotation by the quaternion?\$\endgroup\$ | 677.169 | 1 |
Question Video: Using Relationship between the Sines and Cosines of Complementary Angles to Find the Value of a Trigonometric Function
Mathematics • First Year of Secondary School
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Find csc 𝛽 given 𝛼 and 𝛽 are two complementary angles, where sec 𝛼 = 5/4.
02:04
Video Transcript
Find csc 𝛽 given 𝛼 and 𝛽 are two
complementary angles, where sec 𝛼 is equal to five-quarters.
We begin by recalling that two
complementary angles sum to 90 degrees. In this question, this means that
𝛼 and 𝛽 sum to 90 degrees. Subtracting 𝛼 from both sides of
this equation, we have 𝛽 is equal to 90 degrees minus 𝛼. We are trying to calculate csc of
𝛽. Therefore, this must be equal to
the csc of 90 degrees minus 𝛼. As we are told in the question that
the sec of 𝛼 is five-quarters, we'll need to rewrite our expression using the
reciprocal and cofunction identities.
One of the cofunction identities
states that the sin of 90 degrees minus 𝜃 is equal to the cos of 𝜃. Since the cosecant and secant
functions are the reciprocal of the sine and cosine functions, respectively, we have
csc 𝜃 is equal to one over sin 𝜃 and sec 𝜃 is one over cos 𝜃. This means that the csc of 90
degrees minus 𝜃 is equal to the sec of 𝜃.
In this question, we have the csc
of 90 degrees minus 𝛼, which is equal to sec 𝛼. And as already mentioned, we are
told that this is equal to five-quarters. If 𝛼 and 𝛽 are two complementary
angles where the sec of 𝛼 is five-quarters, then the csc of 𝛽 must also equal
five-quarters. | 677.169 | 1 |
A triangle ABC is drawn to circumscribing a circle of radius 3cm such that segments BD and DC into BC is divided by the point of contact D are of length 9cm. And 3 cm. Respectively. Find the sides AB and AC. | 677.169 | 1 |
Triangle Transformation
First of all draw a right triangle, then cut it in four pieces that are right triangle too. Then it doesn't matter which one you put first because it is the same size, and for me, I made the length of the right triangle 6cm and 8.5cm which that means it is an isosceles triangle. I made 4 tiny triangles and the length is 4.25 and 3 cm.
Letizia from Bangkok Patana in Thailand sent in the following:
To complete this challenging task, I drew an isosceles triangle (with a ruler) on my squared paper to make it easier. Then I cut my triangle out, so I could do the next step. After that, I got my isosceles triangle and cut it in half and then I attached the two pieces back together so now
instead of being an isosceles triangle it was a rectangle.
Here are pictures to show step by step how I completed this process.
These pupils from Broughton Anglican College in Australia sent in their ideas
Adeola
To complete this activity what I did was I found a triangle and ruled then cut it and I then equally halved it then put two sides together to make a rectangle. Remember a rectangle is technically a four-sided shape!
Ebony
The answer is yes because if you get a triangle and cut it in half to make two triangles and put them together it will make a rectangle because two triangles always make a rectangle.
Austin
Cut your triangle into a quadrilateral triangle then cut it right through the middle and then put the to sides back together.
Sam
Cut out a rectangle then gut the rectangle from corner to corner diagonally done.
Laiba from Doha College in Qatar sent in the following excellent work:
There are many different ways to cut and rearrange the pieces of a rectangle into a square. The least amount of pieces that you would need to cut your rectangle into would be two.
If you start with a rectangle with dimensions of the ratio 4:1, you can cut it in half vertically and put one half on top of the other to make a square. This is the only way that you can make a square out of a rectangle by cutting it into only two pieces.
EDSR from the Priestlands School in the United Kingdom sent in the following:
These are the triangles I am working with. The lines inside the triangles are the places you should cut
These are the triangles rearranged; as you can see these rectangles used to be triangles:
Answer to Question: Out of all the Triangles I have worked with, you can always make a rectangle by cutting them fewer than four times.
We had four solutions from Norwich Lower School. Firstly, here's what Poppy and George did:
We found all different triangles the equilateral, the isosceles, the scalene and the right angle. We started with an isosceles and used four small triangles from the big triangles to make a rectangle. The scalene took us quite a lot of time as we tried equal pieces but then realised we needed irregular pieces. We think it is possible as we think we got them
all. The rest were quite similar.
From Tom and Isaac: We started with an isosceles triangle then we cut it into four pieces then we managed to make it into a rectangle.
From Else and William:
First, we cut out a big equilateral triangle and drew a rectangle, cut the rectangle out as well as cutting the rectangle in smaller pieces then put it together.
Finally from William and India:
We started with an isosceles triangle and then we split it into two pieces. We then adjusted it and made it into a small rectangle
From a small school in the very north of Scotland - Scourie Primary - we had these four pictures. You can see then enlargedhere.jpg.
Their teachers said that although they didn't have time to find a method that would work for every triangle, they decided to split up the task so each experimented with a different idea | 677.169 | 1 |
Curved line shortest distance between two points?
In summary, Einstein proposed the concept of geodesics, which states that a curved line may be shorter than a straight line in spaces with curvature. This applies to scenarios such as walking across a curved field or flying on intercontinental routes. This concept was not introduced by Einstein himself, but he did mention it in his theory of general relativity.
Jan 3, 2017
#1
ChrisisC
54
4Imagine the field isn't flat, but 200 yards high. Would you still walk over the grass? This is basically what it is about in spaces with curvature: a giant heap in the middle. Or take the earth. There is no way directly through it. Masses bend spacetime and thus it has a curvature and the shortest way isn't a straight line anymore.
In the book "Hyperspace" by Kaku, Kaku says that Einstein said a curved line is less distance than a straight line. I'm assuming Kaku said this because it refers to GR. I could be wrong.
Could also mean the pseudo-Euclidean line interval in Minkowski space time.
Related to Curved line shortest distance between two points?
1. What is a curved line?
A curved line is a line that deviates from a straight path. It can be represented by a mathematical equation and can take on various shapes such as circles, ellipses, and parabolas.
2. Why is the shortest distance between two points on a curved line important?
The shortest distance between two points on a curved line is important because it allows us to find the most efficient path between those two points. This is useful in many fields including engineering, transportation, and navigation.
3. How is the shortest distance between two points on a curved line calculated?
The shortest distance between two points on a curved line can be calculated using the shortest distance formula, which takes into account the curvature of the line. This formula involves using calculus and can be solved using various methods such as differentiation or integration.
4. Can the shortest distance between two points on a curved line be shorter than the straight-line distance?
Yes, the shortest distance between two points on a curved line can be shorter than the straight-line distance. This is because the curved line may take a more direct path between the two points, even though it may appear to be longer when graphed on a two-dimensional plane.
5. Are there any real-life applications of calculating the shortest distance between two points on a curved line?
Yes, there are many real-life applications of calculating the shortest distance between two points on a curved line. Some examples include finding the most efficient flight path for airplanes, determining the optimal route for a train, and designing curved roads for transportation systems. | 677.169 | 1 |
The Elements of Euclid with Many Additional Propositions and Explanatory Notes
THEOREM. From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it: and there can be but one perpendicular to a plane from a point above the plane.
DEMONSTRATION.
For,
if it be possible, let the two straight lines AB, AC be at right angles to a given plane, from the same point A in the plane, and upon the same side of it. Let a plane pass through BA, AC; the common section of this with the given plane is a straight line passing through A (a): let DAE be their common section: therefore the straight lines AB, AC, DAE are in one plane: and because CA is at right angles to the given plane, it makes
right angles with every straight line meeting it in that plane (6): but DAE, which is in that plane, meets CA; therefore CAE is a right angle: for the same reason, BAE is a right angle; wherefore the angle CAE is equal to the angle BAE (c); and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane; for if there could be two, they would be parallel to one another (d); which is absurd.
PROPOSITION XIV.
THEOREM.-If the same straight line (AB) is perpendicular to each of two planes (CD, EF), they are parallel to one
another.
DEMONSTRATION. If not, they shall meet one another when produced: let them meet; their common section is a straight line
GH, in which take any point K, and join AK, BK. Then, because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK, which is in that plane (a); therefore ABK is a right angle: for the same reason BAK is a right angle; wherefore the two angles ABK, BAK of the triangle ABK, are equal to two right angles; which is impossible (6): therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallel (c.
PROPOSITION XV.
THEOREM.-If two straight lines (AB, BC) meeting one another, be parallel to two other straight lines (DE, EF) which meet one another, but are not in the same plane with the first two, the plane which passes through these is parallel to the plane passing through the others.
DEMONSTRATION.
From the point B, draw BG perpendicular to the plane which passes through DE, EF (a), and let it meet that plane in G; and through G, draw GH parallel to ED, and GK parallel to EF (b). And because BG is perpendicular to the plane through DE, EF, it makes right angles with every straight
K
line meeting it in that plane (c): but the straight lines GH, GK in that plane meet it; therefore each of the angles BGH, BGK is a right angle: and because BA is parallel to GH (d) (for each of them is parallel to DE, and they are not both in the same plane with it), the angles GBA, BGH are together equal to two right angles (e): and BGH is a right angle; therefore also
GBA is a right angle, and GB perpendicular to BA: for the same reason, GB is perpendicular to BC; since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B; GB is perpendicular to the plane through BA, BC (f): and it is perpendicular to the plane through DE, EF (g); therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line is perpendicular, are parallel to one another (h); therefore the plane through AB, BC, is parallel to the plane through DE, EF.
PROPOSITION XVI.
THEOREM.-If two parallel planes (AB, CD) be cut by another plane (EF, GH), their common sections (EF, GH), with it are parallels.
DEMONSTRATION. For, if it is not, EF, GH shall meet if produced either on the side of FH, or EG. First, let them be produced on the side of FH, and meet in the point K: therefore, since EFK is in the plane AB, every point in EFK is in that
plane (a): and K is a point in EFK; therefore K is in the plane AB: for the same reason, K is also in the plane CD; wherefore the planes AB, CD, produced, meet one another: but they do not meet, since they are parallel by the hypothesis; therefore the straight lines EF, GH do not meet when produced on the side of FH: in the same manner it may be
proved, that EF, GH do not meet when produced on the side of EG. But straight lines which are in the same plane, and do not meet, though produced either way, are parallel; therefore EF is parallel to GH.
PROPOSITION XVII.
THEOREM.-If two straight lines be cut by parallel planes, they shall be cut in the same ratio.
DEMONSTRATION. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: as AE is to EB, so shall CF be to FD.
K4
Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel (a): for the same reason, because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel: and because EX is parallel to BD, a side of the triangle ABD; as AE to EB, so is AX to XD (6): again, because XF is parallel to AC, a side of the triangle ADC; as AX to XD, so is CF to FD; and it was proved, that AX is to XD, as AE to EB; therefore, as AE to EB, so is CF to FD (c).
MA
(a) XI. 16. (b) VI. 2. (c) V. 11.
ROPOSITION XVIII.
THEOREM.-If a straight line (AB) be at right angles to a plane (CK), every plane which passes through it shall be at right angles to that plane.
DEMONSTRATION. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG, in the plane DE, at right angles to CE (a): and because AB is perpendicular to the plane CK, therefore it is also perpendicular to every straight line in that plane meeting it (b), and consequently it is perpendicular to CE; wherefore ABF is a right angle; but GFB
is likewise a right angle (c); therefore AB is parallel to FG (d); and AB is at right angles to the plane CK; therefore FG is also at right angles to the same plane (e). But one plane is at right angles to another plane, when the straight lines drawn in one of the planes at right angles to their common section, are also at right angles to the other plane (f); and any straight line FG in the plane DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner it may be proved, that all planes which pass through AB, are at right angles to the plane CK.
PROPOSITION XIX.
THEOREM.-If two planes (AB, CD) which cut one another be each of them perpendicular to a third plane, their common section (BD) shall be perpendicular to the same plane. | 677.169 | 1 |
How to construct an isosceles triangle given the altitude to the base
How to construct an isosceles triangle given the altitude to the base and one of the base angles, where the base is the side that is not congruent to any other side. What would the steps be for this construction and the proof that it would work? | 677.169 | 1 |
Question 111.
a) Sin (A + B) = Sin A Cos B + Cos A Sin B
b) Sin (A – B) = Sin A Cos B – Cos A Sin B
Which is correct in the above statements?
A) a is correct
B) b is correct
C) Both a and b are correct
D) Both a and b are wrong
Answer:
C) Both a and b are correct
Question 112.
a) Sin 30° = 1
b) Cos 30° = \(\frac {1}{2}\)
Which is correct in the above statements?
A) a is correct
B) Both a and b are correct
C) b is correct
D) Both a and b are wrong
Answer:
D) Both a and b are wrong
Question 113.
a) Sin2 θ + Cos2 θ = 1
b) Tan2 θ + Sec2 θ = 1
Which is correct in the above statements?
A) b is correct
B) a is correct
C) Both a and b are correct
D) Both a and b are wrong
Answer:
B) a is correct | 677.169 | 1 |
congruent triangles worksheet grade 9ent Triangles Worksheet #2 With Answer – Triangles are one of the most basic shapes found in geometry. Understanding the concept of triangles is essential for mastering more advanced geometric concepts. In this blog post we will discuss the various kinds of triangles Triangle angles, how to calculate the areas and perimeters of a triangle, and present the examples for each. Types of Triangles There are three types that of triangles are equilateral isosceles, and scalene. … Read more
Congruent Triangles Worksheet #1 Answer Key – Triangles are among the most fundamental designs in geometry. Knowing how triangles work is essential to learning more advanced geometric terms. In this blog it will explain the various types of triangles triangular angles, the best way to calculate the size and perimeter of a triangle and will provide illustrations of all. Types of Triangles There are three kinds of triangles: equilateral isosceles, and scalene. Equilateral triangles consist of three … Read more
Congruence Triangles Worksheet 9th Grade – Triangles are one of the fundamental shapes in geometry. Knowing how triangles work is essential to getting more advanced concepts in geometry. In this blog post we will go over the different types of triangles that are triangle angles. We will also explain how to determine the extent and perimeter of any triangle, and also provide the examples for each. Types of Triangles There are three types to triangles: … Read more | 677.169 | 1 |
In the figure, the circles with centers O and Q are externally tangent at point C, and AB is the common external tangent where points A and B are points of tangency.
If angle OQB measures 112 degrees, calculate the measure of angle BAC. | 677.169 | 1 |
Activities to Teach Students Triangle Angle – Sum Theorem
The Triangle Angle-Sum Theorem is a fundamental concept in geometry that students must understand. It states that the sum of all the angles in a triangle is equal to 180 degrees. Knowing this theorem is essential in solving various problems related to triangles. Therefore, teachers must make sure that their students have a good understanding of it. In this article, we will discuss the various activities that teachers can use to teach students the Triangle Angle-Sum Theorem.
1. The triangle game: In this game, the teacher should draw different triangles on the blackboard or whiteboard and ask individual students to come forward and calculate the angles. To make it more interesting, the teacher can divide the students into teams and make it a competition. Students can earn points for their team by correctly calculating the angles. This game is not only a fun way to learn, but it also helps students develop critical thinking and problem-solving skills.
2. Using manipulatives: Using manipulatives is an excellent way to help students visualize the concept of the Triangle Angle-Sum Theorem. The teacher can use different materials such as straws, Popsicle sticks, and cardboard to make triangles of various shapes and sizes. By manipulating these objects, students can see how the angles change when they move the vertices of the triangle. This activity helps students understand the concept better by making it more tangible and hands-on.
3. Real-world examples: Students often find it challenging to relate mathematical concepts to real-life situations. Therefore, teachers can use examples from the real world to help students connect the Triangle Angle-Sum Theorem to practical applications. For example, the teacher can use pictures of different types of roofs, such as gable, hip, or flat. Students can then analyze the pictures and calculate the angles of the triangles that make up the roofs. This activity not only makes the concept more relatable but also provides students with an opportunity to apply their knowledge in real-life situations.
4. Interactive online activities: With the rise of technology, teachers can use interactive online activities to teach the Triangle Angle-Sum Theorem. Various educational websites offer virtual manipulatives and tools that allow students to create and manipulate triangles to understand the concept visually. Additionally, students can play online games that reinforce the concept, such as math puzzles and quizzes.
In conclusion, teaching the Triangle Angle-Sum Theorem is an important part of geometry that students must understand. By using interactive and engaging activities such as games, manipulatives, real-life examples, and online tools, teachers can help students develop a better understanding of the concept. With these tools, students will not only master the Triangle Angle-Sum Theorem but also develop critical thinking and problem-solving skills necessary for their academic and future | 677.169 | 1 |
angles-in-triangles
ANGLES IN TRIANGLES
The angle formed between two adjacent sides of a triangle is called an interior angle. The sum of all of the measures of the interior angles in a triangle is 180∘.
An angle formed between a side of the triangle and an adjacent side extending outward is called an exterior angle. The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
In the above diagram a, b, and c are the interior angles. x, y, and z are the exterior angles | 677.169 | 1 |
Lesson 7 3 Proving Triangles Similar AA SSS
AA Similarity (Angle-Angle) If 2 angles of one triangle are congruent to 2 angles of another triangle, then the triangles are similar. Given: and Conclusion: 2
SSS Similarity (Side-Side) If the measures of the corresponding sides of 2 triangles are proportional, then the triangles are similar. 5 8 11 10 16 22 Given: Conclusion: 3
SAS Similarity (Side-Angle-Side) If the measures of 2 sides of a triangle are proportional to the measures of 2 corresponding sides of another triangle and the angles between them are congruent, then the triangles are similar. 5 10 11 22 Given: Conclusion: 4
Proving Triangles Similarity is reflexive, symmetric, and transitive. Steps for proving triangles similar: 1. Mark the Given. 2. Mark … Shared Angles or Vertical Angles 3. Choose a Method. (AA, SSS , SAS) Think about what you need for the chosen method and be sure to include those parts in the proof. 5 8 | 677.169 | 1 |
The hexagonal-hexagonal prismantiprismoid or hihipap, also known as the edge-snub hexagonal-hexagonal duoprism or 6-6 prismantiprismoid, is a convex isogonalpolychoron that consists of 6 hexagonal antiprisms, 6 hexagonal prisms, 12 ditrigonal trapezoprisms, and 36 wedges. 1 hexagonal antiprism, 1 hexagonal prism, 2 ditrigonal trapezoprisms, and 3 wedges join at each vertex. It can be obtained through the process of alternating one class of edges of the dodecagonal duoprism so that one ring of dodecagons become ditrigons. However, it cannot be made uniform, as it generally has 4 edge lengths, which can be minimized to no fewer than 2 different sizes.
Using the ratio method, the lowest possible ratio between the longest and shortest edges is 1:7+33+384+174322{\displaystyle {\frac {7+3{\sqrt {3}}+{\sqrt {384+174{\sqrt {3}}}}}{22}}} ≈ 1:1.74436. | 677.169 | 1 |
The vertices
P
and
Q
have the same
x
-coordinates. So, the distance between the two points is the
absolute value
of the difference between their
y
-coordinates.
That is,
P
Q
=
|
6
−
2
|
=
4
.
The vertices
P
and
R
have the same
y
-coordinates. So, the distance between the two points is the absolute value of the difference between their
x
-coordinates.
That is,
P
R
=
|
−
4
−
1
|
=
5
.
Example 2:
The three points marked are the vertices of a rectangle. Find the fourth vertex.
The points
A
and
B
have the same
x
-coordinates, so the side
A
B
¯
is a vertical line. Also, the points
A
and
C
have the same
y
-coordinates and the side
B
C
¯
is a horizontal line.
Opposite sides of a rectangle are congruent. The fourth vertex, say
D
, will have the same
x
-coordinate as
C
, and it will be
A
B
units away from
C
. Also,
D
will have the same
y
-coordinate as
A
and it will be
B
C
units away from
A
. That is, the vertex
D
will have the coordinates
(
6
,
4
)
.
First, plot the vertices and join the adjacent ones to draw the rectangle.
Count the number of blocks between adjacent vertices to find the length and width of the rectangle.
The length of the rectangle is
5
and the width is
5
.
The perimeter of a rectangle is twice the sum of its length and width. Therefore, the perimeter of rectangle
L
M
N
O
is
2
(
5
+
4
)
=
18
sq. units.
Example 4:
The swimming pool and the nearby landscape of an apartment complex are laid out on a square grid. The center of the rectangular shaped swimming pool is at the origin and there is a pathway of uniform width around it. Gardens are there on all the four sides with a palm tree on each corner of the pathway. If the coordinates of the corners of the swimming pool are
(
−
4
,
3
)
,
(
4
,
3
)
,
(
4
,
−
3
)
and
(
−
4
,
−
3
)
and the uniform width of the pathway is 2 blocks, what is the longest distance between two adjacent palm trees?
First, plot the vertices on coordinate plane and join the adjacent ones to draw the rectangle that represents the pool.
Since the pathway is of uniform width of
2
blocks, the border and corners of the path way can be drawn as shown.
The longest distance between two adjacent palm trees is the length of the rectangle formed by the borders of the pathway. The coordinates of the corners are
(
−
6
,
5
)
,
(
6
,
5
)
,
(
6
,
−
5
)
, and
(
−
6
,
−
5
)
.
Thus, the length is
12
blocks. Therefore, the longest distance between two adjacent palm trees is
12
blocks | 677.169 | 1 |
What is the expected distance of any point on Earth and the north pole? Take Earth radius 1.
Clarification: Shortest distance cuts through the sphere, instead of lying on surface.
Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?
Imagine a ring of some thickness, whose distance from N is constant from all points on that ring. Get the average of all such rings.
4/3
2 * sin(x/2)is the distance of north pole from a point on the ring at angle x from the z axis. So, integrate from 0 to pi, (2 * pi * sin(x) * 2 * sin(x/2) dx) and divide by the total area which is 4 * pi.
Answer:4/3
Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole).
Area of ring = 2 * pi * dy
Probability of choosing point on this ring = dy/2
Distance of N & a point on ring = sqrt(2y)
Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3
And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question. | 677.169 | 1 |
What is Paraboloid: Definition and 83 Discussions
In geometry, a paraboloid is a quadric surface that has exactly one axis of symmetry and no center of symmetry. The term "paraboloid" is derived from parabola, which refers to a conic section that has a similar property of symmetry.
Every plane section of a paraboloid by a plane parallel to the axis of symmetry is a parabola. The paraboloid is hyperbolic if every other plane section is either a hyperbola, or two crossing lines (in the case of a section by a tangent plane). The paraboloid is elliptic if every other nonempty plane section is either an ellipse, or a single point (in the case of a section by a tangent plane). A paraboloid is either elliptic or hyperbolic.
Equivalently, a paraboloid may be defined as a quadric surface that is not a cylinder, and has an implicit equation whose part of degree two may be factored over the complex numbers into two different linear factors. The paraboloid is hyperbolic if the factors are real; elliptic if the factors are complex conjugate.
An elliptic paraboloid is shaped like an oval cup and has a maximum or minimum point when its axis is vertical. In a suitable coordinate system with three axes x, y, and z, it can be represented by the equation
z
=
x
2
a
2
+
y
2
b
2
.
{\displaystyle z={\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}.}
where a and b are constants that dictate the level of curvature in the xz and yz planes respectively. In this position, the elliptic paraboloid opens upward.
A hyperbolic paraboloid (not to be confused with a hyperboloid) is a doubly ruled surface shaped like a saddle. In a suitable coordinate system, a hyperbolic paraboloid can be represented by the equation
z
=
y
2
b
2
−
x
2
a
2
.
{\displaystyle z={\frac {y^{2}}{b^{2}}}-{\frac {x^{2}}{a^{2}}}.}
In this position, the hyperbolic paraboloid opens downward along the x-axis and upward along the y-axis (that is, the parabola in the plane x = 0 opens upward and the parabola in the plane y = 0 opens downward).
Any paraboloid (elliptic or hyperbolic) is a translation surface, as it can be generated by a moving parabola directed by a second parabola.
A question of sign. Is the curvature of Flamm's paraboloid positive or negative? If I've gotten the signs correct, it's a negative curvature. This is the opposite of the positive curvature of a sphere, and it implies that that geodesics drawn on Flamm's parabaloid should diverge. I think...
the equation of a parabola that is obtained by taking a cross-section passing through the center of the paraboloid is ##y = ax^2##
breaking the paraboloid into cylinders of height ##(dy)## the volume of each tiny cylinder is given by ##\pi x^2 dy##
since ##y = ax^2## we have ##\pi (y/a) dy##...
How to prove that every quadric surface can be translated and/or rotated so that its equation matches one of the six types of quadric surfaces namely 1) Ellipsoid
2)Hyperboloid of one sheet
3) Hyperboloid of two sheet 4)Elliptic Paraboloid
5) Elliptic Cone 6) Hyperbolic Paraboloid
The...
I am checking the divergence theorem for the vector field:
$$v = 9y\hat{i} + 9xy\hat{j} -6z\hat{k}$$
The region is inside the cylinder ##x^2 + y^2 = 4## and between ##z = 0## and ##z = x^2 + y^2##
This is my set up for the integral of the derivative (##\nabla \cdot v##) over the region...
I am trying to piece together how the parabolic mirror manages to reflect the "red dot" from the focal point to the eye without distortion.
I compare this with a conventional car headlight, which operates almost exactly the same way, except it has a non-transparent backing. Why does the ret dot...
Homework Statement
A bead slides under the influence of gravity on the frictionless interior surface of the paraboloid of revolution z = (x^2+y^2)/2a = r^2/2a Find the speed v_0 at which the bead will move in a horizontal circle of radius r_0 Find the frequency of small radial...
Hello. I solve this problem:
1. Homework Statement
The particles of mass m moves without friction on the inner wall of the axially symmetric vessel with the equation of the rotational paraboloid:
where b>0.
a) The particle moves along the circular trajectory at a height of z = z(0)...
I would like to know if there exist any equations in Cartesian coordinates that describe the shape in three dimensions of Flamm´s paraboloid and if you can write them to me because I have searched for them but I can't find any specific equations of what I want. I suppose that this shape would...
Homework Statement
Sketch the surface of a paraboloid z=9-x2 -92 in 3-dimensional xyz-space
Homework Equations
I assume partial derivatives are involved in some manner
The Attempt at a Solution
[/B]
I attempted to solve by making each variable equal to zero... That didn't work xD. I would...
The shape of the Einstein-Rosen bridge is often visualized/modelized with the Flamm's paraboloid, and many other references have also stated clearly that it's a "surface of revolution of a parabola".
But as far as I can see, when we rotate the parabola w^2 = 8M(r-2M) (in natural units c=G=1)...
Homework Statement
[/B]
The diameter and depth of an antenna that is
shaped like a paraboloid are ##d = 2.0m## and ##s = 0.5m## respectively. The antenna is set up so that its axis of symmetry is at an angle ##\theta = 30^{\circ}## from it's usual vertical orientation. How much can the antenna...
Homework Statement
So I am trying to accomplish the above by using spherical coordinates, I am aware the problem may be solved using dv=dxdydz= zdxdy were z is known but I would like to try it using a different approach (using spherical coordinates). Any help would be greatly appreciated...
Homework Statement
I am only currently in multivariate calculus, so i haven't even touched differential geometry yet, but a question that i had while learning about gradients came up and it led me to the topic of geodesics and differential geometry, so here goes:
Class problem: Find the...
Homework Statement
The angular velocity is ω, R is the radius of the vessel. at rest the water has depth H.
The face of the water form a paraboloid y=Ax2. find R for which the maximum height h of the water above the bottom doesn't depend on ω.
Homework Equations
Centripetal force: ##F=m...
Evaluate the triple integral ∫∫∫E 5x dV, where E is bounded by the paraboloid
x = 5y2+ 5z2 and the plane x = 5.
My work so far:
Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y2+5z2 ≤x ≤ 5 ----->...
Homework Statement
[/B]
Find the coordinates of the point P on the surface of the paraboloid z=6x2+6y2-(35/6) where the normal line to the surface passes through the point (25/6, (25√22)/6, -4). Note that a graphing calculator may be used to solve the resulting cubic equation.
Homework...
Homework Statement
Let F = <x, z, xz> evaluate ∫∫F⋅dS for the following region:
x2+y2≤z≤1 and x≥0
Homework Equations
Gauss Theorem
∫∫∫(∇⋅F)dV = ∫∫F⋅dS
The Attempt at a Solution
This is the graph of the entire function:
Thank you Wolfram Alpha.
But my surface is just the half of this...
I will state the specifics to this problem if necessary.
I need to find the parametric equations for the the tan line at point, P(x1,y1,z1) on the curve formed from paraboloid intersection with ellipsoid.
The parametric equations for the level surfaces that make up paraboloid and ellipsoid...
Need a 2nd opinion on my solution.
Homework Statement
A point mass moves frictionlessly in a circle inside a parabolic cup, with the radius at the top being R. The particle's position vector makes an angle theta wrt the center of symmetry (generatrix going from -z to +z).Homework Equations...
Homework Statement
A surface S in three dimensional space may be specified by the equation
f(x, y, z) = 0, where f(x, y, z) is a real function. Show that a unit vector nˆ normal to
the surface at point (x0, y0, z0) is given by
Homework Equations
The Attempt at a Solution
r...
Hi Folks,I have come across some text where f(x,y)=c_1+c_2x+c_3y+c_4xy is used to define the corner pointsf_1=f(0,0)=c_1
f_2=f(a,0)=c_1+c_2a
f_3=f(a,b)=c_1+c_2a+c_3b+c_4ab
f_4=f(0,b)=c_1+c_3bHow are these equations determined? F_1 to F_4 starts at bottom left hand corner and rotates counter...
Homework Statement
Find the distance from the paraboloid z = X2 + 2Y2 to the plane
2X + 8Y + Z = -8.
Homework Equations
The partial derivatives with respect to X, And Y for the paraboloid.
The Attempt at a Solution
My professor said we need to find the point where the...
If the hyperbolic paraboloid z=(x/a)^2 - (y/b)^2
is rotated by an angle of π/4 in the +z direction (according to the right hand rule), the result is the surface
z=(1/2)(x^2 + y^2) ((1/a^2)-((1/b^2)) + xy((1/a^2)-((1/b^2))
and if a= b then this simplifies to
z=2/(a^2) (xy)
suppose...
Homework Statement
Use stokes theorem to elaluate to integral \int\int_{s} curlF.dS where F(x,y,z)= x^2 z^2 i + y^2 z^2 j + xyz k and s is the part of the paraboliod z=x^2+ y^2 that lies inside the cylinder x^2 +y^2 =4 and is orientated upwards
Homework Equations
The Attempt at a...
Homework Statement
a elliptic paraboloid is x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h. Its apex occurs at the point (0,0,h). Suppose a>=b. Calculate the volume of that part of the paraboloid that lies above the disc x^2+y^2<=b^2.:confused:
2. The attempt at a solution
We normally do the...
Homework Statement
a elliptic paraboloid is x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h, show that the horizontal cross-section at height z, is an ellipse
Homework Equations
The Attempt at a Solution
i don't know how to prove this, i only know that the standard ellipse is...
Homework Statement
Verify Stokes' Theorem for F(x,y,z)=(3y,4z,-6x) where S is part of the paraboloid z=9-x2-y2 that lies above the xy-plane, oriented upward.
Homework Equations
Stokes' Theorem is ∫F*ds=∫∫scurl(F)*ds
Where curl(F)=∇*F
The Attempt at a Solution
I got...
Homework Statement
Find the equation of the surface that is equidistant from the plane x=1, and the point (-1,0,0).
The Attempt at a Solution
Okay, if I set the distance from the surface to the point, and the distance from the surface to the plane as being equal, I should have the...
Homework Statement
Find the moment of inertia of a paraboloid f(x,y)=x^2+y^2 whose density function is ρ(r)=cr=dm/dv. use mass M and height H to express your answer
2. The attempt at a solution
I took the double integral ∫∫r^2 ρ r dr dθ to find the I of a single disk as a function of r...
Homework Statement
Find the volume of the solid E bounded by z = 3+x2 +y2 and z = 6.
Homework Equations
The Attempt at a Solution
I'm going to use cylindrical coordinates. So, I have,
z = 3 + r2
Clearly, my bounds on z are 3 and 6. If I project the intersection of the...
Homework Statement
Where does the normal line to the paraboloid z=x^2+y^2 at the point (1,1,2) intersect the paraboloid a second time?
Homework Equations
The Attempt at a Solution
I found the normal line to be 0=2x+2y-1, but I'm not sure what to do next.
Hi there, I have to compute the surface area for
V:\{ -2(x+y)\leq{}z\leq{}4-x^2-y^2 \}
I have a problem on finding the surface area for the paraboloid limited by the plane. I've parametrized the plane in polar coordinates, I thought it would be easier this way, but also tried in cartesian...
Homework Statement
Find the volume bounded by the paraboloid z= 2x2+y2 and the cylinder z=4-y2. Diagram is included that shows the shapes overlaying one another, with coordinates at intersections. (Will be given if necessary)
Homework Equations
double integral? function1-function2...
The Problem: I have a paraboloid open along the positive z-axis, starting at the origin and ending at z = 100. At z=100, the horizontal surface is a circle with a radius of 20. Water is flowing through the paraboloid with the velocity F = 2xzi - (1100 + xe^-x^2)j + z(1100 - z)k. I'm asked to...
I just want to see if my logic is sound here. If we have the paraboloid z=x2+y2 from z=0 to z=1, and I wanted a parametric form of that I think this should work for polar coordinates:
\vec{r}(u,v)=(vcosu,vsinu,v^{2})
u:[0..2\pi],v:[0..1]
Does this make sense?
Homework Statement
Hi, I am trying to solve the following problem, and seem to just be going in circles.
A sphere of radius=4 is "dropped" into a paraboloid with equation z=(x^2)+(y^2).
Find the distance "a" from the origin to the center of the sphere at the point where it will
"get stuck" or...
Homework Statement
The volume of the solid below the plane: z=2x and above the paraboloid z=x^2 + y^2.
I need help setting this one up, I can handle the evaluating.
The Attempt at a Solution
I just don't know.
Homework Statement
Find an equation of the form Ax2+By2+Cz2+Dxy+Exz+Fyz+Gx+Hy+Jz+K=0 Satisfied by the set of all points in space, (x,y,z), whose distance to the origin is equal to their distance to the plane x+y+z=3. Based on what you know about parabolas, what does this collection of points... | 677.169 | 1 |
--------------- Draw a circle with a sector subtending 120 degrees at the center. Then draw an inscribed circle. Let its raius be r Now the r will be tangent to the radius of the bigger circle leading to 30-60-90 triangle. sin (60) = r/10-r since sin 60 = sqrt(3)/2 [ 1.75 MiB | Viewed 3301 times ]as VeritasPrepKarishma has pointed out it is a 30-60-90 triangle.. Another way to know that - the center of the inscribed angle has to be on the bisector of this sector , so half of 120 = 60 degree .. Also the two radius containing this sector will be tangential to the circle so 90 degree..
If I am short of time, i would use the choices for my benefit.. the center of the inscribed angle has to be on the bisector of this sector, which itself is equal to the radius .. radius is 10, and the centre of the inscribed circle will be closer to the arc, so the radius will be LESS than 10/2 or 5 ONLY C is less than 5, therefore our answerHi Karishma I understood that 10 is the radius of the sector so that is why it is 3‾√x+2x=10I did not get how did you take 3‾√x+2x=10 . So far I understand that it is because of the c^2=a^2+b^2, but it is not mentioned that C^2 will be 10. I am missing something.
Look at the diagram above. DO + AO is the radius of the sector. This is given to be 10. So \(2x + \sqrt{3}x = 10\) what is the radius (in cm) of the biggest possible circle [#permalink] | 677.169 | 1 |
Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions
Find the value of sin² 45 + cos² 45.
01:21
Video Transcript
Find the value of sine squared 45 plus cos squared 45.
To help us find the value of sine squared 45 plus cos squared 45, we're actually gonna use one of our Pythagorean trigonometric identities. I'm just gonna show you three of them and then I'll show you the one that we're actually gonna be using.
And now to actually solve this problem, find the values of sine squared 45 plus cos squared 45, we're actually gonna be using this top one, which is sine squared 𝜃 plus cos squared 𝜃 is equal to one. And this is actually sometimes known as the fundamental Pythagorean trigonometric identity.
So if you see how we're going to use it, well actually the way we gonna use it is that our 𝜃 is gonna be equal to 45. Cause as you can see in the identity, we've got sine squared 𝜃 plus cos squared 𝜃 equals one, so still the same 𝜃 in each part, where in ours we've got 45 in each part.
So therefore, we can say that sine squared 45 plus cos squared 45 must be equal to one. And we've actually achieved that because we've used our top Pythagorean trigonometric identity, which is sin squared 𝜃 plus cos squared 𝜃 equals one.
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congruent triangle proofs mixed worksheet answers
Triangle Congruence Theorems Worksheet Answers – Triangles are among the most fundamental geometric shapes in geometry. Understanding triangles is crucial for understanding more advanced geometric principles. In this blog, we will cover the various types of triangles, triangle angles, how to determine the dimensions and the perimeter of a triangle, and show illustrations of all. Types of Triangles There are three kinds of triangles: equilateral isoscelesand scalene. Equilateral triangles consist of three equal sides and three … Read more | 677.169 | 1 |
Pythagoras was a Greek mathematician and philosopher, born on the island of Samos (ca. 582 BC). He founded a number of schools, one in particular in a town in southern Italy called Crotone, whose members eventually became known as the Pythagoreans. The inner circle at the school, the Mathematikoi, lived at the school, rid themselves of all personal possessions, were vegetarians, and observed a strict vow of silence. They studied mathematics, philosophy, and music, and held the belief that numbers constitute the true nature of things, giving numbers a mystical or even spiritual quality.
Today, nothing is known of Pythagoras's writings, perhaps due to the secrecy and silence of the Pythagorean society. However, one of the most famous theorems in all of mathematics does bear his name, the Pythagorean Theorem.
PYTHAGOREAN THEOREM
Let c represent the length of the hypotenuse, the side of a right triangle directly opposite the right angle (a right angle measures 90º) of the triangle. The remaining sides of the right triangle are called the legs of the right triangle, whose lengths are designated by the letters a and b.
The relationship involving the legs and hypotenuse of the right triangle, given by
\[a^2 + b^2 = c^2 \label{1} \]
is called the Pythagorean Theorem.
Note that the Pythagorean Theorem can only be applied to right triangles. Let's look at a simple application of the Pythagorean Theorem (Equation \ref{1}).
Example \(\PageIndex{2}\)
Given that the length of one leg of a right triangle is 4 centimeters and the hypotenuse has length 8 centimeters, find the length of the second leg.
Let's begin by sketching and labeling a right triangle with the given information. We will let x represent the length of the missing leg.
Figure 1. A sketch make things a bit easier
Here is an important piece of advice.
TIP \(\PageIndex{3}\): Hypotenuse
The hypotenuse is the longest side of the right triangle. It is located directly opposite the right angle of the triangle. Most importantly, it is the quantity that is isolated by itself in the Pythagorean Theorem (Equation \ref{1}). Always isolate the quantity representing the hypotenuse on one side of the equation. The "legs" go on the other side of the equation.
So, taking the tip to heart, and noting the lengths of the legs and hypotenuse in Figure 1, we write
\(4^2+x^2 = 8^2\).
Square, then isolate x on one side of the equation.
\(16+x^2 = 64\)
\(x^2 = 48\)
Normally, we would take plus or minus the square root in solving this equation, but x represents the length of a leg, which must be a positive number. Hence, we take just the positive square root of 48.
\(x = \sqrt{48}\)
Of course, place your answer in simple radical form.
\(x = \sqrt{16}\sqrt{3}\)
\(x = 4\sqrt{3}\)
If need be, you can use your graphing calculator to approximate this length. To the nearest hundredth of a centimeter, \(x \approx 6.93\) centimeters.
Proof of the Pythagorean Theorem
It is not known whether Pythagoras was the first to provide a proof of the Pythagorean Theorem. Many mathematical historians think not. Indeed, it is not even known if Pythagoras crafted a proof of the theorem that bears his name, let alone was the first to provide a proof.
There is evidence that the ancient Babylonians were aware of the Pythagorean Theorem over a 1000 years before the time of Pythagoras. A clay tablet, now referred to as Plimpton 322 (see Figure 2), contains examples of Pythagorean Triples, sets of three numbers that satisfy the Pythagorean Theorem (such as 3, 4, 5).
Figure 2. A photograph of Plimpton 322.
One of the earliest recorded proofs of the Pythagorean Theorem dates from the Han dynasty (206 BC to AD 220), and is recorded in the Chou Pei Suan Ching (see Figure 3). You can see that this figure specifically addresses the case of the 3, 4, 5 right triangle. Mathematical historians are divided as to whether or not the image was meant to be part of a general proof or was just devised to address this specific case. There is also disagreement over whether the proof was provided by a more modern commentator or dates back further in time.
Figure 3. A figure from the Chou Pei Suan Ching
However, Figure 3 does suggest a path we might take on the road to a proof of the Pythagorean Theorem. Start with an arbitrary right triangle having legs of lengths a and b, and hypotenuse having length c, as shown in Figure 4(a).
Next, make four copies of the triangle shown in Figure 4(a), then rotate and translate them into place as shown in Figure 4(b). Note that this forms a big square that is c units on a side.
Further, the position of the triangles in Figure 4(b) allows for the formation of a smaller, unshaded square in the middle of the larger square. It is not hard to calculate the length of the side of this smaller square. Simply subtract the length of the smaller leg from the larger leg of the original triangle. Thus, the side of the smaller square has length b − a.
Figure 4. Proof of the Pythagorean Theorem.
Now, we will calculate the area of the large square in Figure 4(b) in two separate ways.
First, the large square in Figure 4(b) has a side of length c. Therefore, the area of the large square is
\(Area = c^2\).
Secondly, the large square in Figure 4(b) is made up of 4 triangles of the same size and one smaller square having a side of length b−a. We can calculate the area of the large square by summing the area of the 4 triangles and the smaller square.
The area of the smaller square is \((b−a)^2\).
The area of each triangle is \(\frac{ab}{2}\). Hence, the area of four triangles of equal size is four times this number;
i.e., \(4(\frac{ab}{2})\). Thus, the area of the large square is
Area = Area of small square + \(4 \cdot\) Area of triangle
=\((b−a)^2+4(\frac{ab}{2})\).
We calculated the area of the larger square twice. The first time we got \(c^2\); the second time we got \((b−a)^2+4(\frac{ab}{2})\). Therefore, these two quantities must be equal.
\(c^2 = (b−a)^2+4(\frac{ab}{2})\).
Expand the binomial and simplify.
\(c^2 = b^2−2ab+a^2 +2ab\)
\(c^2 = b^2+a^2\)
That is,
\(a^2+b^2 = c^2\),
and the Pythagorean Theorem is proven.
Applications of the Pythagorean Theorem
In this section we will look at a few applications of the Pythagorean Theorem, one of the most applied theorems in all of mathematics. Just ask your local carpenter.
The ancient Egyptians would take a rope with 12 equally spaced knots like that shown in Figure 5, and use it to square corners of their buildings. The tool was instrumental in the construction of the pyramids.
Figure 5. A basic 3-4-5 right triangle for squaring corners.
The Pythagorean theorem is also useful in surveying, cartography, and navigation, to name a few possibilities.
Let's look at a few examples of the Pythagorean Theorem in action.
Example \(\PageIndex{4}\)
One leg of a right triangle is 7 meters longer than the other leg. The length of the hypotenuse is 13 meters. Find the lengths of all sides of the right triangle.
Let x represent the length of one leg of the right triangle. Because the second leg is 7 meters longer than the first leg, the length of the second leg can be represented by the expression x + 7, as shown in Figure 6, where we've also labeled the length of the hypotenuse (13 meters).
Figure 6. The second leg is 7 meters longer than the first.
Remember to isolate the length of the hypotenuse on one side of the equation representing the Pythagorean Theorem. That is,
\(x^2+(x+7)^2 = 13^2\).
Note that the legs go on one side of the equation, the hypotenuse on the other. Square and simplify. Remember to use the squaring a binomial pattern.
\(x^2+x^2+14x+49 = 169\)
\(2x^2 +14x+49 = 169\)
This equation is nonlinear, so make one side zero by subtracting 169 from both sides of the equation.
\(2x^2+14x+49−169 = 0\)
\(2x^2 +14x−120 = 0\)
Note that each term on the left-hand side of the equation is divisible by 2. Divide both sides of the equation by 2.
Because length must be a positive number, we eliminate −12 from consideration. Thus, the length of the first leg is x = 5 meters. The length of the second leg is x+7, or 12 meters.
Check. Checking is an easy matter. The legs are 5 and 12 meters, respectively, and the hypotenuse is 13 meters. Note that the second leg is 7 meters longer than the first. Also,
\(5^2+12^2 = 25+144 = 169\),
which is the square of 13.
The integral sides of the triangle in the previous example, 5, 12, and 13, are an example of a Pythagorean Triple.
PYTHAGOREAN TRIPLE
A set of positive integers a, b, and c, is called a Pythagorean Triple if they satisfy the Pythagorean Theorem; that is, if
\(a^2+b^2 = c^2\).
If the greatest common factor of a, b, and c is 1, then the triple (a, b, c) is called a primitivePythagorean Triple.
Thus, for example, the Pythagorean Triple (5, 12, 13) is primitive. Let's look at another example.
Example \(\PageIndex{5}\)
If (a,b,c) is a Pythagorean Triple, show that any positive integral multiple is also a Pythagorean Triple.
Thus, if the positive integers (a, b, c) is a Pythagorean Triple, we must show that(ka, kb, kc), where k is a positive integer, is also a Pythagorean Triple.
However, we know that
\(a^2+b^2 = c^2\).
Multiply both sides of this equation by \(k^2\).
\(k^{2}a^2+k^{2}b^2 = k^{2}c^2\)
This last result can be written
\((ka)^2 + (kb)^2 = (kc)^2\).
Hence, (ka, kb, kc) is a Pythagorean Triple.
Hence, because (3, 4, 5) is a Pythagorean Triple, you can double everything to get another triple (6, 8, 10). Note that \(6^2 + 8^2 = 10^2\) is easily checked. Similarly, tripling gives another triple (9, 12, 15), and so on.
In Example 5, we showed that (5, 12, 13) was a triple, so we can take multiples to generate other Pythagorean Triples, such as (10, 24, 26) or (15, 36, 39), and so on.
Formulae for generating Pythagorean Triples have been know since antiquity.
Example \(\PageIndex{6}\)
The following formula for generating Pythagorean Triples was published in Euclid's (325–265 BC) Elements, one of the most successful textbooks in the history of mathematics. If m and n are positive integers with m > n, show
\(a = m^2−n^2\),
b = 2mn, (7)
\(c = m^2+n^2\),
generates Pythagorean Triples.
We need only show that the formulae for a, b, and c satisfy the Pythagorean Theorem. With that is mind, let's first compute \(a^2+b^2\).
\(a^2+b^2 = (m^2−n^2)^2+(2mn)^2\)
= \(m^4−2m^{2}n^{2}+n^4+4m^{2}n^2\)
= \(m^4+2m^{2}n^2+n^4\)
On the other hand,
\(c^2 = (m^2+n^2)^2\)
= \(m^4+2m^{2}n^2+n^4\).
Hence, \(a^2+b^2 = c^2\), and the expressions for a, b, and c form a Pythagorean Triple.
It is both interesting and fun to generate Pythagorean Triples with the formulae from Example 6. Choose m = 4 and n = 2, then
\(a = m^2−n^2 = (4)^2−(2)^2 = 12\),
\(b = 2mn = 2(4)(2) = 16\),
\(c = m^2+n^2 =(4)^2+(2)^2 = 20\).
It is easy to check that the triple (12, 16, 20) will satisfy \(12^2+16^2 = 20^2\). Indeed, note that this triple is a multiple of the basic (3, 4, 5) triple, so it must also be a Pythagorean Triple.
It can also be shown that if m and n are relatively prime, and are not both odd or both even, then the formulae in Example 6 will generate a primitive Pythagorean Triple. For example, choose m = 5 and n = 2. Note that the greatest common divisor of m = 5 and n = 2 is one, so m and n are relatively prime. Moreover, m is odd while n is even. These values of m and n generate
\(a = m^2−n^2 = (5)^2−(2)^2 = 21\),
\(b = 2mn = 2(5)(2) = 20\),
\(c = m^2+n^2 = (5)^2+(2)^2 = 29\).
Note that
\(21^2+20^2 = 441+400 = 841 = 29^2\).
Hence, (21, 20, 29) is a Pythagorean Triple. Moreover, the greatest common divisor of 21, 20, and 29 is one, so (21, 20, 29) is primitive.
The practical applications of the Pythagorean Theorem are numerous.
Example \(\PageIndex{8}\)
A painter leans a 20 foot ladder against the wall of a house. The base of the ladder is on level ground 5 feet from the wall of the house. How high up the wall of the house will the ladder reach?
Consider the triangle in Figure 7. The hypotenuse of the triangle represents the ladder and has length 20 feet. The base of the triangle represents the distance of the base of the ladder from the wall of the house and is 5 feet in length. The vertical leg of the triangle is the distance the ladder reaches up the wall and the quantity we wish to determine.
Figure 7. A ladder leans against the wall of a house.
Applying the Pythagorean Theorem,
\(5^2+h^2 = 20^2\).
Again, note that the square of the length of the hypotenuse is the quantity that is isolated on one side of the equation.
Next, square, then isolate the term containing h on one side of the equation by subtracting 25 from both sides of the resulting equation.
\(25+h^2 = 400\)
\(h^2 = 375\)
We need only extract the positive square root.
\(h = \sqrt{375}\)
We could place the solution in simple form, that is, \(h = 5\sqrt{15}\), but the nature of the problem warrants a decimal approximation. Using a calculator and rounding to the nearest tenth of a foot,
\(h \approx 19.4\).
Thus, the ladder reaches about 19.4 feet up the wall.
The Distance Formula
We often need to calculate the distance between two points P and Q in the plane. Indeed, this is such a frequently recurring need, we'd like to develop a formula that will quickly calculate the distance between the given points P and Q. Such a formula is the goal of this last section.
Let P(x1, y1) and Q(x2, y2) be two arbitrary points in the plane, as shown in Figure 8(a) and let d represent the distance between the two points.
Figure 8. Finding the distance between the points P and Q.
To find the distance d, first draw the right triangle △PQR, with legs parallel to the axes, as shown in Figure 8(b). Next, we need to find the lengths of the legs of the right triangle △PQR.
The distance between P and R is found by subtracting the x coordinate of P from the x-coordinate of R and taking the absolute value of the result. That is, the distance between P and R is \(|x_{2}−x_{1}|\).
The distance between R and Q is found by subtracting the y-coordinate of R from the y-coordinate of Q and taking the absolute value of the result. That is, the distance between R and Q is \(|y_{2}−y_{1}|\).
THE DISTANCE FORMULA
Let P (x1, y1) and Q(x2, y2) be two arbitrary points in the plane. The distance d between the points P and Q is given by the formula
\(d = \sqrt{(x_{2}−x_{1})^2+(y_{2}−y_{1})^2}\). (9)
The direction of subtraction is unimportant. Because you square the result of the subtraction, you get the same response regardless of the direction of subtraction (e.g.\((5 − 2)^2 = (2 − 5)^2\)). Thus, it doesn't matter which point you designate as the point P, nor does it matter which point you designate as the point Q. Simply subtract x- coordinates and square, subtract y-coordinates and square, add, then take the square root.
Let's look at an example.
Example \(\PageIndex{10}\)
Find the distance between the points P (−4, −2) and Q (4, 4).
It helps the intuition if we draw a picture, as we have in Figure 9. One can now take a compass and open it to the distance between points Pand Q. Then you can place your compass on the horizontal axis (or any horizontal gridline) to estimate the distance between the points Pand Q. We did that on our graph paper and estimate the distance \(d \approx 10\).
Figure 9. Graphing the distance between P (−4, −2) and Q (4, 4).
Let's now use the distance formula to obtain an exact value for the distance d. With \((x_{1}, y_{1})\) = P (−4, −2) and \((x_{2}, y_{2})\) = Q (4, 4),
\(d = \sqrt{(x_{2}−x_{1})^2+(y_{2}−y_{1})^2}\)
= \(\sqrt{(4−(−4))^2+(4−(−2))^2}\)
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10.
It's not often that your exact result agrees with your approximation, so never worry if you're off by just a little bit.
Exercise
In Exercises 1-8, state whether or not the given triple is a Pythagorean Triple. Give a reason for your answer.
Exercise \(\PageIndex{1}\)
(8, 15, 17)
Answer
Yes, because \(8^2 + 15^2 = 17^2\)
Exercise \(\PageIndex{2}\)
(7, 24, 25)
Exercise \(\PageIndex{3}\)
(8, 9, 17)
Answer
No, because \(8^2+9^2 \ne 17^2\)
Exercise \(\PageIndex{4}\)
(4, 9, 13)
Exercise \(\PageIndex{5}\)
(12, 35, 37)
Answer
Yes, because \(12^2 + 35^2 = 37^2\)
Exercise \(\PageIndex{6}\)
(12, 17, 29)
Exercise \(\PageIndex{7}\)
(11, 17, 28)
Answer
No, because \(11^2 + 17^2 \ne 28^2\)
Exercise \(\PageIndex{8}\)
(11, 60, 61)
In Exercises 9-16, set up an equation to model the problem constraints and solve. Use your answer to find the missing side of the given right triangle. Include a sketch with your solution and check your result.
Exercise \(\PageIndex{9}\)
Answer
4
Exercise \(\PageIndex{10}\)
Exercise \(\PageIndex{11}\)
Answer
\(4\sqrt{2}\)
Exercise \(\PageIndex{12}\)
Exercise \(\PageIndex{13}\)
Answer
\(2\sqrt{2}\)
Exercise \(\PageIndex{14}\)
Exercise \(\PageIndex{15}\)
Answer
\(5\sqrt{3}\)
Exercise \(\PageIndex{16}\)
In Exercises 17-20, set up an equation that models the problem constraints. Solve the equation and use the result to answer the question. Look back and check your result.
Exercise \(\PageIndex{17}\)
The legs of a right triangle are consecutive positive integers. The hypotenuse has length 5. What are the lengths of the legs?
Answer
The legs have lengths 3 and 4.
Exercise \(\PageIndex{18}\)
The legs of a right triangle are consecutive even integers. The hypotenuse has length 10. What are the lengths of the legs?
Exercise \(\PageIndex{19}\)
One leg of a right triangle is 1 centimeter less than twice the length of the first leg. If the length of the hypotenuse is 17 centimeters, find the lengths of the legs.
Answer
The legs have lengths 8 and 15 centimeters.
Exercise \(\PageIndex{20}\)
One leg of a right triangle is 3 feet longer than 3 times the length of the first leg. The length of the hypotenuse is 25 feet. Find the lengths of the legs.
Exercise \(\PageIndex{21}\)
Pythagoras is credited with the following formulae that can be used to generate Pythagorean Triples.
a = m
\(b = \frac{m^2−1}{2}\),
\(c = \frac{m^2+1}{2}\)
Use the technique of Example 6 to demonstrate that the formulae given above will generate Pythagorean Triples, provided that m is an odd positive integer larger than one. Secondly, generate at least 3 instances of Pythagorean Triples with Pythagoras's formula.
Exercise \(\PageIndex{22}\)
Plato (380 BC) is credited with the following formulae that can be used to generate Pythagorean Triples.
\(a = 2m\)
\(b = m^2 − 1\),
\(c = m^2 + 1\)
Use the technique of Example 6 to demonstrate that the formulae given above will generate Pythagorean Triples, provided that mis a positive integer larger than 1. Secondly, generate at least 3 instances of Pythagorean Triples with Plato's formula.
In Exercises 23-28, set up an equation that models the problem constraints. Solve the equation and use the result to answer. Look back and check your answer.
Exercise \(\PageIndex{23}\)
Fritz and Greta are planting a 12-foot by 18Answer
21.63 ft
Exercise \(\PageIndex{24}\)
Angelina and Markos are planting a 20-foot by 28Exercise \(\PageIndex{25}\)
The base of a 36-foot long guy wire is located 16 feet from the base of the telephone pole that it is anchoring. How high up the pole does the guy wire reach? Approximate your answer to within 0.1.
Answer
32.25 ft
Exercise \(\PageIndex{26}\)
The base of a 35-foot long guy wire is located 10 feet from the base of the telephone pole that it is anchoring. How high up the pole does the guy wire reach? Approximate your answer to within 0.1 feet.
Exercise \(\PageIndex{27}\)
A stereo receiver is in a corner of a 13-foot by 16-foot rectangular room. Speaker wire will run under a rug, diagonally, to a speaker in the far corner. If 3 feet of slack is required on each end, how long a piece of wire should be purchased? Approximate your answer to within 0.1 feet.
Answer
26.62 ft
Exercise \(\PageIndex{28}\)
A stereo receiver is in a corner of a 10-foot by 15-foot rectangular room. Speaker wire will run under a rug, diagonally, to a speaker in the far corner. If 4 feet of slack is required on each end, how long a piece of wire should be purchased? Approximate your answer to within 0.1 feet.
In Exercises 29-38, use the distance formula to find the exact distance between the given points.
Exercise \(\PageIndex{29}\)
(−8, −9) and (6, −6)
Answer
\(\sqrt{205}\)
Exercise \(\PageIndex{30}\)
(1, 0) and (−9, −2)
Exercise \(\PageIndex{31}\)
(−9, 1) and (−8, 7)
Answer
\(\sqrt{37}\)
Exercise \(\PageIndex{32}\)
(0, 9) and (3, 1)
Exercise \(\PageIndex{33}\)
(6, −5) and (−9, −2)
Answer
\(\sqrt{234} = 3\sqrt{26}\)
Exercise \(\PageIndex{34}\)
(−5, 6) and (1, 4)
Exercise \(\PageIndex{35}\)
(−7, 7) and (−3, 6)
Answer
\(\sqrt{17}\)
Exercise \(\PageIndex{36}\)
(−7, −6) and (−2, −4)
Exercise \(\PageIndex{37}\)
(4, −3) and (−9, 6)
Answer
\(\sqrt{250} = 5\sqrt{10}\)
Exercise \(\PageIndex{38}\)
(−7, −1) and (4, −5)
In Exercises 39-42, set up an equation that models the problem constraints. Solve the equation and use the result to answer the question. Look back and check your result.
Exercise \(\PageIndex{39}\)
Find k so that the point (4, k) is \(2\sqrt{2}\) units away from the point (2, 1).
Answer
k = 3, −1.
Exercise \(\PageIndex{40}\)
Find k so hat the point (k, 1) is \(2\sqrt{2}\) units away from the point (0, −1)
Exercise \(\PageIndex{41}\)
Find k so hat the point (k, 1) is \(\sqrt{17}\) units away from the point (2, −3)
Answer
k = 1, 3.
Exercise \(\PageIndex{42}\)
Find k so that the point (−1, k) is \(\sqrt{13}\) units away from the point (−4, −3).
Exercise \(\PageIndex{43}\)
Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Plot the points P (0, 5) and Q (4, −3) on your coordinate system.
a) Plot several points that are equidistant from the points P and Q on your coordinate system. What graph do you get if you plot all points that are equidistant from the points P and Q? Determine the equation of the graph by examining the resulting image on your coordinate system.
b) Use the distance formula to find the equation of the graph of all points that are equidistant from the points P and Q. Hint: Let (x, y) represent an arbitrary point on the graph of all points equidistant from points P and Q. Calculate the distances from the point (x, y) to the points P and Q separately, then set them equal and simplify the resulting equation. Note that this analytical approach should provide an equation that matches that found by the graphical approach in part (a).
Answer
a) In the figure that follows, XP = XQ.
b) \(y = \frac{1}{2}x\)
Exercise \(\PageIndex{44}\)
Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Plot the point P (0, 2) and label it with its coordinates. Draw the line y = −2 and label it with its equation.
a) Plot several points that are equidistant from the point P and the line y = −2 on your coordinate system. What graph do you get if you plot allpoints that are equidistant from the points P and the line y = −2.
b) Use the distance formula to find the equation of the graph of all points that are equidistant from the points P and the line y = −2. Hint: Let(x, y) represent an arbitrary point on the graph of all points equidistant from points P and the line y = −2. Calculate the distances from the point(x,y) to the points P and the line y = −2 separately, then set them equal and simplify the resulting equation.
Exercise \(\PageIndex{45}\)
Copy the following figure onto a sheet of graph paper. Cut the pieces of the first figure out with a pair of scissors, then rearrange them to form the second figure. Explain how this proves the Pythagorean Theorem.
Exercise \(\PageIndex{46}\)
Compare this image to the one that follows and explain how this proves the Pythagorean Theorem | 677.169 | 1 |
Truncated Tetrahedron Calculator
Calculations at a regular truncated tetrahedron. A truncated tetrahedron is constructed by cutting off the vertices of a tetrahedron in a way, so that every edge has the same length. Its dual body is the triakis tetrahedron. Enter one value and choose the number of decimal places. Then click Calculate.
The truncated tetrahedron is an Archimedean solid. Edge length and radius have the same unit (e.g. meter), the area has this unit squared (e.g. square meter), the volume has this unit to the power of three (e.g. cubic meter). A/V has this unit -1.
Archimedean solids are regular polyhedra. Their regularity is less than that of the Platonic solids. The five Platonic solids tetrahedron, cube, octahedron, dodecahedron and icosahedron consist of the same regular polygons. The requirement for the Archimedean solids is only that they consist of regular polygons and all corners are the same. Prisms and antiprisms are excluded, otherwise their number would be infinite. There are 13 (or 15) Archimedean solids. Most consist of two different regular polygons; for the truncated cuboctahedron, rhombicosidodecahedron and truncated icosidodecahedron there are three different polygons. There are two versions of the snub cube and the snub dodecahedron that are mirror images of each other but otherwise have the same dimensions. Hence the ambiguity in the number of Archimedean solids. These 13 or 15 are all Archimedean solids there are, no more polyhedra with these properties exist. They are sorted by the number of their sides, the truncated tetrahedron being the simplest of them. Presumably all Archimedean bodies were discovered by Archimedes of Syracuse himself. | 677.169 | 1 |
The properties of the parallelogram are simply those things that are true about it. These properties concern its sides, angles, and diagonals.
The parallelogram has the following properties:
Opposite sides are parallel by definition.
Opposite sides are congruent.
Opposite angles are congruent.
Consecutive angles are supplementary.
The diagonals bisect each other.
If you just look at a parallelogram, the things that look true (namely, the things on this list) are true and are thus properties, and the things that don't look like they're true aren't properties.
If you draw a picture to help you figure out a quadrilateral's properties, make your sketch as general as possible. For instance, as you sketch your parallelogram, make sure it's not almost a rhombus (with four sides that are almost congruent) or almost a rectangle (with four angles close to right angles). If your parallelogram sketch is close to, say, a rectangle, something that's true for rectangles but not true for all parallelograms (such as congruent diagonals) may look true and thus cause you to mistakenly conclude that it's a property of parallelograms. Capiche?
Imagine that you can't remember the properties of a parallelogram. You could just sketch one (as in the above figure) and run through all things that might be properties. (Note that this parallelogram does not come close to resembling a rectangle of a rhombus.)
The following questions concern the sides of a parallelogram (refer to the preceding figure).
Do the sides appear to be congruent?Yes, opposite sides look congruent, and that's a property. But adjacent sides don't look congruent, and that's not a property.
Do the sides appear to be parallel?Yes, opposite sides look parallel (and of course, you know this property if you know the definition of a parallelogram).
The following questions explore the angles of a parallelogram (refer to the figure again).
Do any angles appear to be congruent?Yes, opposite angles look congruent, and that's a property. (Angles A and C appear to be about 45°, and angles B and D look like about 135°).
Do any angles appear to be supplementary?Yes, consecutive angles (like angles A and B ) look like they're supplementary, and that's a property. (Using parallel lines! angles A and B are same-side interior angles and are therefore supplementary.)
Do any angles appear to be right angles?Obviously not, and that's not a property.
The following questions address statements about the diagonals of a parallelogram
Do the diagonals appear to be congruent?Not even close (in the above figure, one is roughly twice as long as the other, which surprises most people) — not a property.
Do the diagonals appear to be perpendicular?Not even close; not a property.
Do the diagonals appear to be bisecting each other?Yes, each one seems to cut the other in half, and that's a property.
Do the diagonals appear to be bisecting the angles whose vertices they meet?No. | 677.169 | 1 |
The Angles of a Triangle
Elementary+
You are given the lengths for each side on a triangle.
You need to find all three angles for this triangle.
If the given side lengths cannot form a triangle (or form a degenerated triangle),
then you must return all angles as 0 (zero).
The angles should be represented as a list of integers in
ascending order
.
Each angle is measured in degrees and rounded to the nearest integer number (Standard mathematical rounding). | 677.169 | 1 |
what is congruence of triangle
Congruence Of Triangles Worksheet – Triangles are one of the most fundamental shapes of geometry. Understanding triangles is crucial to understanding more advanced geometric principles. In this blog post we will look at the various kinds of triangles Triangle angles, how to determine the areas and perimeters of a triangle, and show details of the various. Types of Triangles There are three types for triangles: Equal, isosceles, as well as scalene. Equilateral triangles include three equally sides, and … Read more | 677.169 | 1 |
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Trigonometry is a branch of mathematics concerned with relationships between angles and side lengths of triangles. Replace "LHS" and "RHS" below with the left-hand side and the right-hand side of the identity you are trying to verify. In particular, the trigonometric functions … Verifying trig identities | Desmos. Practice your math skills and learn step by step with our math … Verify trigonometric identities with our Trigonometric Identity Calculator.
And if you lose it, you may need to perform an IRS PIN retrieval to get back the PIN. The process wil. If the left-hand side and the right-hand side of the identity are equivalent, then they have the same graph and the same values at a given input. Practice your math skills and learn step by step with our math … Verify trigonometric identities with our Trigonometric Identity Calculator. If the left-hand side and the right-hand side of the identity are equivalent, then they have the same graph and the same values at a given input. cscθcotθ = cosθ.
In today's digital age, online security has become more important than ever. If the left-hand side and the right-hand side of the identity are equivalent, then they have the same graph and the same values at a given input. In today's digital age, where personal information is easily accessible and can be misused, it is crucial to take necessary steps to protect your identity. ….
Jun 21, 2024 · This trig calculator finds the values of trig functions and solves right triangles using trigonometry. Google is going to start displaying a blue checkmark next to s. cscθcotθ = cosθ.
This solution was automatically generated by our smart calculator: $\frac {1} {\cos\left … cscθcotθ = cosθ. You and your spouse or children can request e-filing PIN numbers online, by phone, in writing or by visiting your local IRS office. | 677.169 | 1 |
how to find exterior angles of a triangle
Find The Exterior Angle Of A Triangle Worksheet – Triangles are among the fundamental shapes in geometry. Understanding triangles is vital to learning more advanced geometric terms. In this blog it will explain the different kinds of triangles that are triangle angles. We will also explain how to determine the size and perimeter of a triangle, and provide specific examples on each. Types of Triangles There are three kinds of triangles: equal isosceles, as well … Read more | 677.169 | 1 |
Now that you have learned all about
scale, test your new skills with the following activities.
On the two maps below, use the scale to find the distance from point
A to point B and from point B to point C. Fill in the blanks with your
answer. Click on the maps to see them enlarged. | 677.169 | 1 |
4 2 study guide and intervention angles of triangles
Find the missing measure in each triangle. Then classify the triangle as acute, right, or obtuse. 1. 2. 3. Classify each triangle by its angles and by its sides. 4. 5. 6. 40˚ 110 ˚30 60˚ 50˚ 70˚ 45˚ 40˚ x˚ 43˚ x˚ 82˚ 75˚ x˚ Triangles can be classified by the measures of their angles. An acute trianglehas three acute angles. An ...Study Guide and Intervention Angles of Triangles Triangle Angle-Sum Theorem If the measures of two angles of a triangle are known, the measure of the third angle can always be found. ... 4-2 Study Guide And Intervention Degrees And Radians Of Trig. Objectives: This is your review of trigonometry: angles, six trig. functions, Conversion between ...
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4 2 Study Guide And Intervention Angles Of Triangles Helping Children Learn Mathematics - National Research Council 2002-07-31 Results from national and international assessments indicate that school children in the United States are not learning mathematics well enough. Many students cannot correctly apply computational algorithms to solve ...CDC - Blogs - Genomics and Precision Health – Toward More Precision in Implementation Science in the Age of COVID-19 - Genomics and Precision Health Blog Implementation science (IS...of the chapter. Vocabulary … NAME DATE PERIOD 4-2 Study Guide and Intervention WEBChapter 4 11 Glencoe Geometry Study Guide and Intervention Angles of Triangles Triangle Angle-Sum Theorem If the measures of two angles of a triangle are known, the measure of the third angle can always be found. Triangle Angle Sum Theorem The …
View Test prep - 2.2.1 Study The Angles of a Triangle from MATHEMATIC Math 137 at Ivy Tech Community College, Indianapolis. 2.2.1 Study: The Angles of a Triangle Study Sheet Geometry Sem 1. ... P Conol - 2.2 Study Guide + Checkup .pdf. Waiakea High School. MATH MGX1150H. 52. triangle. 60. Sum Theorem. 38.NAME _____ DATE _____ PERIOD _____ Chapter 5 5 Glencoe Geometry 5-1 Study Guide and Intervention Bisectors of Triangles Perpendicular Bisectors A perpendicular bisector is a line, segment, or ray that is perpendicular to the given segment and passes through its midpoint. Some theorems deal with perpendicular bisectors.Here's where traders and investors who are not long AAPL could go long. Employees of TheStreet are prohibited from trading individual securities. Despite the intraday reversal ...triangle is greater than the length of the third side. The measures of two sides of a triangle are 5 and 8. Find a range. for the length of the third side. By the Triangle Inequality, all three of the following inequalities must be true. must be between 3 and 13. triangle. Write. 1.
4 2 Study Guide And Intervention Angles Of Triangles 4-2-study-guide-and-intervention-angles-of-triangles 2 Downloaded from imgsrv.amazonservices.com on 2019-12-21 by guest versions of a selection of papers from the Proceedings of the 13th International Conference on Technology in Mathematics TeachingThe first step in writing a coordinate proof is to place a figure on the coordinate plane and label the vertices. Use the following guidelines. 1. Use the origin as a vertex or center of the figure. 2. Place at least one side of the polygon on an axis. 3. Keep the figure in the first quadrant if possible. 4.
sks synma Lesson Resources: 4.1 Triangles and Angles 4.2 Congruence and Triangles 4.3 Proving Triangles are Congruent: SSS and SAS 4.4 ... Study Guide And Intervention Answer Key 4-3 - localexam 1 day ago · 3. ... BD is the angle bisector of lABC and lADC.. is there a arbysks sakhnh triangle as long as you know the other two lengths. However, now we can relate angle measures to the right triangle side lengths. This allows us to find both missing side lengths when we only know one length and an acute angle measure. The acute angle measures in a right triangle can be found based on any two side lengths. athkar alsbah When people hear of market intervention, sometimes they assume the stock market has been rigged and you have to be high up in a corporate structure to benefit. But that's rarely th...Study Guide and Intervention Workbook 000i_GEOSGIFM_890848.indd 10i_GEOSGIFM_890848.indd 1 66/26/08 7:49:12 PM/26/08 7:49:12 PM. ... 5-6 Inequalities in Two Triangles .....69 6-1 Angles of Polygons .....71 6-2 Parallelograms .....73 6-3 Tests for Parallelograms ... joey mitinerariocitta.aspsyksy sghar Unformatted text preview: Chapter 4 6 Glencoe Geometry4-1 Study Guide and Intervention (continued) Angles of Triangles Exterior Angle Theorem At each vertex of a triangle, the angle formed by one side and an extension of the other side is called an exterior angle of the triangle. For each exterior angle of a triangle, the remote interior angles are the interior angles that are not adjacent to ... dollar750 cash app on radio A right triangle has one 90° angle, which is often marked with the symbol shown in Figure \(\PageIndex{6}\). Figure \(\PageIndex{6}\) If we know that a triangle is a right triangle, we know that one angle measures 90° so we only need the measure of one of the other angles in order to determine the measure of the third angle.NAME DATE PERIOD. Lesson 4-3. 2nd Pass. Chapter 419 Glencoe Geometry. Congruence and Corresponding Parts. Triangles that have the same size and same shape are congruent triangles. Two triangles are congruent if and only if all three pairs of corresponding angles are congruent and all three pairs of corresponding sides are congruent. leccion 1 lesson testlyrics to itgln lookup Of the interior angleg Of those n — 2 triangles. Polygon Interior Angle Tho sum of the interior angle measures of an n.sided convex polygon (n — 2) 180. Sum Theorem Study Guide and Intervention Parallelograms Sides and Angles of Parallelograms A quadrilateral with both pairs Of opposite sides parallel is a parallelogram. Here are four | 677.169 | 1 |
In \triangle A B C, A B=3, B C=4, C A=5. Circle \omega intersects \overline{A B} at E and B, \overline{B C} at B and D, and \overline{A C} at F and G. Given that E F=D F and \frac{D G}{E G}=\frac{3}{4}, length D E=\frac{a \sqrt{b}}{c}, where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. Find a+b+c. | 677.169 | 1 |
Three different points from the 16 points of this 4 × 4 grid are to be chosen as vertices of a triangle. How many different triangles can be drawn?
I already found all the right triangles (200) by finding all the rectangles in the grid. But since there's also the obtuse and acute triangles that I forgot when solving this. So is there any easy way to find the other triangles?
I'm only asking about the obtuse and acute triangles, since I already found the right triangles. (I got the question wrong because I didn't find the other triangles.)
0 users composing answers..
Yes there is an easier way. Assuming that triangles are still distinct even through rotations/reflections, etc, then the total number of points you could choose is 16 choose 3. However that would get some straight lined triangles, so we have to eliminate those. There are 16 + 16 + 4 + 4 = 40 of these cases counting three consecutive, 2 consecutive and 1 two apart, then considering the diagonal cases. Thus there are 16 choose 3 - 40 = 520 total triangles that are non-degenerate. (assuming that rotating the grid is not going to eliminate cases) | 677.169 | 1 |
Understanding Angles | Exploring the Definition, Measurement, and Properties of Angles in Mathematics
angle
In mathematics, an angle is a geometric figure formed by two rays or lines that share a common endpoint, called the vertex
In mathematics, an angle is a geometric figure formed by two rays or lines that share a common endpoint, called the vertex. The rays extending from the vertex are referred to as the sides of the angle. Angles are typically measured in degrees, although they can also be measured in radians.
Here are some important terms related to angles:
1. Vertex: The shared endpoint of the two rays or lines that form an angle.
2. Sides: The two rays or lines that form an angle. They are also referred to as arms or legs of the angle.
3. Angle Measure: The magnitude or size of an angle, typically measured in degrees. In a standard position, where one side lies along the x-axis, angles are measured counterclockwise from the positive x-axis.
4. Degrees: The unit of measurement for angles. A full circle is divided into 360 degrees, with each degree further divided into minutes and seconds.
5. Radians: An alternative unit of measurement for angles, often used in advanced mathematics and physics. A radian is defined as the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.
6. Acute Angle: An angle that measures less than 90 degrees.
7. Right Angle: An angle that measures exactly 90 degrees. One of its sides is perpendicular to the other.
8. Obtuse Angle: An angle that measures more than 90 degrees but less than 180 degrees.
10. Reflex Angle: An angle that measures more than 180 degrees but less than 360 degrees.
11. Complementary Angles: Two angles are complementary if the sum of their measures is equal to 90 degrees. For example, angles measuring 30 degrees and 60 degrees are complementary.
12. Supplementary Angles: Two angles are supplementary if the sum of their measures is equal to 180 degrees. For example, angles measuring 120 degrees and 60 degrees are supplementary.
Understanding angles and their properties is essential in various mathematical applications, such as geometry, trigonometry, and physics. They play a crucial role in determining shapes, measuring distances, and calculating the relationships between different objects or | 677.169 | 1 |
Hint: We are given a diagram here in which there are two parallelograms as stated in the question. We are going to use the property of parallelogram that the opposite sides of a parallelogram are parallel and equal to each other. After using that we will tick the correct option out of the given four options. You need to be aware about the properties of a parallelogram for this question to be done correctly.
Complete step by step answer: We have the following figure given in the question:
Note here that BDEF is a parallelogram. So, using the property of parallelogram that the opposite sides of a parallelogram are equal, we can write the fact that: FE is parallel to BD and also it can be said that: FE=BD…………$\left(1\right)$ Now, FDCE is also a parallelogram. So again using the same property, we can say that: FE is parallel to CD and also we can say that: FE=CD…………$\left(2\right)$ From$\left(1\right)$and$\left(2\right)$, we can say that BD=CD. Looking at the options we find that it matches option $\left(d\right)$. So, the correct answer is "Option d".
Note: Do not look directly at the figure and then point out the correct option. You cannot just look at a line segment and assume its length to be equal to that of any other line segment. Always prove the results in a concrete manner first then tick the correct option. Moreover, the basic properties of a parallelogram should always be kept in mind and using those you can figure out the correct answer by yourself. | 677.169 | 1 |
by
Alexander, Daniel C.; Koeberlein, Geralyn M.
Answer
Work Step by Step* Prove that $\triangle DAB\cong\triangle CAE$
1) $\overline{DB}\bot\overline{BC}$ and $\overline{CE}\bot\overline{ED}$ (Given)
2) $\angle DBA$ and $\angle CEA$ are right $\angle$s (if 2 lines are perpendicular with each other, then the angles that they make up are right angles)
3) $\angle DBA\cong\angle CEA$ (2 corresponding right angles are congruent)
4) $\overline{AB}\cong\overline{AE}$ (Given)
5) $\angle DAB\cong\angle CAE$ (2 vertical angles are congruent)
So now 2 angles and the included side of $\triangle DAB$ are congruent with 2 corresponding angles and the included side of $\triangle CAE$.
5) $\triangle DAB\cong\triangle CAE$ (ASA)
6) $\overline{DB}\cong\overline{CE}$ and $\overline{DA}\cong\overline{CA}$ (CPCTC)
* Prove that $\triangle DCB\cong\triangle CDE$
We see that $\overline{DA}\cong\overline{CA}$ (proved above) and $\overline{AE}\cong\overline{AB}$ (given)
So $\overline{DA}+\overline{AE}\cong\overline{CA}+\overline{AB}$
That means $\overline{DE}\cong\overline{CB}$
7) $\overline{DE}\cong\overline{CB}$ (proved above)
8) $\overline{CE}\cong\overline{DB}$ (proved in 6)
9) $\angle CED\cong\angle DBC$ (proved in 3)
So now 2 sides and the included angle of $\triangle DEC$ are congruent with 2 corresponding sides and the included angle of $\triangle CBD$.
10) $\triangle DEC\cong\triangle CBD$ (SAS) | 677.169 | 1 |
Understanding the Conical Shape in Design
The concept of the conical shape in design is both fundamental and fascinating. This three-dimensional geometric figure tapers smoothly from a flat base to a point, known as the apex or vertex. While many might associate this form with simple objects like ice cream cones, its applications in design are diverse and profound.
The versatility of the cone stems from its varied base shapes – not restricted solely to circles, the base can be any polygon, broadening its utilization. When a cone is crafted with a right circular base and its apex is aligned perpendicularly above the center, it's termed a right circular cone. Conversely, if the apex does not align with the base center, the shape is classified as an oblique cone.
Designers often turn to the conical shape for its structural strength and aesthetic appeal. The sleek lines that taper to a single point can create a sense of movement and directionality within a design, guiding the viewer's eyes and giving an object a dynamic presence.
In terms of functionality, its geometric properties provide a sturdy base while minimizing material usage towards the apex, making it a cost-effective option. This efficiency is not only practical but also aligns with sustainability efforts in design, reducing waste without compromising on stability.
When considering the integration of conical shapes in design, it's important to acknowledge the mathematical elements that influence its creation. These include key measurements such as the radius, height, and slant height, which are critical in defining the shape's dimensions and ensuring the accurate execution of a design vision.
Key Dimension
Explanation
Radius (r)
The distance from the center of the base to any point along its edge.
Height (h)
The vertical distance from the base to the apex of the cone.
Slant Height (l)
The length of the line segment running from the apex to the edge of the base along the surface of the cone.
These physical attributes of conical designs are often calculated with precise formulas, ensuring that designers can create functional, safe, and aesthetically pleasing objects. As we will explore throughout this article, the conical shape is not just a design choice but also a reflection of an object's purpose and the message it aims to convey.
The Evolution of Conical Shapes in Modern Aesthetics
The evolution of conical shapes within modern aesthetics has been characterized by innovative adaptations and multidisciplinary applications. Originally rooted in practicality, with historical artifacts like pottery and tools utilizing conical forms for stability and efficiency, contemporary design has embraced cones for their symbolic and visual potency.
In the realm of architecture, the cone has transcended its primary function as a sturdy structure, morphing into a symbol of modernity and technological advancement. Designers and architects have reimagined this shape to give a sense of aspiration and progress, often reflected in the skylines of burgeoning cities.
Within the fashion industry, the allure of the conical shape has been particularly evident. From the dramatic silhouettes of haute couture gowns to striking accessories, designers leverage the conical form for its ability to juxtapose elegance with edginess, thereby making a memorable visual statement.
Graphic design also showcases the versatility of conical shapes, utilizing them to direct focus or convey downward or upward momentum. This is evident in logo designs and visual branding, where cone-inspired imagery can suggest growth, stability, or forward-thinking depending on its orientation and context.
Moving into the digital realm, conical shapes have found a place in user interface (UI) and user experience (UX) design. Their pointed nature makes for excellent directional cues and can guide users visually through a digital space, promoting a seamless interaction with a website or application.
As we delve further into the age of technology, even the field of virtual reality and gaming has observed a surge in the use of conical shapes. In these digital landscapes, cones assist in creating depth and dimension, contributing to a more immersive and engaging visual experience.
Architectural Skylines
Fashion and Couture
Brand Identity in Logos
User Interface Design
Virtual Reality Depth
In summary, the conical shape's journey in design aesthetics has been transformative. It has seamlessly integrated into various disciplines, proving that a basic geometric form can have a powerful and enduring impact on modern design trends.
Key Benefits of Using Conical Shapes in Design
Delving into the key benefits of using conical shapes in design, we uncover a myriad of advantages that extend beyond their aesthetic appeal. One primary benefit is the inherent durability that cones offer. Due to their geometric properties, conical shapes can distribute stress evenly throughout their structure, making them inherently robust and ideal for use in high-stress environments.
Another significant advantage is the efficiency of conical designs in terms of space utilization. With a wider base gradually tapering to a point, cones can occupy less vertical space while still providing a solid foundation. This is especially useful in areas where space is at a premium and maximizing usability is essential.
Conical shapes are also known for their aerodynamic properties, which reduce air resistance and enable better airflow around objects. This characteristic is particularly advantageous in automotive and aerospace design, where reducing drag is crucial for improving speed and fuel efficiency.
The acoustic benefits of conical shapes should not be overlooked. The shape effectively channels sound waves, making cones an excellent choice for designing speaker systems, auditoriums, and other sound-centric environments, where controlling the direction and quality of sound is important.
From an optical standpoint, the conical form is ideal for concentrating light, which can be utilized in designing lenses, lighting fixtures, and even natural lighting solutions within buildings, to enhance illumination and energy efficiency.
The conical shape offers tangible benefits to designers and engineers, allowing them to optimize structures for strength, space, aerodynamics, acoustics, and lighting.
To summarize, cones offer multifaceted design benefits that make them an invaluable shape in a diverse array of fields. From structural soundness to optimized space usage and improved functional performance, the conical shape is much more than a visual element—it's a strategic choice for innovative and sustainable design solutions.
Incorporating Conical Shapes into Various Design Spheres
Incorporating conical shapes into various design spheres showcases their utility across a spectrum of creative and scientific fields. In interior design, for example, conical forms are often chosen for their ability to blend function with decorative appeal. Furniture pieces with conical bases offer stability and style, becoming focal points within living spaces.
Industrial design leverages the strength and simplicity of cones to create machinery and equipment that exhibit both sleekness and functionality. The use of conical components in devices can facilitate precise movements, guide materials through machinery, or provide pivotal connection points.
In the world of landscape architecture, conical elements are employed to create visual interest and direct foot traffic. They serve as guiding posts or even as trees sculpted into conical topiaries, integrating natural beauty with geometric precision.
The entertainment industry also finds value in conical shapes for set designs and stage construction. Their dramatic form can add depth and perspective to a scene, creating a dynamic backdrop for performances. Additionally, conical props or costumes can contribute to a narrative, symbolizing power, alertness, or stability depending on their context and orientation.
Even in the sectors of marketing and display design, cones make powerful tools for attracting attention. As part of a display, they can guide the customer's gaze to a product or information, influencing the flow of foot traffic and optimizing the consumer experience.
Interior Design: Furnishing and Decor
Industrial Design: Machinery and Equipment
Landscape Architecture: Visual Guidance and Aesthetics
Entertainment: Set and Costume Design
Marketing: Display and Consumer Engagement
Ultimately, the act of incorporating conical shapes across these diverse spheres harnesses their intrinsic properties for strategic applications. Whether it is to stabilize, guide, or captivate, the cone emerges as a versatile player in the design world.
The Psychology Behind the Popularity of Conical Shapes
The popularity of conical shapes is not only due to their physical benefits and versatility but also to the psychological impact they exert on observers. The cone itself is a powerful symbol that can evoke a range of emotions and associations, playing into our innate desire for balance and order.
On an emotional level, cones can represent growth and aspiration, with their narrowing shape pointing upwards, suggesting movement towards a peak or goal. This is why they are often used in imagery and objects meant to inspire and motivate.
Conical shapes also convey a sense of direction and precision. Their pointed tips can focus attention, drawing the eye to a particular area or aspect of a design, and effectively guiding one's gaze or even one's movements within a physical space.
Add to this, the stability of a cone's wide base conveys a feeling of steadfastness and reliability, creating a subliminal message of trustworthiness when used in product design or branding. This psychological anchoring can influence consumer behavior, making conical shapes potent tools in marketing and advertising.
Additionally, the symmetry and balance inherent in conical shapes are aesthetically pleasing to the human eye. We are naturally drawn to symmetrical forms, and cones satisfy this preference, while also adding a dynamic three-dimensional aspect that can be more engaging than other symmetrical forms.
Conical shapes can subtly influence our perceptions and behaviors, making them not just a design choice but a psychological one as well.
In essence, the appeal of conical shapes transcends the practical, reaching into the realm of the psychological. By understanding the subconscious impact of cones on emotions, behavior, and preference, designers can craft experiences that not only serve a functional purpose but also connect with users on a deeper level.
Material Considerations for Conical Design Elements
Selecting the appropriate materials is a pivotal aspect when incorporating conical design elements into any project. Material choice can significantly affect not only the durability and functionality of the cone but also its aesthetic quality and psychological impact.
For outdoor applications, such as in architecture or landscape design, materials like stone and concrete are often favored for their robustness and ability to withstand the elements. These materials provide a sense of permanence and strength, echoing the stability that conical shapes naturally suggest.
In product design, plastics and metals are common choices, allowing for precision in manufacturing and versatility in application. Plastics can be molded into a conical form fairly easily and can be made translucent to diffuse light, serving both functional and decorative roles. Metals, on the other hand, are chosen for their structural integrity and reflective properties, which can be used to create a sleek, modern appearance.
For interior design elements, such as lighting fixtures or decorative pieces, glass and ceramics are popular materials. A glass cone can create interesting light refractions, while ceramics offer a classic look and feel, with the added benefit of customizability through glazes and paints.
In the context of acoustics, designers may opt for specific woods or composites known for their sound-enhancing qualities. The cone's shape, combined with such materials, can effectively channel and amplify sound in musical instruments or high-fidelity speakers.
Textiles also come into play, particularly in fashion and upholstery. The choice of fabric can influence the drape and flow of a conical shape, affecting its movement and interaction with the human form or furniture pieces.
Outdoor Durability: Stone and Concrete
Product Precision: Plastics and Metals
Interior Elegance: Glass and Ceramics
Acoustic Perfection: Wood and Composites
Textile Fluidity: Fabrics for Fashion and Furniture
Ultimately, whether the goal is to innovate, to ensure longevity, or to make an aesthetic statement, the materials chosen for conical design elements play a crucial role in their success. A thoughtful approach to material selection can optimize the cone's utility and align it with the broader design vision.
Innovative Uses of Conical Shapes in Contemporary Architecture
Contemporary architecture has embraced conical shapes to push the boundaries of design and functionality. One innovative application of conical shapes is in the design of roofs and towers. The tapered form of a cone allows these structures to withstand high winds and support substantial weight at the base while minimizing materials toward the top.
In energy-efficient building designs, conical shapes are ingeniously implemented to enhance temperature regulation. The surface area of a conical roof, for instance, can be optimized to reduce heat absorption, promoting a cooler interior environment without the reliance on excessive artificial cooling.
Conical structures also serve as natural light wells, channeling sunlight into the interiors of buildings. This not only reduces dependency on artificial lighting but also creates visually striking patterns of light and shadow that change throughout the day, adding a dynamic element to static spaces.
Furthermore, the aerodynamic profile of cones is utilized in designing buildings that aim to blend with or harness the environment. Specialized coatings or materials applied to conical features can reduce wind resistance or collect rainwater, contributing to the building's sustainability efforts.
Waterfront properties often incorporate conical shapes to provide panoramic views and to reflect nautical themes. The tapering form can also be strategic in managing water runoff, preventing flooding and water damage during storm surges.
Structural Durability: Roofs and Towers
Climate Control: Energy Efficiency
Daylight Harvesting: Light Wells
Environmental Integration: Aerodynamics and Sustainability
Expansive Visibility: Waterfront Design
These innovative applications highlight how conical shapes are more than just visual elements in architecture; they are functional tools that respond to both human needs and environmental conditions, shaping the spaces in which we live, work, and play in a responsible and responsive manner.
Conical Shapes in Product Design: Function Meets Form
In product design, the integration of conical shapes is a testament to the harmony of function and form. Designers often utilize conical forms to guide users intuitively, particularly in the realm of ergonomic products, where the shape can dictate the way an object is held or used, enhancing user comfort and efficiency.
The dynamic of conical shapes in directing motion is also exploited in items like funnels or megaphones, where the tapering form naturally channels liquids or amplifies sound, harnessing the physics of the cone to achieve a specific functional outcome.
In the case of lighting fixtures, conical designs serve to focus and control the direction of light, creating targeted illumination or a desired atmospheric effect. The material transparency and finish can be expertly chosen to further modulate the light quality, such as diffusing a soft glow or projecting a sharp beam.
For household items such as vases or decorative pieces, the cone's sleek silhouette can accentuate the elegance of an arrangement while providing a stable base, blending the piece into the decor with a subtle, yet distinct presence.
Cones are also prominent in technological gadgets, such as speakers or microphones, where they play an essential role in sound production and quality. Their acoustically beneficial shape makes them ideal for devices meant to project or capture audio clearly.
Ergonomic Handheld Devices
Liquid and Sound Direction
Targeted Lighting Solutions
Stylish Home Decor Items
High-Fidelity Audio Equipment
Whether enhancing the user experience, achieving a specific utility, or creating visual allure, conical shapes in product design demonstrate a perfect marriage of form following function. Designers leverage the unique properties of the cone to craft innovative, practical, and aesthetically pleasing products that resonate with consumers.
Sustainability and Conical Shapes: A Synergistic Approach
Embracing sustainability in design has become a defining feature of our times, and conical shapes have played a significant role in this movement. The synergy between sustainability and conical forms can be observed in a multitude of design aspects, from reducing material consumption to enhancing energy efficiency.
The streamlined profile of a cone necessitates less material towards its apex, which can lead to a marked reduction in resource use. This quality makes conical shapes particularly suitable for sustainable product design, where material efficiency is just as critical as the end product's functionality.
Conical forms also benefit sustainable building practices. Their shape is conducive to rainwater collection, allowing for efficient water harvesting systems that can reduce a structure's environmental footprint. This design choice aligns with green building certifications and contributes to an edifice's overall resource consciousness.
In urban planning, the conical shape's ability to deflect wind and manage sunlight can be used to create more comfortable and energy-efficient public spaces. By mitigating wind tunnel effects and optimizing shading, cones can enhance the microclimate of urban environments.
Educational and community projects have adopted conical features to create interactive and eco-friendly learning environments. These spaces often incorporate conical structures made from recycled or sustainable materials, serving as hands-on demonstrations of green design principles.
Material Efficiency in Products
Rainwater Harvesting in Buildings
Microclimate Optimization in Urban Design
Sustainable Learning Environments
The adoption of conical shapes in pursuit of sustainability is a testament to the adaptability and foresight inherent in modern design. By aligning form with eco-friendly functions, designers can not only create visually appealing and innovative solutions but also contribute to the preservation and well-being of our environment.
The Future of Conical Shapes in Design Trends
As we gaze into the future of conical shapes within design trends, it's clear that they are poised to continue their influence in innovative and transformative ways. Their timeless nature, coupled with adaptability to new technologies and materials, suggests that cones will endure as a staple in design.
Emerging technologies such as 3D printing offer exciting prospects for conical shapes. With the ability to create complex, intricate forms with precision and speed, designers are exploring new possibilities for conical applications that may have been too challenging or costly to produce in the past.
Smart materials and responsive design elements point to a future where conical shapes could adapt to environmental stimuli, such as temperature or light, changing their functionality or appearance in real-time. These advancements will allow for more dynamic and interactive designs that respond to user needs and preferences.
The interplay between digital experiences and physical spaces may also be enriched by the strategic use of conical shapes. Augmented reality (AR) and virtual reality (VR) could leverage cones to direct users' attention or to navigate virtual environments more intuitively, blending the digital with the tangible.
Biomimicry, which involves drawing inspiration from natural forms and systems, is likely to expand the conical form's applications even further. The field's advancements could lead to new architectural marvels and product designs with improved performance, modeled after nature's own use of cones for efficiency and function.
3D Printing Innovations
Adaptive Smart Materials
AR and VR Integration
Biomimetic Applications
The trajectory of conical shapes in design is one of growth and evolution. By embracing both technological advancements and sustainable practices, designers will continue to discover new dimensions to this classic form, ensuring its relevance and desirability for years to come.
Maintaining Balance: The Challenge of Designing with Cones
While conical shapes offer numerous benefits in design, achieving balance presents a unique set of challenges. The key to successful cone utilization lies in the careful consideration of proportion, aesthetics, and functionality to maintain equilibrium in the final design.
One challenge is ensuring structural stability, especially for conical objects or buildings with a narrow base. It's crucial to engineer these designs to prevent toppling, which might involve complex calculations and innovative materials to provide the necessary support without compromising the design's integrity.
Balancing visual impact is also essential. A conical shape can be a dominant feature within a design; thus, it must be integrated in a way that complements rather than overpowers other design elements. This often involves meticulous spatial planning and the strategic use of colors and textures.
Another consideration is the functional aspect of conical designs. As they taper to a point, there is less space available at the apex, which can limit functionality if not thoughtfully addressed. Designers must cleverly maximize the usable space and consider potential limitations when incorporating cones into their work.
From a user experience perspective, the tactile and interactive qualities of conical forms must be optimized to ensure comfort and accessibility. Conical handles, grips, or touchpoints need to be designed with human ergonomics in mind to facilitate ease of use.
Designing with cones demands a harmonious blend of stability, usability, and visual appeal, which, when achieved, can lead to distinctive and memorable outcomes.
In conclusion, while conical shapes hold immense potential in various design disciplines, their implementation requires a meticulous approach. By overcoming the challenges of balance, designers can elevate the function and aesthetic of their creations, fully harnessing the power of the conical form.
Case Studies: Successful Conical Shape Implementations
Examining case studies of successful conical shape implementations reveals the impactful and innovative ways in which designers have leveraged this geometric form. These real-world examples highlight the effectiveness and versatility of cones in various design applications.
One illustrative case is the use of a conical shape in renowned architect Frank Gehry's design of the Museum of Pop Culture in Seattle. The building features a conical dome that not only creates an iconic silhouette but also contributes to the building's acoustic excellence, making it a landmark for both its appearance and functionality.
Another example can be seen in consumer electronics, such as the Bang & Olufsen BeoLab 90 loudspeakers. These conical speakers are celebrated for their advanced sound technology and their striking design, which seamlessly merges the cones' form with cutting-edge audio engineering to deliver unmatched sound quality.
In the automotive industry, the Tesla Supercharger network utilizes the conical shape for its charging stations. This design choice serves a practical function, protecting the connectors from the elements, while also establishing a futuristic and visually cohesive brand aesthetic across the grid.
The implementation of conical shapes in product packaging also demonstrates success in engaging consumers. Take, for instance, the distinctive cone-shaped packaging of Toblerone chocolate, which has become iconic and synonymous with the brand's identity, showcasing the shape's potential in marketing and product recognition.
Museum Architecture: Acoustic and Aesthetic Design
Audio Equipment: Sound Quality and Visual Appeal
Electric Charging Stations: Functionality and Branding
Product Packaging: Consumer Engagement and Brand Identity
These case studies illustrate that when conical shapes are implemented thoughtfully and creatively, they can enhance the experience of a product or space, making them memorable and effective in both their function and their message to the audience.
Design Tips: Working with Conical Shapes for Beginners
For beginners venturing into designing with conical shapes, there are several tips that can guide the creative process toward success. Understanding the fundamentals of working with this dynamic form can make a significant difference in the outcome of a design project.
Firstly, it's essential to recognize the importance of proportions. When working with cones, the ratio between the base and the height can dramatically alter the perception of the shape. Experiment with different dimensions to find the right balance for your specific design goals.
Keep in mind the function and context of your design; for every application, a conical shape can offer different advantages. In packaging, for instance, focus on how the shape will impact the user's unboxing experience or on shelf presence. For architectural elements, consider how the cone's shape will influence the structural integrity and overall aesthetic of the building.
Material choices can greatly affect the visual and tactile qualities of a conical design. Play with textures, colors, and finishes to discover how they can enhance or subdue the shape's impact. Lighter materials may give a sense of elegance and simplicity, while heavier ones emphasize solidity and prominence.
It's also important to consider the cone's orientation in your design. An upward-pointing cone can have an entirely different meaning and impact than one that points downward or lies on its side. Think about the message you wish to convey and how the cone's direction can reinforce that narrative.
Assessing Proportions for Visual Balance
Aligning Function with Conical Advantages
Exploring Material Impact on Conical Forms
Considering Orientation and Symbolism
Designing with conical shapes offers boundless opportunities to innovate and captivate. By starting with these fundamental tips, beginners can harness the power of cones to create designs that are both functional and visually engaging.
Conical Shapes and Brand Identity: Creating Visual Impact
Conical shapes can significantly contribute to a brand's identity by creating a distinctive visual impact that resonates with consumers. The strategic use of cones in branding can evoke a sense of innovation, dynamism, and progression that aligns with a company's values and messaging.
When incorporating conical shapes into a brand's visual language, it's important to consider how these forms align with the brand's personality. A sharply tapered cone might suggest cutting-edge precision suited for a tech company, whereas a more gently sloping cone could represent stability and reliability, ideal for financial institutions.
Consistency across all mediums is crucial for building a strong brand identity. Whether it's the packaging, storefront design, or the logo itself, the repeated motif of the conical shape can create a sense of cohesion that strengthens brand recognition and customer recall.
Furthermore, conical forms can be employed in marketing campaigns to focus attention on new products or services. The directionality and pointed nature of cones can act as arrows or beacons, guiding the viewer to the most important information or features.
Alignment with Brand Personality
Cohesion in Visual Branding
Cones as Directional Marketing Tools
In leveraging conical shapes for brand identity, businesses can craft a visual statement that is both memorable and meaningful. The proper integration of these geometric forms can transform the abstract essence of a brand into a concrete, visual form that effectively communicates with its intended audience.
Conclusion: Why Conical Shapes Continue to Captivate
In conclusion, conical shapes hold a timeless allure in the design world, their popularity undiminished by changing trends and technologies. They encapsulate a perfect blend of beauty, utility, and symbolism that can elevate an ordinary design to extraordinary.
The elegance of the cone's tapering form, able to convey movement and energy, has established it as a go-to shape for designers seeking to imbue their work with a sense of aspiration and upward momentum. Moreover, their efficiency in material use aligns seamlessly with the growing demand for sustainability in design practices.
Cones also possess an inherent versatility that enables them to cross disciplines and integrate into various contexts, from architecture to consumer products. Their ability to direct attention and signify direction or precision makes them invaluable in communication and user experience design.
Additionally, the conical shape's potential to be both structurally robust and visually striking ensures that it will remain a favorite for designers looking to strike that delicate balance between form and function.
From their roots in ancient practice to their role in contemporary design, conical shapes continue to captivate our imaginations and serve as pillars of innovation and creativity.
Their enduring appeal is a testament to their adaptability; whether we are conscious of it or not, conical shapes touch every facet of our designed environment. As we look to the future, the cone will undoubtedly continue to be a fundamental, yet ever-evolving, element in the designer's toolkit.
What are the unique benefits of using conical shapes in modern design?
Conical shapes offer unique benefits in modern design, including structural strength due to even stress distribution, space efficiency with a smaller vertical profile, aerodynamic properties that reduce air resistance, and acoustic benefits as they can channel sound effectively. Their form is also conducive to natural light directing, adding functionality and aesthetic appeal to various design applications.
How do conical shapes contribute to sustainability in design?
Conical shapes align with sustainable design by reducing material consumption towards the apex, aiding in rainwater collection for efficient harvesting, and improving structures' aerodynamics to benefit energy efficiency. They embody eco-friendly principles through less resource-intensive production and operational processes.
In which design fields are conical shapes commonly used?
Conical shapes are prevalent in various design fields, including architecture for structural and aesthetic purposes, fashion for creating dynamic silhouettes, graphic design for focus and movement directionality, product design for functionality and ergonomics, and user interface design for guiding user interaction.
What considerations should be taken into account when using conical shapes in design?
Integral considerations when implementing conical shapes include the choice of materials for durability and expressive potential, assessing proportions for visual balance, understanding the impact of the orientation on perception, and ensuring a practical balance between form and function to achieve user-friendly applications.
How can conical shapes influence brand identity and visual communication?
Conical shapes can significantly influence brand identity and communication by creating memorable visual statements that imply innovation, dynamism, and directionality. They can be used to establish a cohesive branding approach, symbolize company values, and focus consumer attention on specific marketing goals.
Article Summary
The conical shape is a versatile design element with applications ranging from structural strength to aesthetic appeal, and its dimensions are critical for accurate execution. Its evolution in modern aesthetics spans various disciplines, symbolizing progress and innovation while offering practical benefits like durability, space efficiency, aerodynamics, acoustics, and optical concentration. | 677.169 | 1 |
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Notes Area of Triangles and Parallelograms
Transcript Notes Area of Triangles and Parallelograms
Notes #71 Area of Triangles and
Parallelograms (11.1)
Postulate 24
Area of a square Postulate
The area of a square is the square of the
length of its sides.
A = s2
Postulate 25
Area Congruence Postulate
If two polygons are congruent, then they
have the same area.
Postulate 26
Area Addition Postulate
The area of a region is the sum of the areas
of its non-overlapping parts.
Theorem 11.1
Area of a Rectangle
The area of a rectangle is the product of its
base and height.
A = bh
Theorem 11.2
Area of a Parallelogram
The area of a parallelogram is the product
of a base and its corresponding height.
A=bh
Theorem 11.3
Area of a Triangle
The area of a triangle is one half the
product of a base and its corresponding height
A = ½ bh
Examples
Find the area of the polygons.
1.
2.
3.
Examples
4. Find the value of x.
Examples
5. Find the perimeter and area of the triangle.
Hypotenuse.: 17 ft.; leg: 8 ft.
Examples
6. Find the area of the quadrilateral.
Example
7. Find the area of the shaded region.
Independent Practice.
#1 Find the value of x.
Independent Practice.
#2 Find the perimeter and area of the triangle.
Hypotenuse.: 53 in.; leg: 45 in.
Independent Practice.
#3 Find the area of the quadrilateral.
Independent Practice.
#4 Find the area of the shaded region.
Independent Practice.
#5 In
ABCD, 𝐴𝐷 base
is 25 units and 𝐴𝐵 is 12
units. Find the height and area of ABCD if
m∠A is 30 | 677.169 | 1 |
Consider the points A(0,12), B(10,9), C(8,0), and D(-4,7). There is a unique square \mathcal{S} such that each of the four points is on a different side of \mathcal{S}. Let K be the area of \mathcal{S}. Find the remainder when 10 K is divided by 1000 . | 677.169 | 1 |
Projections of solids (Cylinder, Cone, Pyramid and Prism) along with frustum with its inclination to one reference plane and with two reference planes
6.3
Section of such solids and the true shape of the section
6.4
Development of surfaces
6.1 Classification of Solids
Solids may be divided into two main groups:
(1) Polyhedra
(2) Solids of revolution.
Polyhedra: A polyhedron is defined as a solid bounded by planes called faces. When all faces are equal and regular, the polyhedron is said to be regular. There are seven regular polyhedra which may be defined as stated below:
(i) Tetrahedron (fig. 1): It has four equal faces, each an equilateral triangle.
(ii) Cube or hexahedron (fig. 2): It has six faces, all equal squares.
(vi) Prism: This is a polyhedron having two equal and similar faces called its ends or bases, parallel to each other and joined by other faces which are parallelograms. The imaginary line joining the centres of the bases is called the axis. A right and regular prism (fig. 6) has its axis perpendicular to the bases. All its faces are equal rectangles.
(vii) Pyramid: This is a polyhedron having a plane figure as a base and many triangular faces meeting at a point called the vertex or apex. The imaginary line joining the apex with the centre of the base is its axis.
A right and regular pyramid (fig. 7) has its axis perpendicular to the base which is a regular plane figure. Its faces are all equal isosceles triangles.
Oblique prisms and pyramids have their axes inclined to their bases. Prisms and pyramids are named according to the shape of their bases, as triangular, square, pentagonal, hexagonal etc.
Solids of revolution
(i) Cylinder (fig. 8): A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides which remains fixed. It has two equal circular bases. The line joining the centres of the bases is the axis. It is perpendicular to the bases.
(ii) Cone (fig. 9): A right circular cone is a solid generated by the revolution of a right-angled triangle about one of its perpendicular sides which is fixed.
It has one circular base. Its axis joins the apex with the centre of the base to which it is perpendicular. Straight lines drawn from the apex to the circumference of the base-circle are all equal and are called generators of the cone. The length of the generator is the slant height of the cone.
(iii) Sphere (fig. 10): A sphere is a solid generated by the revolution of a semi-circle about its diameter as the axis. The mid-point of the diameter is the centre of the sphere. All points on the surface of the sphere are equidistant from its centre.
(iv) Frustum: When a pyramid or a cone is cut by a plane parallel to its base, thus removing the top portion, the remaining portion is called its frustum (fig. 11).
(v) Truncated: When a solid is cut by a plane inclined to the base it is said to be truncated
6.2 Projections of solids (Cylinder, Cone, Pyramid and Prism)
A solid has three dimensions, viz. Length, breadth and thickness. To represent a solid on a flat surface having only length and breadth, at least two orthographic views are necessary. Sometimes, additional views projected on auxiliary planes become necessary to make the description of a solid complete.
Projections of solids in simple positions
A solid in simple position may have its axis perpendicular to one reference plane or parallel to both. When the axis is perpendicular to one reference plane, it is parallel to the other. Also, when the axis of a solid is perpendicular to a plane, its base will be parallel to that plane. When a plane is parallel to a reference plane, its projection on that plane shows its true shape and size.
Therefore, the projection of a solid on the plane to which its axis is perpendicular, will show the true shape and size of its base.
Hence, when the axis is perpendicular to the ground, i.e. to the H.P., the top view should be drawn first, and the front view projected from it.
When the axis is perpendicular to the V.P., beginning should be made with the front view. The top view should then be projected from it.
When the axis is parallel to both the H.P. And the V.P., neither the top view nor the front view will show the actual shape of the base. In this case, the projection of the solid on an auxiliary plane perpendicular to both the planes, viz. The side view must be drawn first. The front view and the top view are then projected from the side view. The projections in such cases may also be drawn in two stages.
Axis perpendicular to the H.P.:
Problems:
1. Draw the projections of a triangular prism, base 40 mm side and axis 50 mm long, resting on one of its bases on the H.P. With a vertical face perpendicular to the V.P.
(i) As the axis is perpendicular to the ground i.e. the H.P. Begin with the top view. It will be an equilateral triangle of sides 40 mm long, with one of its sides perpendicular to xy. Name the corners as shown, thus completing the top view. The corners d, e and fare hidden and coincide with the top corners a, b and c respectively
(ii) Project the front view, which will be a rectangle. Name the corners. The line b'e' coincides with a'd'.
2. Draw the projections of a pentagonal pyramid, base 30 mm edge and axis 50 mm long, having its base on the H.P. And an edge of the base parallel to the V.P. Also draw its side view.
(i) Assume the side DE which is nearer the V.P., to be parallel to the V.P. As shown in the pictorial view.
(ii) In the top view, draw a regular pentagon abcde with ed parallel to and nearer xy. Locate its centre o and join it with the corners to indicate the slant edges.
(iii) Through o, project the axis in the front view and mark the apex o', 50 mm above xy. Project all the corners of the base on xy. Draw lines o'a', o'b' and o'c' to show the visible edges. Show o'd' and o'e' for the hidden edges as dashed lines.
(iv) For the side view looking from the left, draw a new reference line x1y1 perpendicular to xy and to the right of the front view. Project the side view on it, horizontally from the front view as shown. The respective distances of all the points in the side view from x1y1, should be equal to their distances in the top view from xy. This is done systematically as explained below:
(v) From each point in the top view, draw horizontal lines up to x1y1. Then draw lines inclined at 45° to x1y1 (or xy) as shown. Or, with q, the point of intersection between xy and x1y1 as centre, draw quarter circles. Project up all the points to intersect the corresponding horizontal lines from the front view and complete the side view as shown in the figure. Lines o1d1 and o1c1 coincide with o1e1 and o1a1 respectively.
Axis perpendicular to the V.P.:
A hexagonal prism (fig. 14) has one of its rectangular faces parallel to the H.P. Its axis is perpendicular to the V.P. And 3.5 cm above the ground. Draw its projections when the nearer end is 2 cm in front of the V.P. Side of base 2.5 cm long; axis 5 cm long.
(i) Begin with the front view. Construct a regular hexagon of 2.5 cm long sides with its centre 3.5 cm above xy and one side parallel to it.
(ii) Project down the top view, keeping the line for nearer end, viz. 1-4, 2 cm below xy.
2. A square pyramid fig. 15, base 40 mm side and axis 65 mm long, has its base in the V.P. One edge of the base is inclined at 30° to the H.P. And a corner contained by that edge is on the H.P. Draw its projections.
(i) Draw a square in the front view with the corner d' in xy and the side d'c' inclined at 30° to it. Locate the centre o' and join it with the corners of the square.
(ii) Project down all the corners in xy (because the base is in the V.P.). Mark the apex o on a projector through o'. Draw lines for the slant edges and complete the top view.
Axis parallel to both H.P. And the V.P.
A triangular prism1 base 40 mm side and height 65 mm is resting on the H.P. On one of its rectangular faces with the axis parallel to the V.P. Draw its projections.
As the axis is parallel to both the planes, begin with the side view.
(i) Draw an equilateral triangle representing the side view, with one side in xy.
(ii) Project the front view horizontally from this triangle.
(iii) Project down the top view from the front view and the side view, as shown.
Problems:
Draw the projections of the following solids, situated in their respective positions, taking a side of the base 40 mm long or the diameter of the base 50 mm long and the axis 65 mm long.
1. A hexagonal pyramid, base on the H.P. And a side of the base parallel to and 25 mm in front of the V.P.
2. A square prism, base on the H.P., a side of the base inclined at 30° to the V.P. And the axis 50 mm in front of the V.P.
3. A triangular pyramid, base on the H.P. And an edge of the base inclined at 45° to the V.P.; the apex 40 mm in front of the V.P.
Projections of solids with axis inclined to one of the reference planes and parallel to the other:
When a solid has its axis inclined to one plane and parallel to the other, its projections are drawn in two stages.
(1) In the initial stage, the solid is assumed to be in simple position, i.e. its axis perpendicular to one of the planes.
If the axis is to be inclined to the ground, i.e. the H.P., it is assumed to be perpendicular to the H.P. In the initial stage. Similarly, if the axis is to be inclined to the V.P., it is kept perpendicular to the V.P. In the initial stage.
Moreover
(i) if the solid has an edge of its base parallel to the H.P. Or in the H.P. Or on the ground, that edge should be kept perpendicular to the V.P.; if the edge of the base is parallel to the V.P. Or in the V.P., it should be kept perpendicular to the H.P.
(ii) If the solid has a corner of its base in the H.P. Or on the ground, the sides of the base containing that corner should be kept equally inclined to the V.P.; if the corner is in the V.P., they should be kept equally inclined to the H.P.
Axis inclined to the VP and parallel to the HP:
1. Draw the projections of a pentagonal prism, base 25 mm side and axis 50 mm long, resting on one of its faces on the H.P., with the axis inclined at 45° to the V.P.
In the simple position, assume the prism to be on one of its faces on the ground with the axis perpendicular to the V.P.
Draw the pentagon in the front view with one side in xy and project the top view [fig. 17].
The shape and size of the figure in the top view will not change, so long as the prism has its face on the H.P. The respective distances of all the corners in the front view from xy will also remain constant.
Method I:
(i) Alter the position of the top view, i.e. reproduce it so that the axis is inclined at 45° to xy. Project all the points upwards from this top view and horizontally from the first front view, e.g. a vertical from a intersecting a horizontal from a' at a point a'1.
(ii) Complete the pentagon a'1b'1c'1d'1e'1 for the fully visible end of the prism. Next, draw the lines for the longer edges and finally, draw the lines for the edges of the other end. Note carefully that the lines a'1 1'1, 1'12'1 and 1'15'1 are dashed lines. e'1 5'1 is also hidden but it coincides with other visible lines.
Method II:
(i) Draw a new reference line x1y1, making 45° angle with the top view of the axis, to represent an auxiliary vertical plane.
(ii) Draw projectors from all the points in the top view perpendicular to x1y1 and on them, mark points keeping the distance of each point from x1y1 equal to its distance from xy in the front view. Join the points as already explained. The auxiliary front view and the top view are the required projections.
Axis inclined to the H.P. And parallel to the V.P.
Problems:
1. A hexagonal pyramid, base 25 mm side and axis 50 mm long, has an edge of its base on the ground. Its axis is inclined at 30° to the ground and parallel to the V.P. Draw its projections. In the initial position assume the axis to be perpendicular to the H.P.
Draw the projections with the base in xy and its one edge perpendicular to the V.P. Fig. 18 (i)
If the pyramid is now tilted about the edge AF (or CD) the axis will become inclined to the H.P. But will remain parallel to the V.P. The distances of all the corners from the V.P. Will remain constant.
The front view will not be affected except in its position in relation to xy. The new top view will have its corners at same distances from xy, as before.
Method I: [fig. 18 (ii)]:
(i) Reproduce the front view so that the axis makes 30° angle with xy and the point a' remains in xy.
(ii) Project all the points vertically from this front view and horizontally from the first top view. Complete the new top view by drawing (a) lines joining the apex o'1 with the corners of the base and (b) lines for the edges of the base.
The base will be partly hidden as shown by dashed line a1b1, e1f1 and f1a1. Similarly, o1f1 and o1a1 are also dashed lines.
Method II:
(i) Through a' draw a new reference line x1y1 inclined at 30° to the axis, to represent an auxiliary inclined plane.
(ii) From the front view project, the required top view on x1y1, keeping the distance of each point from x1y1 equal to the distance of its first top view from xy, viz. e1q = eb' etc.
2. Draw the projections of a cone, base 75 mm diameter and axis 100 mm long, lying on the H.P. On one of its generators with the axis parallel to the V.P.
(i) Assuming the cone to be resting on its base on the ground, draw its projections.
(ii) Re-draw the front view so that the line o'7' (or o'1 ') is in xy. Project the required top view as shown. The lines from o1 should be tangents to the ellipse.
The top view obtained by auxiliary-plane method is shown in fig. 13-24(ii). The new reference line x1y1 is so drawn as to contain the generator o'1' instead of o'7' (for sake of convenience). The cone is thus lying on the generator o'1 '. Note that 1 '1 1 = 1 '1, o'o1 = 4'o etc. Also note that the base is fully visible in both the methods.
Projections of the solids with axis inclined to both the H.P. And the V.P.
The projections of a solid with its axis inclined to both the planes are drawn in three stages:
(i) Simple position
(ii) Axis inclined to one plane and parallel to the other
(iii) Final position.
The second and final positions may be obtained either by the alteration of the positions of the solid, i.e. the views, or by the alteration of reference lines.
Problem:
1. A square prism, base 40 mm side and height 65 mm, has its axis inclined at 45° to the H.P. And has an edge of its base, on the H.P and inclined at 30° to the V.P. Draw its projections.
i) Assuming the prism to be resting on its base on the ground with an edge of the base perpendicular to the V.P., draw its projections. Assume the prism to be tilted about the edge which is perpendicular to the V.P., so that the axis makes 45° angle with the H.P.
(ii) Hence, change the position of the front view so that the axis is inclined at 45° to xy and f' (or e') is in xy. Project the second top view. Again, assume the prism to be turned so that the edge on which it rests, makes an angle of 30° with the V.P., keeping the inclination of the axis with the ground constant. The shape and size of the second top view will remain the same; only its position will change. In the front view, the distances of all the corners from xy will remain the same as in the second front view.
(iii) Therefore, reproduce the second top view making f1g1 inclined at 30° to xy. Project the final front view upwards from this top view and horizontally from the second front view, e.g. a vertical from a1 and a horizontal from a' intersecting at a'1. As the top end is further away from xy in the top view it will be fully visible in the front view. Complete the front view showing the hidden edges by dashed lines.
(iv) The second top view may be turned in the opposite direction as shown. In this position, the lower end of the prism, viz. e'lf'1g'1h'1 will be fully visible in the front view.
Method II:
(i) Draw the top view and front view in simple position.
(ii) Through f', draw a new reference line x1y1 making 45° angle with the axis. On it, project the auxiliary top view.
(iii) Draw another reference line x2y2 inclined at 30° to the line fig 1. From the auxiliary top view, project the required front view, keeping the distance of each point from x2y2, equal to its distance (in the first front view) from x1y1 i.e. a'1q1 = a'q etc.
6.3 Section of such solids and the true shape of the section
. In this chapter sections of different solids are explained in stages by means of typical problems as follows:
1. Sections of prisms
2. Sections of pyramids
3. Sections of cylinders
4. Sections of cones
5. Sections of spheres.arehence, the section will be a line in the top view coinciding with the H.T.of the section plane.
(i) Draw a line H.T. In the top view (to represent the section plane) parallelto section(i) Draw the projections of the pyramid in the required position and show a line V.T. For the section plane, parallel to andProblem 9 (fig. 11): A cylinder of 40 mm diameter, 60 mm height and havingits axis vertical, is cut by a section plane, perpendicular to the V.P., inclined at 45° tothe H.P. And intersecting the axis32 mm above the base. Drawits front view, sectional top view,sectional side view and true shape of the section. greater throughDraw the sectional side view by projecting the points on corresponding generators, as shown.Section plane parallel to a generator of the cone:
Problem (fig. 17): The cone in same position as in fig. 12, is cut bayand axis 55 mm long is resting on the H.P. On its base. It is cut by a section plane,perpendicular to both the H.P. And the V.P. And 6 mm away from the axis. Drawits front view, top view and sectional side view.
The section will be a line, perpendicular to xy, in both the front viewand the top view. The side view will show the true shape of the section. The widthof the section at any point, say c', will be equal to cc1 obtained by the circlemethod [fig. 18(i)].
(i) Draw the side view of the cone.
(ii) Project the points (on the section) in the side view taking the widths from thetoptrue shape of the section is always a circle.
The sphere in fig. 19 is cut by a horizontal sectionplane. The true shape of the section (seen in the top view) isa circle of diameter a'a'. The width of the section at any pointsay b', is equal to the length of the chord bb.
(2) Section plane parallel to the V.P.: When the sphere iscut by a section plane parallel to the V.P. (fig. 20), the trueshape of the section, seen in the front view, is a circle ofdiameter cc. The width of section at any point d is equal tothe length of the chord d'd'.
(3) Section plane perpendicular to the V.P. And inclined to the H.P.:
Problem (fig. 21): A sphere of 50 mm diameter is cut by a sectionplaneof 10 mm radius drawn with o' as center. Mark many points on this line.
Method I:
(i) Find the width of section at each point in the top view as shown infig. 19. For example, the chord cc is the width of section at the point c'.
(ii) Draw a curve through the points thus obtained. It will be an ellipse. Thetrue shape of the section will be a circle of diameter a'g'. theThe widths of the section of the sphere at various points are obtained from the semi-circle drawn in the top view.
Problems:
1. A cube of 50 mm long edges is resting on the H.P. With a vertical face inclined at 30° to the V.P. It is cut by a section plane, perpendicular to the V.P., inclined at 30° to the H.P. And passing through a point on the axis, 38 mm above the H.P. Draw the sectional top view, true shape of the section and development of the surface of the remaining portion of the cube.
2. A hexagonal prism, side of base 35 mm and height 75 mm is resting on one of its corners on the H.P. With a longer edge containing that corner inclined at 60° to the H.P. And a rectangular face parallel to the V.P. A horizontal section plane cuts the prism in two equal halves.
(i) Draw the front view and sectional top view of the cut prism.
(ii) Draw another top view on an auxiliary inclined plane which makes an angleof 45° with the H.P.
6.4 Development of surfaces
Imagine that a solid is enclosed in a wrapper of thin material, such as paper. If this covering is opened out and laid on a flat plane, the flattened-out paper is the development of the solid. Thus, when surfaces of a solid are laid out on a plane, the figure obtained is called its development.
Fig. 1 shows a square prism covered with paper in process of being opened out. Its development (fig. 2) consists of four equal rectangles for the faces and two similar squares for its ends. Each figure shows the true size and shape of the corresponding surface of the prism. The development of a solid, thus represents the actual shape of all its surfaces which, when bent or folded at the edges, would form the solid.
Hence, it is very important to note that every line on the development must bathe true length of the corresponding edge on the surface.
The knowledge of development of surfaces is essential in many industries such as automobile, aircraft, ship building, packaging and sheet-metal work. In construction of boilers, bins, process-vessels, hoppers, funnels, chimneys etc., the plates are marked and cut according to the developments which, when folded, form the desired objects. The form of the sheet obtained by laying all the outer surfaces of the solid with suitable allowances for the joints is known as pattern.
Only the surfaces of polyhedra (such as prisms and pyramids) and single curved surfaces (as of cones and cylinders) can be accurately developed. Warped and double-curved surfaces are undevelopable. These can however be approximately developed by dividing them up into many parts.
This chapter deals with the following topics:
1. Methods of development.
2. Developments of lateral surfaces of right solids.
3. Development of transition pieces.
4. Spheres (approximate method).
Methods of Development:
The following are the principal methods of development:
1. Parallel-line development:
It is employed in case of prisms and cylinders in which stretch-out-line principle issued. Lines A-A and A1-A1in fig.2 are called the stretch-out lines.
2. Radial-line development:
Itis used for pyramids and cones in which the true length of the slant edge or the generator is used as radius.warpedof the side of the squares being equal to the length of the edge of the cube.
Problem:
1. Draw the development of the surface of the part P of the cube,the front view of which is shown in fig. 3(i).Name all the corners of the cube and the points at which the edges are cut.
(i) Draw the stretch-out lines A-A and E-E directly in line with the front view, andassuming the cube to be whole, draw four squares for the vertical faces, onesquare for the top and another for the bottom as shown in fig. 3(ii).
(ii) Name all the corners. Draw a horizontal line through 1' to cut AE at 1 andDH at 4. a' b' is the true length of the edge. Hence, mark a point 2 onAB and 3 on CD such that A 2 = a' 2' and C 3 = c' 3'. Mark the point3 on CD in the top square also.
(iii) Draw lines 1-2, 2-3, 3-4 and 4-1, and complete the development as shown.Keep lines for the removed portion, viz. A1, A2, 3D, D4 and DA thin and fainter.
2. Draw the development of the surface of the part P of the cubeshown in two views in fig. 4(i).
(i) Draw horizontal lines through points 1 ', 2' and 5' to cut AE in 1, BF in 2and DH in 5 respectively. Lines b'c' and c'd' do not show the true lengths of the edges. The sides of the square in the top view show the true length. Therefore, mark points 3 in BC and 4 in CD such that 83 = b3 and C4 = c4.(i) Obtain all the points except 5 and 6 by drawing horizontal lines. Note thatpoints 3 and 8 lie on vertical lines drawn through the mid-points of BC and EF.
The development of the lateral surface of a cylinder is a rectangle having one side equal to the circumference of its base-circle and the other equal to its length.
Problem
1. Develop the lateral surface of the truncated cylinder shown in fig. 7(i).
(i) Divide the circle in the top view into twelve equal parts. Project the division points to the front view and draw the generators. Mark points a', b' andbshorterant edge of the pyramid, draw an arc of the circle. With radius equal to the true length of the side of the base, step-off (on this arc) the same number of divisions as the number of sides of the base.
(ii) Draw lines joining the division-points with each other in correct sequence and with the centre for the arc. The figure thus formed (excluding the arc) is the development ofas explained above. On 01 mark a point A such that OA = o'a'. o'2' (withwhich o'3' coincides) is not the true length of the slant edge.
(ii) Hence, through b', draw a line parallel to the base and cutting o' a' at b".o'b" is the true length of o'b' as well as o'c'. Mark a point B in 02 andC in 03 such that OB = OC = o'b".
(iii) Draw lines AB, BC and CA and complete the required development as shown.Keep the arc and the lines for the removed part fainter.
2. Draw the development of the lateral surface of the frustum ofthe square pyramid shown in fig. 9 (i).
(i) Determine the position of the apex. None of the lines in the front viewshows the true length of the slant edge. Therefore, draw the top view andmake any one line (for the slant edge) horizontal, i.e. parallel to xy anddetermine the true length o'1 '1. Through a', draw a line parallel to the baseand obtain the true length o'a".
(ii) With O as centre and radius o'1 '1, draw an arc and obtain the developmentof(i) Mark the mid-point P of CD and Q of A1B1. Draw a line joining P and Q andcutting CC1 at R and BB1 at S. Transfer these points to the front view and thetop view. For example, with o' as centre and radius o'R, draw an arc cuttingo' A1 at R1. Through R1, draw a line parallel to the base and cutting c'c'1 at r'.Project rsand sq which will show the top view of the line PQ. p'r's'q' will be thepath of the line PQ in the front view.
Cone
The development of the curved surface of a cone is a sector of a circle, the radius and the length of the arc of which are respectively equal to the slant height and the circumference of the base-circle of the cone.
Problem:
1. Draw the development of the lateral surface of the truncated cone shown in fig. 11 (i).the the cone shown in fig. 12
Draw the development as explained in problem [fig. 12 (ii)]. For the points at which the base of the cone is cut, mark points A and A1 on the arcs2-3 and 11-12 respectively, such that A2 = A1 12 = a2. Draw the curve passing through the points A, B, C etc. The figure enclosed between this curve and the arcA-A1 is the required development.
3. Draw the projections of a cone resting on the ground on its base and show on them, the shortest path by which a point P, starting from a point on the circumference of the base and moving around the cone will return to the same point. Base of cone 61 mm diameter; axis 75 mm long.
(i) Draw the projections and the development of the surface of the cone showing all twelve generators (fig. 13). The development may be drawn attached to o'1 '.
(ii) Assume that P starts from the point 1 (i.e. point 1' in the front view). Draw a straight line 1 '1' on the development. This line shows the required shortest path. Let us take a point P4 at which the path cuts the generator o'4. Mark a point P"4 on o'1' such that o'P"4 = o'P4. This can be done by WithDevelopment of transition pieces:
Pipes are used in many industries to convey hot or cold fluids. When two different sizes and shapes of pipes are joined using special pipe joint which is known as transition piece. In most cases, transition pieces are composed of plane surfaces and conical surfaces, the latter being developed by triangulation.
Problem:
1. In air-conditioning system a rectangular duct of 100 mm x 50 mm connects another rectangular duct of 50 mm x 25 mm through the transition piece as shown in fig. 14(i). Neglecting thickness of a metal sheet, develop the lateral surface of the transition piece as shown in fig. 14(ii).
The transition piece is a frustum of a rectangular pyramid.
(i) Determine the position of the apex of the pyramid by extending a'p' and b'q'as shown. None of the lines in the front view shows the true length of the slant edge. Therefore, draw the top view and make any slant line parallel toxy and determine its true length o'b".
(ii) With O as centre and radius o'b", draw an arc and obtain the development of a whole pyramid as shown.
(iii) With O as centre and radius o'q", draw an arc cutting oa, ob, oc at points p, q, etc. respectively. Join them in sequence and complete the development as shown.
2. An air-conditioning duct of a square cross-section70 mm x 70 mm connects a circular pipe of 40 mm diameter through the transition piece. Draw the projections and develop the lateral surface of the transition piece.
(i) Draw the front view and the top view as shown in fig. 15 (i).
(ii) Divide the top view of circle into some convenient divisions, say 16 parts as shown.
(iii) Note that the transition piece is composed of four isosceles triangles and four conical surfaces. The seam is a long line 1-P.
(iv) Begin the development from the seam line 1-P (1 '-P'). As shown in fig. 15(ii)draw the right-angle triangle 1-P-b, whose base pb is equal to half the side ab and whose hypotenuse 1-b is equal to the true length 1 '-b' of side 1-b.
(v) The conical surfaces are developed by the triangulation method as follows.(vi) In the top view, join division of the circle 1, 2, 3 etc. with the corner a,b, c and d. Project them in the front view as shown. Obtain the true length of sides of each triangle as shown.
(vii) With b as centre and 2'b' (true length) radius draw an arc, cutting the arc drawn with 1' as centre and 1 '2' as radius. Similarly, obtain the points 3', 4', 5' etc. Join them in the proper order as shown. | 677.169 | 1 |
Did Pythagoras really come up with his Theorem or did he murder to get it? Such is one of life's mysteries which are discussed in this awesome video! I even do some role play to help get the point across.
When I'm not talking about murder (alleged) I am telling you how the rule was discovered and can be used to determine if a triangle has a right angle or not.
I ponder how Pyramids were built and about the formula for Pythagoras' Theorem and the wonderful subject of Pythagorean triples and how they can be found. We also look at similar triangles and how they can be used to find Pythagorean Triples.
I even do a little bit of Michael Jackson! It really doesn't get much better than this.
There are no current errors with this video ... phew!
There are currently no lesson notes at this time to download. I add new content every week.
Video tags
right trianglesright trianglepythagoras theorempythagorean theorempythagorean theorem right triangleright triangles and the pythagorean theoremwhat is pythagoras theorem in mathswhat is pythagoras theoremHow do I tell if a triangle has a right angleright angled triangleright angled triangle problemspythagorean triplesmichael jackson | 677.169 | 1 |
FAQs on Trigonometric Functions Class 11 Notes Maths Chapter 3
Ans. The main trigonometric functions are sine (sin), cosine (cos), and tangent (tan). These functions relate the angles of a triangle to the ratios of its sides.
2. How are trigonometric functions used in real-life applications?
Ans. Trigonometric functions are used in various real-life applications such as navigation, engineering, physics, and astronomy. They help in calculating distances, angles, heights, and trajectories in these fields.
3. What is the unit circle and its relationship with trigonometric functions?
Ans. The unit circle is a circle with a radius of 1 unit. It is centered at the origin of a coordinate plane. The trigonometric functions can be defined using the coordinates of points on the unit circle, where the angle is measured from the positive x-axis.
4. How are trigonometric functions related to right-angled triangles?
Ans. Trigonometric functions are primarily used to calculate the ratios of sides in a right-angled triangle. For example, sine (sin) is the ratio of the length of the side opposite an angle to the hypotenuse, cosine (cos) is the ratio of the length of the adjacent side to the hypotenuse, and tangent (tan) is the ratio of the length of the opposite side to the adjacent side.
5. How can trigonometric functions be used to solve problems involving angles and sides in triangles?
Ans. Trigonometric functions can be used to solve problems involving angles and sides in triangles using trigonometric identities and formulas. By knowing the values of certain trigonometric functions and one side or angle of a triangle, we can calculate the other unknown sides or angles. This is commonly done using the sine, cosine, and tangent functions, along with their inverses.
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Students of JEE can study Revision Notes: Trigonometric Function alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Revision Notes: Trigonometric Function Trigonometric Function is prepared as per the latest JEE syllabus. | 677.169 | 1 |
2
A
286 sq. units
B
$$\sqrt{286}$$ sq. units
C
300 sq. units
D
$$\sqrt{300}$$ sq. units
3
MHT CET 2021 23th September Morning Shift
MCQ (Single Correct Answer)
+2
-0
The distance between the parallel lines $$\frac{x-2}{3}=\frac{y-4}{5}=\frac{z-1}{2}$$ and $$\frac{x-1}{3}=\frac{y+2}{5}=\frac{z+3}{2}$$ is
A
$$\frac{1}{\sqrt{38}}$$ units
B
$$\sqrt{\frac{333}{38}}$$ units
C
$$\sqrt{\frac{300}{37}}$$ units
D
$$\sqrt{\frac{300}{35}}$$ units
4
MHT CET 2021 23th September Morning Shift
MCQ (Single Correct Answer)
+2
-0
The coordinates of the foot of the perpendicular drawn from the origin to the plane $$2 x+y-2 z=18$$ are | 677.169 | 1 |
$008 / 2 / 11(a)$ $A B C D$ is a parallelogram. $
Updated on
Sat Dec 23 2023
Question
008/2/11(a)008//2//11(a) ABCDABCD is a parallelogram. XX is the point on BCBC such that BX:XC≡2:1BX:XC-=2:1. AB→=2pvec(AB)=2p and AD→=3qvec(AD)=3q.
Find, in terms of pp and qq.
(a) AC→vec(AC),
Answer (a)AC→≡(a) vec(AC)-=
III
(b) Ax→vec(Ax),
Answer (b) AX→≡vec(AX)-=
[I]
Best Answer
I apologize, but the first question is incomplete. The number 008 is not written properly and there is no instruction or equation to solve.
For the second question:
(a) AC→vec(AC)
Answer: AC→vec(AC) is equal to AD→+DC→vec(AD)+ vec(DC). Since ABCDABCD is a parallelogram and AB→=DC→vec(AB)= vec(DC), then DC→=−2pvec(DC)=-2p. Thus, AC→=AD→+DC→=3q+(−2p)=3q−2pvec(AC)= vec(AD)+ vec(DC)=3q+(-2p)=3q-2p.
(b) AX→vec(AX)
Answer: Since BX:XC=2:1BX:XC=2:1, then BX→=23BC→vec(BX)=(2)/(3) vec(BC) and XC→=13BC→vec(XC)=(1)/(3) vec(BC). Since BC→=CD→+DB→=−AD→+AB→=2p−3qvec(BC)= vec(CD)+ vec(DB)=- vec(AD)+ vec(AB)=2p-3q, then BX→=23(2p−3q)=43p−2qvec(BX)=(2)/(3)(2p-3q)=(4)/(3)p-2q and XC→=13(2p−3q)=23p−qvec(XC)=(1)/(3)(2p-3q)=(2)/(3)p-q. Thus, AX→=AB→+BX→=2p+43p−2q=103p−2qvec(AX)= vec(AB)+ vec(BX)=2p+(4)/(3)p-2q=(10)/(3)p-2q. | 677.169 | 1 |
Hint: We need to understand the condition given in the problem and then we have to use appropriate formulas to find the coordinates of the points which trisect \[AB\]. We have to use the section formula for internal division to calculate the coordinates of the points which trisect \[AB\]. Formula used: Section formula for internal division: Coordinates of the point $P(x,y,z)$ which divides line segment joining $A({x_1},{y_1},{z_1})$ and $B({x_2},{y_2},{z_2})$ internally in the ratio $m:n$ are given by, \[P(x,y,z) = (\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}},\dfrac{{m{z_2} + n{z_1}}}{{m + n}})\]
Note: Coordinate of $Q$ can also be calculated using midpoint formula for segment $PB$ , as from the diagram we can see that $Q$ is the midpoint of segment $PB$ and can find the coordinate of P by the midpoint formula after finding the coordinates of point Q as P behaves as the midpoint of AQ. The midpoint formula for 3-D coordinates $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)$ | 677.169 | 1 |
Triangle Congruence Worksheet Answer Key
Triangle Congruence Worksheet Answer Key. She attracts each so that two sides are four in. Useful for revision, classwork and homework.. Proving triangles congruent worksheet answer key. One purpose why so much consideration is given to congruent triangles.
Determine the missing congruence property in a pair of triangles to substantiate the concept. Acute triangle right triangle obtuse triangle all acute angles one right angle Draw two triangles that match each part of the venn diagram below. The reply key's routinely generated and is positioned on the second.
American Credit Acceptance Repo Time
Congruent Triangles Asa And Aas Answers from… If we enlarge one shape to make it larger or smaller, then the shapes are said to be similar. The corresponding sides of comparable shapes have to be in the same proportion and the …
Check whether two triangles PQR and STU are congruent. Check whether two triangles PQR and CDE are congruent. Check whether two triangles PQR and ABC are congruent.
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A really nice exercise for allowing college students to know the ideas of the Congruence of Triangles. Some of the worksheets for this idea are […] These worksheets, created by specialists, can be used on a regular basis by your teen. In similar triangles, the ratio of the corresponding sides are equal.
This checks the scholars capacity to grasp Congruence of Triangles. This tests the scholars capacity to know Congruent Triangles. Members have exclusive amenities to obtain an individual worksheet, or a complete level.
Proving Triangles Congruent Worksheet Reply Key Triangle
Calculate the perimeter of triangle ABC. Since this could be a right-angled triangle, we will use Pythagoras! Algebra finance simple and compound interest evaluate worksheet quiz solving linear equations easy.
View worksheet Independent Practice 1 A actually great activity for allowing students to grasp the ideas of Similar & Congruent Figures. View worksheet Independent Practice 2 Students classify figures as congruent or related. The answers may be found under.
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This exercise was designed for a high school stage geometry class. If a second triangle is successfully fashioned you may be requested if they're congruent. This obtain contains 6 totally different questions units for students to practice identifying the 5 major congruence types.
Students can use math worksheets to master a math ability by way of apply, in a examine group or for peer tutoring. Use the buttons under to print, open, or download the PDF version of the Identifying Shapes math worksheet. The measurement of the PDF file is bytes.
Each question set has 6 problems. The completely different versions make this very versatile in order that academics could use as an exit ticket or evaluation instead of simply independent practice. The question units every match on a half sheet of paper, so as to reduce quantity of paper wanted when printing.
Select the very best one in your course. This construction shows how to draw the perpendicular bisector of a given line section with compass and straightedge or ruler. This both bisects the phase , and is perpendicular to it.
Make positive you're employed is the proof usinf the poatule , or theorem, feom the important thing on the las two pages. This is a coloring exercise for sixteen problems. Worksheet congruent triangles answer key (QSTION.CO) – If a second triangle is successfully shaped you'll be asked if they're congruent.
Which is larger the moons interval of rotation or its period of revolution b. Attempt to prove these triangles congruent if you cannot due to a ignorance its time to take a detour three. Delightful to find a way to the blog, on this explicit event i'm going to offer you relating to proving triangles congruent worksheet solutions.
Congruence of Triangles • Congruent triangles are triangles which have the same size and shape. This implies that the corresponding sides are equal and the corresponding angles are equal • In the above diagrams, the corresponding sides are a and d; b and e ; c and f. • The corresponding angles are x and s; y and t; z and u.
Learn cpctc congruent with free interactive flashcards. Choose from 31 different sets of cpctc congruent flashcards on Quizlet. Give Thanks Turkey Coloring PageHave you ever seen as many turkey coloring pages as what we've got here?!
Check whether or not two triangles ABC and DEF are comparable. Triangle PQR and triangle WXY are right triangles. Because they both have a proper angle.
If three corresponding angles of two triangles are equal then triangles are congruent. The congruent figure super impose one another fully. The high and bottom faces of a kaleidoscope are congruent.
The stepwise mechanism of those worksheets helps students turn out to be properly versed with concepts, as they move on to extra complicated questions. Benefits of Congruent Triangles Worksheets. Angles of Triangles and Congruent Triangles.
On this page you'll find a way to learn or download reply key triangle congruence answer key gina wilson in pdf format. Procedure for missing diagram …. Circle the determine that is congruent to the first determine in the sequence.
Learn concerning the appearance backdrop acclimated to analyze the names of altered accredited and aberrant polygons, corresponding to triangles or hexagons. The similarity and congruence worksheets ask them to discover out whether pairs of triangles are congruent in increasingly troublesome geometrical situations. This work pack also comes with handy reply sheets to speed up marking work.
Worksheet WS four.4D 2004 Name Per Date CONGRUENT TRIANGLES #2 From the markings on the figures, decide if the triangles are congruent by SSS or SAS. Get your pupils in mathematical shape by letting them measurement up this fantastic worksheet on Congruence! The question of Congruence is also applied to shapes such as triangles, the place a collection of strategies are used to ….
Pdf, 327.3 KB Congruent triangles KS3 KS4 non-claculator. Useful for revision, classwork and homework.. If the second triangle is efficiently constructed, you may be requested if it suits.
Answers may in fact differ . A triangle congruence theorem like sss,. If three sides in a single triangle are congruent to three sides in one other.
Related posts of "Triangle Congruence Worksheet Answer Key"Balancing Act Worksheet Answers. During this introductory lesson, students learn the ideas behind balancing chemical equations. This unit incorporates most of the classroom lessons listed in the next part.. ClickHERE for a PPT with a link to the digital worksheet for faculty students and an answer key you need to use in class. Students additionally...
Molar Mass Practice Worksheet. A chemical reaction may be symbolically depicted by way of a chemical equation. Warm soda tastes flat as a end result of there isn't as a lot CO2 dissolved in it as there's in cold soda. The box first accelerates with a negative acceleration till it hits the water. These methods... | 677.169 | 1 |
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The sum of the space of the 2 focal factors to the sting is constant for each point within the curve. The distance between the center. A sphere is a 3-dimensional completely rounded object, whose distance from the middle is the same at any point or level. Cones have a circular face with curved sides which roll up to a point at the top. Any point is understood as the radius. An ellipsoid is a three-dimensional determine whose plane sections are circles or ellipses. Will you be one of those people who will get tripped up on the difference between ellipse and ellipsoid or trapezium and trapezoid? Somebody could require new lungs for a a range of causes, one among which is cystic fibrosis. Sadly, attributable to lack of donor lungs, solely a portion of people with cystic fibrosis who need a lung transplant really obtain one. A frustum is a shape that is still after a portion of a strong pyramid or cone has been minimize off, parallel to the bottom. A crescent is circular disk with a portion of one other circle removed from the edge. A circle is a closed shape which all factors within the boundary are equidistant from a hard and fast point, which is the center.
An ellipse is a curve in a aircraft which surrounds two focal points. A star is a self-intersecting and equilateral polygon which is defined as "an equiangular polygon created by connecting one vertex of a polygon to another." Most stars are seen with 5 points, however extra factors are potential. An equilateral triangle is a type of triangle which is equal in size on all three sides. It has additionally been described as a rectangle whose adjoining sides are equal in length. Traditional rectangles have two sides which are longer than the other two, but all angles are nonetheless equal to 90 levels. Robots have the potential to enhance the quality of life. Turkeys: Some critics stated the subject matter of this piece was ridiculous, but this fowl painting shows a uncommon example of animal life in a Monet work. Before woodworkers can work with smaller items, they typically use an ax-like tool called an adze to form and lower larger pieces.
A heptagon is a seven-sided polygon and will be both common (in which all sides and angles are equal) or irregular (by which not all sides or angles are equal). Otherwise from exhausting biometrics, nonetheless, comparable to fingerprints or DNA, it can't be used in high-stakes settings or to uniquely identify topics amongst very giant teams, e.g., greater than 1,00010001,0001 , 000 people. Golden raisins (U.S.) are the much more poetic-sounding sultanas (U.Okay.). The alternative angles are additionally equal. A proper-angled triangle is a triangle by which one of many angles is a proper angle. Which one is it? One particular isosceles triangle is a golden triangle. Since all triangles must add as much as 180 levels, it's inconceivable for a triangle to have multiple obtuse angle. Earlier than that, writers may need drawn inspiration from folktales, but they didn't truly record them. In that case, a pupil may consider work-examine opportunities within the host nation. Subsequently, exploring interaction effects between the gender of the participant and the gender of robotic within the context of gender-process match could be promising as effectively. | 677.169 | 1 |
Radians to degrees
To change from radians to degrees, you need to multiply the number of radians by 180/π. This number will help you switch between the two units. For example, if you multiply π/2 radians by 180/π, you will get 90 degrees.Created by Sal Khan and Monterey Institute for Technology and Education. | 677.169 | 1 |
Ans. The properties of inverse trigonometric functions are as follows:
- The domain of an inverse trigonometric function is the range of the corresponding trigonometric function.
- The range of an inverse trigonometric function is the domain of the corresponding trigonometric function.
- The inverse trigonometric functions are all one-to-one functions.
- The inverse trigonometric functions can be used to find the angle measures when given the ratios of the sides of a right triangle.
- The inverse trigonometric functions have a restricted range in order to ensure that they have unique values.
2. How are inverse trigonometric functions denoted?
Ans. Inverse trigonometric functions are denoted using the prefix "arc" or "a" followed by the abbreviation of the corresponding trigonometric function. For example, the inverse sine function is denoted as "arcsin" or "asin", the inverse cosine function as "arccos" or "acos", and the inverse tangent function as "arctan" or "atan".
3. What is the relationship between trigonometric functions and their inverses?
Ans. The relationship between trigonometric functions and their inverses is that they "undo" each other. When a trigonometric function is applied to an angle, the resulting value can be used as an input for the corresponding inverse trigonometric function, which will give back the original angle. For example, if we take the sine of an angle and then take the arcsine of the resulting value, we will obtain the original angle.
4. How are inverse trigonometric functions used in real-life applications?
Ans. Inverse trigonometric functions are used in various real-life applications, such as:
- Calculating the angle of elevation or depression in geometry and physics.
- Solving problems related to navigation and surveying, such as determining distances and angles.
- Analyzing waveforms in electrical engineering and signal processing.
- Modeling and analyzing periodic phenomena in fields like physics, engineering, and biology.
- Solving problems involving triangles and angles in trigonometry and geometry.
5. What are the principal values of inverse trigonometric functions?
Ans. The principal values of inverse trigonometric functions are the values that lie within a certain range and are used to ensure the functions have unique values. The principal values of inverse trigonometric functions are generally defined as follows:
- For arcsine (arcsin), the principal values lie between -π/2 and π/2, inclusive.
- For arccosine (arccos), the principal values lie between 0 and π, inclusive.
- For arctangent (arctan), the principal values lie between -π/2 and π/2, exclusive.
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Students of JEE can study Properties of Inverse Trigonometric Functions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Properties of Inverse Trigonometric Inverse Trigonometric Functions is prepared as per the latest JEE syllabus. | 677.169 | 1 |
Investigating Symmetry Task
There are two shapes on the app. The left shape is the original shape with where the possible line of symmetry falls on the triangle. The right shape is the exact same shape, but it folds the shape across the possible line of symmetry.
The red point on the line allows you to drag the line around the shape. Drag the red point in a circle around the shape to check for all possible lines of symmetry.
After you figure out how many lines of symmetry there are, put your answer in the top right yellow box. When you push "Enter", you will get a pop-up box that tells you whether you are correct. If you are incorrect, it will give you a hint to help you try again.
Once you get it correct, use the slider in the top left corner to go to the next shape. Try all 14 shapes! | 677.169 | 1 |
At this level, students apply their knowledge of Pythagoras' theorem to solve problems involving angles of elevation and depression and direction. They explore how direction can be indicated using true bearings and compass points.
The use of concrete materials and/or dynamic geometry software can assist students in visualising 3D objects.
Students will use this knowledge in future study when working in three dimensions.
Teaching and learning summary:
Revise application of Pythagoras' theorem.
Revise the trigonometrical ratios.
Apply knowledge of trigonometrical ratios to find the lengths and angles in contextual problems.
Students will:
identify the opposite and adjacent sides of a right-angled triangle for a given angle
state the three trigonometrical ratios
use the correct trigonometrical ratio to solve a problem involving the side lengths and angles in a right-angle triangle
use appropriate terminology, diagrams and symbols in mathematical contexts
select and use the appropriate strategy to solve problems
provide reasoning to support conclusions that are appropriate to the context
identify the angle of elevation or depression.
Some students may:
not appreciate the nature of sine, cosine and tangent – these are ratios, not numbers.
misidentify the sides of the triangle in relation to the angle given.
not use the correct trigonometrical ratio and try to apply to non-right-angled triangles.We are learning how to write a formal proof of a geometric property using the correct mathematical language and notation.
We are learning how to apply geometrical properties to solve geometry problems.
Why are we learning about this?
There are many areas where knowledge of shapes and angles is useful.
It is important that you can communicate your ideas using the correct mathematical terminology and notation.
If you are going on to study mathematics at a higher level, knowledge of formal proof is important.
What to do
The spider and the fly A spider is sitting in the middle of one of the smallest walls in my living room and a fly is resting by the side of the window on the opposite wall, 1.5 m above the ground and 0.5 m from the adjacent wall. The room is 5 m long, 4 m wide and 2.5 m high. What is the shortest distance the spider would have to crawl to catch the fly?
To solve this problem, draw a picture (diagram) of the room, labelling the dimensions. What pathways can the spider take?
Finding the height of an object like a mountain is not as easy as it sounds. The great Adirondack surveyor explains how Verplank Colvin solved this problem in the 1870s using Pythagoras and trigonometry. The methods he devised are still used in surveying today.
Success criteria
I can identify the opposite and adjacent sides of a right-angled triangle for a given angle.
I can state the three trigonometrical ratios.
I can use the correct trigonometrical ratio to solve a problem involving the side lengths and angles in a right-angled triangle.
I can use appropriate terminology, diagrams and symbols in mathematical contexts.
I can select and use the appropriate strategy to solve problems.
I can provide reasoning to support conclusions that are appropriate to the context.
I can identify the angle of elevation or depression.
Please note: This site contains links to websites not controlled by the Australian Government or ESA. More information here | 677.169 | 1 |
Appendix C. Moments of Composite Areas
C.1 Centroid
This appendix is concerned with the geometric properties of cross sections of a member. These plane area characteristics have special significance in various relationships governing stress and deflection of beams, columns, and shafts. Geometric properties for most areas encountered in practice are listed in numerous reference works [Ref. C.1]. Table C.1 presents several typical cases.
Table C.1. Properties of Some Plane Areas
The first step in evaluating the characteristics of a plane area is to locate the centroid of the area. The centroid is the point in the plane about which the area is equally ...
Get Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition now with the O'Reilly learning platform.
O'Reilly members experience books, live events, courses curated by job role, and more from O'Reilly and nearly 200 top publishers. | 677.169 | 1 |
fact families multiplication and division worksheets
Fact Triangles Multiplication And Division Worksheets – Triangles are among the most basic shapes found in geometry. Understanding triangles is vital to studying more advanced geometric concepts. In this blog, we will cover the various types of triangles and triangle angles, as well as how to determine the area and perimeter of a triangle, and present specific examples on each. Types of Triangles There are three types of triangles: equal, isoscelesand scalene. Equilateral triangles have … Read more | 677.169 | 1 |
Converting Radians to Degrees
Radians to Degrees conversion calculator is a free online tool that easily makes you Converting Radians
to Degrees in no time. Just enter an angle in radians and click the calculate button to get the desired
output along with show work.
Enter Radians
Converting Radians to Degrees: Are you struggling in converting complex radians to
degrees? Make use of this online converting radians to degrees tool and easily find out the converted
value. Usually, we consider both degrees and radians to measure the angles in geometry. Want to learn
more about the conversion between radians and degrees? Make a look at the modules like definitions,
formula, steps on how to convert, solved examples, etc.
Definitions of Radians and Degrees
radian is as
complete rotation is 2π radians.
equal parts is called a degree. The symbol of degree is
degrees then the instrument used is a protractor. | 677.169 | 1 |
JAVA program that checks if a triangle is scalene, isosceles, equilateral, or not a triangle
I am trying to write java program to see if a triangle is scalene, isosceles, equilateral or not a triangle. With the integers I used it is supposed to be not a triangle (1, 1, 30). But I keep getting scalene and not a triangle together. Any help is appreciated! Thank you! | 677.169 | 1 |
question. 27 people found it helpful. facundo3141592. We want to solve the Trigonometry Maze. So we need to remember some rules: Sin (θ) = (opposite …Level: Year 9. Language: English (en) ID: 1053958. 02/06/2021. Country code: AU. Country: Australia. School subject: Math (1061955) Main content: Trigonometry (2011238) Students are required to apply their knowledge of trigonometry to find both missing sides and angles in right-angled triangles.Scanned by CamScanner. 450 44î Special Right Triangles I Directions: Find each missing side. Write all answers in simplest radical form. Use your solutions to navigate through the maze. Staple all work to this paper! 12 600 36 18 450 30. 24N5. 30 15 450 Start! 450 9dî 450 14 600 24 600 15. 14 12vG 600 28 300 End! 2N6ð 450 450.Jan
Description. In this maze activity, students will evaluate all 6 trigonometric functions using the unit circle values. There are two versions: Version 1: Degrees. Version 2: Radians. ----. Note: This resource is for one teacher only. If you would like to share this resource with your colleagues, please redirect them to this page for this freebie.JEE Main answer key 2024 official - NTA has released the final JEE Main answer key for Paper 2 (B.Arch/B.Planning) session 1 on March 4.Candidates can download the JEE Mains paper 2 answer key pdf on this page. JEE Main answer key 2024 comprises correct answers to the questions framed in the JEE Mains exam.Moreover, …Free math problem solver answers your trigonometry homework questions with step-by-step explanations.
Web special right triangles maze answer key in this triangle one angle is 90 degrees, and the other two angles are 45 degrees. Web #special right triangles maze version 1 answers free; Source: shotwerk.blogspot.com. A number of the worksheets for this idea are title bear in mind 450 450 300. Web #special …Introduction to Trigonometric Identities and Equations; 9.1 Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions; 9.2 Sum and Difference Identities; 9.3 Double-Angle, Half-Angle, and Reduction Formulas; 9.4 Sum-to-Product and Product-to-Sum Formulas; 9.5 Solving Trigonometric EquationsWorksheets are special right triangles, special right triangles practice, right triangle reference,. This product may need to practice arc length of our new. Web special right triangles worksheet answers. Web the area is √3 × opposite. Web special right triangles coloring activity answer key pdf.
chime hack 2023 Find step-by-step solutions and answers to Trigonometry - 9780136679431, as well as thousands of textbooks so you can move forward with confidence. ... Now, with expert-verified solutions from Trigonometry 12th Edition, you'll learn how to solve your toughest homework problems. Our resource for Trigonometry includes answers to chapter …4.9. (270) $4.00. PDF. … showtimes scottsdale azrolling stone muckrack PDF Compositions Of Trig Functions Maze Answer Key. figure 1 can be used to simplify compositions of trigonometric functions such as sin (tan-1 x). Compositions of Inverse Functions Previously we learned that in f (x) and f-1 (x) were inverses, then f (f-1 (x)) = x and f-1 (f (x)) = x. The same is true for trigonometric functions with an …Displaying all worksheets related to - Right Triangles And Trigonometry Gina Wilson. Worksheets are Trigonometry quiz gina wilson, Gina wilson all things algebra right angles and trigonometry, Right triangle trigonometry, Test review right triangle trigonometry answer key, Right triangle trig missing sides and angles, Gina wilson all things algebra … rural king online Tesla's Full Self-Driving beta version 11 is still facing delays, but early testers have given a glimpse of some new features. Tesla has started rolling out version 11 of its Full ... 281. Q&A. 1. More from All Things Algebra. Description. This is a set of four trigonometry mazes to practice finding missing side and angles measures in right triangles using the sine, cosine, and tangent ratios. Students use their solutions to navigate through the maze. nws durangoeyemart express pocatellosharon sharon facebook JanTrigonometry 1 Answers – A-Level Maths - Curriculum Press. Home. Resources. Trigonometry 1 Answers – A-Level Maths. Samples and Free Resources. … john wick 4 showtimes near tinseltown okc A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1, and use the Pythagorean Theorem. Figure 1. a 2 + b 2 = c 2 5 2 + 12 2 = c 2 169 = c 2. Now, we need to find out the length that, when squared ... urban vista crossword clue2 handed axe osrschase notary hours Use sine, cosine, or tangent to find a missing side of a right triangle in this trigonometry maze. Both print and digital options available. Answer key is included. The correct path through the maze requires students to solve nine problems. Displaying top 8 worksheets found for - Gina Wilson Answer Key. Some of the worksheets for this concept are Factoring polynomials gina wilson work, Two step equations maze gina wilson answers, Pdf gina wilson algebra packet answers, Algebra antics answers key, Unit 3 relations and functions, Gina wilson unit 8 quadratic equation answers pdf, Loudoun county public schools, Solve for assume that ... | 677.169 | 1 |
Exterior Angle of Triangle
The exterior angles of a triangle are those angles that are formed outside it. In other words, the exterior angle of a triangle is the angle that is formed between one of its sides and its adjacent extended side. Let us learn more about the exterior angle of triangle in this article.
What is the Exterior Angle of Triangle?
When any side of a triangle is extended, the angle that is formed with this side and its adjacent side is called the exterior angle of a triangle. There are three exterior angles in a triangle. It should be noted that each exterior angle forms a linear pair with its corresponding interior angle. We know that the interior angle of a triangle is formed inside it where the sides meet at a vertex. Observe the following figure to distinguish between the interior angles and the exterior angles of a triangle.
We can see that each interior angle forms a linear pair with its corresponding exterior angle. This means that the sum of each exterior angle and its respective interior angle is equal to 180°.
Exterior Angle of Triangle Properties
There are three basic properties of the exterior angles of a triangle.
In a triangle, each exterior angle and its corresponding interior angle form a linear pair of angles. This means that the sum of the interior and exterior angle is equal to 180°.
The exterior angle of a triangle is equal to the sum of the two opposite interior angles (remote interior angles). This is also known as the Exterior Angle theorem.
The sum of all the exterior angles of a triangle is 360°.
Exterior Angle of Triangle Formula
Based on the properties of the exterior angles, the following formulas can be used to find the exterior angles of a triangle. Referring to the triangle given below, the formulas can be understood in a better way.
Example 2: If one interior angle of a triangle is 56°, find the measure of its corresponding exterior angle.
Solution: According to the properties of the exterior angle of a triangle, each interior angle forms a linear pair with its respective exterior angle. This means, Exterior angle + Interior angle = 180°. In the question, one interior angle is given as 56°. Therefore, the corresponding exterior angle can be calculated using the formula: Each Exterior angle = 180° - Interior angle. Substituting the values in the formula, Exterior angle = 180 - 56 = 124°.
Practice Questions on Exterior Angle of Triangle
FAQS on Exterior Angle of Triangle
What is the Exterior Angle of Triangle?
The exterior angle of a triangle is the angle that is formed with one side and the adjacent extended side of a triangle. There are 3 exterior angles in a triangle and the sum of the exterior angles of a triangle is always equal to 360°.
How to Find the Exterior Angle of Triangle?
The value of the exterior angle of a triangle can be calculated using various formulas depending on the other known angles. The following formulas can be used according to the given parameters.
Each Exterior angle = 180° - Interior angle. This formula can be used if the corresponding interior angle is given.
Exterior angle = Sum of Interior opposite angles. This formula can be used to find the exterior angle when its remote interior opposite angles are given.
The sum of all the exterior angles of a triangle is 360°. This formula can be used to find the unknown value of an exterior angle when the other two exterior angles are given.
What is the Sum of Exterior Angles of a Triangle?
The sum of the exterior angles of a triangle is always equal to 360°. This property applies to all polygons, which means that the sum of the exterior angles of all polygons is 360°.
How to find the Missing Exterior Angle of a Triangle?
A missing exterior angle of a triangle can be calculated using any of the following formulas. This depends on the angles that are given in the question.
The first formula to find the exterior angle can be used if the corresponding interior angle is given. Each Exterior angle = 180 - Interior angle.
The second formula can be used to find the exterior angle when its interior opposite angles are given. Exterior angle = Sum of Interior opposite angles.
The third formula can be used to find the unknown value of an exterior angle when the other two exterior angles are given. The sum of all the exterior angles of a triangle is 360°.
Are the Exterior Angles of a Triangle Equal to 360°?
Yes, the sum of the exterior angles of a triangle is always equal to 360°.
Are the Exterior Angles of a Triangle Always Obtuse?
No, the exterior angles of a triangle may not always be obtuse (more than 90°). However, the sum of all the three exterior angles should always be 360°. For example, if two exterior angles of a triangle are 165° (obtuse) and 141° (obtuse), the third one is 54° (acute).
What is the Measure of Each Exterior Angle of an Equilateral Triangle?
The measure of each exterior angle of an equilateral triangle is 120°. An equilateral triangle is a triangle in which all the sides are equal in length and all the 3 interior angles are of equal measure. This means each interior angle of an equilateral triangle is 60° because the sum of the interior angles is 180°. Now, if each interior angle of an equilateral triangle is 60°, its corresponding exterior angle will be 120°. This is because in a triangle, the exterior angle and its corresponding interior angle form a linear pair of angles. This means that the sum of the interior and exterior angle is equal to 180°.
What is the Exterior Angle Theorem of a Triangle?
According to the exterior angle theorem, the measure of an exterior angle is equal to the sum of the interior opposite angles (remote interior angles). This means if we need to find the exterior angle of a triangle, and its remote interior angles are known, then the value of the exterior angle will be the sum of those two interior opposite angles. | 677.169 | 1 |
I thought about the problem a little harder, even drawing unit circles on the X,Y, X,Z and X,Y,Z planes and realized why my attempt of doing sin(math.rad(offsetAngle)) was giving me the result it was. As i approached the distance of a circle (in radians), the Y position was static because I wasn't involving i at all in the Y positions calculation.
The solution (at least seems to be): sin(i) * sin(rad(offsetAngle)) which makes sense in hindsight. sin(i) is just giving you the Y value of the path…
Edit: some relevant background info on trig would've been helpful in understanding this lol. If you imagine walking around the Unit Circle, the distance to cover the entire circle is TAU (2*math.pi) or ~6.2832.
Just pi alone is half of a circle, pi/2 is a quarter of a circle, etc. This is why the loop is for i = 0, tau, increment do and the increment is divisible by TAU.
math.cos is used to get the X value and math.sin is used to get the Y value (both return a number between -1 and 1).
Edit: As for why cos and sin return a number between -1 and 1 is because of this:
If you look at the image above, the brown arrow points to where you start "walking" on the unit circle. Because math.cos is just giving you the X value, you're at a distance of 1 unit away from the origin (the center). If you printed math.cos(0), you'd get 1. If you printed math.cos(math.pi) you'd get -1. Because math.pi is half the distance of a full circle, so it's at (-1, 0) in the picture above. Cos is just reading that X value of -1.
You can imagine if you do this:
local tau = 2*math.pi -- mathematical constant
local increment = tau/60 -- for the loop
for i = 0, tau, increment do -- start at 0, go the full length of a circle, at an increment of a 60th of a circle each iteration
print(math.cos(i), math.sin(i))
task.wait()
end
math.cos(i) and math.sin(i) will give you the correct X and Y values to place the part to. To get a part to follow a circular path and point at the origin you could do:
local part = script.Parent
local tau = 2*math.pi -- mathematical constant
local increment = tau/360 -- for the loop
local origin = part.Position
local radius = 5 -- how far away the orbit is
while true do -- loop infinitely
for i = 0, tau, increment do -- main for loop like in the example
local pos = Vector3.new(math.cos(i), 0, math.sin(i)) * radius
part.CFrame = CFrame.new(origin + pos, origin)
task.wait()
end
end | 677.169 | 1 |
ArcTan2- Returns the angle between the X-axis and a segment of a line
Function that returns the arctangent of a pair of coordinates, angle expressed in radians.
- ArcTan2(X,Y): float
Parameters:
- Float X: The X coordinate of the segment from which the angle is to be taken;
- Float Y: The Y coordinate of the segment from which to derive the angle;
Description:
Use the ArcTan2 function to derive the angle between the X-axis and a line segment from the 0-0 origin to a pair of X-Y coordinates. The resulting angle is expressed in radians with a value between -PiGreek/2 and PiGreek/2.
To express the arc tangent in degrees, multiply the result by 180/PiGreek or use the conversion instruction. | 677.169 | 1 |
INTRODUCTION
Hey guys, jaisa ki aap jante hai if we want to draw a line we need at least two points. Hum angles ke bare bhi previous classes me padh chuke hai sath hi hum parallel lines ke bare bhi padh chuke hai. In this chapter we will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points. Intersecting line ko transversal bhi kehte hai.
Line Segment
A line which has two end points is called a line-segment. Line segment AB is denoted by ¯¯¯¯¯¯AB, and its length is denoted by AB.
Ray
A part of a line with one end point is called a ray. The ray AB is denoted by −−→AB.
Collinear Points
If three or more points lie on the same line, they are called collinear points, otherwise they are called non-collinear points.
Angle
An angle formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle.
Acute angle
An angle measures between 0∘ and 90∘ is called acute angle.
Right angle
An angle which is exactly equal to 90∘ is called right angle.
Obtuse angle
An angle greater than 90∘ but less than 180∘ is called an obtuse angle.
Straight line angle
An angle equal to 180∘ is called straight line angle.
Reflex angle
An angle which is greater than 180∘ but less than 360∘ is called a reflex angle.
Complementary angle
Two angles whose sum is 90∘ are called complementary angles. For example: 30∘+60∘ .
Supplementary angle
Two angles whose sum is 180∘ are called supplementary angles. For example 120∘+60∘ .
Adjacent angle
Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm means if we have two angles which touches each other on single ray but on different sides.
Linear pair
If sum of adjacent angles is 180∘ then they are called as linear pair .
Vertically opposite angle
When two line intersect each other then angle form opposite to each other are called vertically opposite angles.
Intersecting Lines and Non-intersecting Lines
Man lo we have two lines PQ and RS. Now we extend them on both sides, if the lengths of the common perpendiculars at different points on these lines is different then both lines will intersect each other, jise intersecting lines kehte hai. If the lengths of the common perpendiculars at different points on these lines is the same then both lines will never intersect each other and are called parallel lines. Iska matlab hai ki, the distance between two parallel lines will be the same at every point .
Axiom 6.1
If a ray stands on a line, then the sum of two adjacent angles so formed is 180∘.
Here ∠AOC and ∠BOC are adjacent angle, therefore ∠AOC+∠BOC=∠AOB If ∠AOC and ∠BOC is a linear pair of angles then ∠AOB=180∘ therefore we can write ∠AOC+∠BOC=180∘
Axiom 6.2
If the sum of two adjacent angles is 180∘, then the non-common arms of the angles form a line.
Theorem 6.1
If two lines intersect each other, then the vertically opposite angles are equal.
Let AB and CD be two lines intersecting at O. (i) ∠AOC and ∠BOD (ii) ∠AOD and ∠BOC will form two pairs of vertically opposite angles. We need to prove that ∠AOC=∠BOD and ∠AOD=∠BOC. Now, ray OA stands on line CD. Therefore, ∠AOC+∠AOD=180∘ ................(1) [Straight line angles] and ray OD stands on line AB. Therefore, ∠AOD+∠BOD=180∘ ................(2) [straight line angles] from (1) and (2) ∠AOC+∠AOD=∠AOD+∠BOD This implies that ∠AOC=∠BOD .................(3) Now CD is a straight line so ∠COA+∠AOD=180∘ ..............(4) similarly ∠COB+∠BOD=180∘ .............(5)
From (4) and (5) ∠COA+∠AOD=∠COB+∠BOD From (3) we have ∠AOC=∠BOD ∠AOD=∠COB .................(6) (3) and (6) prove vertically opposite angles are equal.
Parallel Lines and a Transversal
A line which intersects two or more lines at distinct points is called a transversal. Figure ki help se hum ye dekh sakte hai ki line l intersects lines m and n at points P and Q respectively. Therefore, line l is a transversal for lines m and n. Yha humne sare angles ki marking kuch is prakar kar li hai i.e. ∠1,∠2,∠3,∠4,∠5,∠6,∠7and∠8 Yha hum sare angles ko two different group me divide kar sakte hai i.e. Exterior angles : ∠1,∠2,∠7and∠8 Interior angles : ∠3,∠4,∠5and∠6 Here, we have some special pairs of angles which are as follows. (a) Corresponding angles : (i) ∠1and∠5 (ii) ∠2and∠6 (iii) ∠4and∠8 (iv) ∠3and∠7 (b) Alternate interior angles : (i) ∠4and∠6 (ii) ∠3and∠5 (c) Alternate exterior angles : (i) ∠1and∠7 (ii) ∠2and∠8 (d) Interior angles on the same side of the transversal : (i) ∠4and∠5 (ii) ∠3and∠6 Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles. Axiom 6.3 : If a transversal intersects two parallel lines, then each pair of corresponding angles is equal. If m is parallel to n then ∠1=∠5,∠2=∠6,∠4=∠8 and ∠3=∠7 This referred as corresponding angle's axiom. Axiom 6.4 : If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other. This means if ∠1=∠5 or ∠2=∠6 or ∠4=∠8 or ∠3=∠7 then m is parallel to n.
Theorem 6.2
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Theorem 6.4
If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Let transversal PS intersects parallel lines AB and CD at points Q and R Jaisa ki theorem 6.2 me humne dekha hai ∠BQR=∠QRC and ∠AQR=∠QRD ...............(1) Now AQB form a straight line so ∠AQB=180∘ Similarly ∠AQR+∠RQB=180∘
since ∠AQR=∠QRD So, ∠QRD+∠RQB=180∘.............(2) and ∠AQR+∠QRC=180∘ [from (1) and (2)]
Therefore we can conclude that if a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Theorem 6.5
If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.
Let transversal PS intersects lines AB and CD at points Q and R such that ∠QRD+∠RQB=180∘ and ∠AQR+∠QRC=180∘ ∠AQR+∠QRC=180∘ ...........(1) (given) CD form a straight line so ∠CRD=180∘ Similarly ∠CRQ+∠QRD=180∘ .............(2) from (1) and (2) ∠AQR+∠QRC=∠CRQ+∠QRD ∴∠AQR=∠QRD Which are pair of alternate interior angles. From theorem 6.3 hum ye keh sakte hai, if alternate interior angles is equal then the two lines are parallel. Therefore, AB∣∣CD
Theorem 6.6
Lines which are parallel to the same line are parallel to each other.
Let a line t transversal for the lines l, m and n. It is given that line m∣∣ line l and line n∣∣ line l Since line m∣∣ line l so, ∠1=∠2 ...........(1) [Corresponding angles axiom] also line n∣∣ line l so, ∠1=∠3 ............(2) [Corresponding angles axiom] from (1) and (2) we get, ∠2=∠3 Therefore, Line m∣∣ Line n [Converse of corresponding angles axiom]
Angle Sum Property of a Triangle
So guys, jaisa ki hum jante hai, sum of all the angles of a triangle is 180∘ to chalo kuch theorems and axioms ki help se ise prove bhi krke dekhenge.
Theorem 6.7
The sum of the angles of a triangle is 180∘.
Let PQR be triangle and ∠1,∠2 and ∠3, △PQR ke internal angles hain. Construction : Draw a line XPY parallel to QR through the opposite vertex P Now, XPY is a line, Therefore, ∠4+∠1+∠5=180∘ .........(1)[Angles on a staright line] But XPY∣∣QR, sath hi sath PQ and PR are two different transversals. So, ∠4=∠2and∠5=∠3 [Pairs of alternate angles] Substituting ∠4 and ∠5 in (1), we get ∠2+∠1+∠3=180∘ Or, ∠1+∠2+∠3=180∘
So hum ye conclude kar sakte hai, the sum of the angles of a triangle is 180∘.
Theorem 6.8
If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.
Let PQR be triangle and ∠1,∠2 and ∠3 are the angles of △PQR. Side QR is produced to point S Yha hum dekh sakte hai ki QRS is line so, ∠3+∠4=180∘ ...........(1) [Linear Pair] Also, from theorem 6.7 ∠1+∠2+∠3=180∘ ...(2) From (1) and (2) ∠3+∠4=∠1+∠2+∠3 ∴∠4=∠1+∠2 Therefore exterior angle ∠4 is equal to sum of interior opposite angles ∠1+∠2.
Example 4
In the given figure, AB∣∣CD. Find the values of x, y and z.
As we can see from the figure 75∘=y∘ [Alternate interior angles] ⇒y=75. Now, ∠EGF+∠EGD=180∘[Straight line] ⇒∠EGF+125∘=180∘⇒∠EGF=55∘. Now, z∘+y∘+55∘=180∘[Sum of angles of a triangle] ⇒z∘+75∘+55∘=180∘⇒z=50. Now, x∘+y∘=180∘[Angles on straight line] ⇒x+75=180⇒x=105. ∴x=105,y=75 and z=50.
Summary
Is chapter me humne following points ke bare detail me padha hai.
1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. This property is called as the Linear pair axiom. 2. If two lines intersect each other, then the vertically opposite angles are equal. 3. If a transversal intersects two parallel lines, then (i) each pair of corresponding angles is equal, (ii) each pair of alternate interior angles is equal, (iii) each pair of interior angles on the same side of the transversal is supplementary. 4. If a transversal intersects two lines such that, either (i) any one pair of corresponding angles is equal, or (ii) any one pair of alternate interior angles is equal, or (iii) any one pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel. 5. Lines which are parallel to a given line are parallel to each other. 6. The sum of the three angles of a triangle is 180°. 7. If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Let's Revise
Line Segment : A line which has two end points is called a line-segment. Line segment AB is denoted by ¯¯¯¯¯¯AB , and its length is denoted by AB Ray : A part of a line with one end point is called a ray. The ray AB is denoted by −−→AB. Is chapter me hum symbols ka use nahi karte huye simply ise AB se hi represent krenge. Collinear Points : If three or more points lie on the same line, they are called collinear points, otherwise they are called non-collinear points. Angle : An angle formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle. Acute angle : An angle measures between 0∘ and 90∘ is called acute angle. Right angle : An angle which is exactly equal to 90∘ is called right angle. Obtuse angle : An angle greater than 90∘ but less than 180∘ is called an obtuse angle. Straight line angle : An angle equal to 180∘ is called straight line angle. Reflex angle : An angle which is greater than 180∘ but less than 360∘ is called a reflex angle. Complementary angle : Two angles whose sum is 90∘ are called complementary angles. For example: 30∘+60∘ Supplementary angle : Two angles whose sum is 180∘ are called supplementary angles. For example 120∘+60∘ Adjacent angle : Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm , means if we have two angles which touches each other on single ray but on different sides. Linear pair : If sum of adjacent angles is 180∘ i.e. supplementary angle so then adjacent angles are called linear pair of angles. Note : All linear pairs are supplementary angles but all supplementary angles are not linear pair. Vertically opposite angle : When two line intersect each other then angle form opposite to each other are called vertically opposite angles.
The distance between two parallel lines will be the same at every point. Axiom 6.1 : If a ray stands on a line, then the sum of two adjacent angles so formed is 180∘. Axiom 6.2 : If the sum of two adjacent angles is 180∘, then the non-common arms of the angles form a line. This two axioms together is called the Linear Pair Axiom. Theorem 6.1 If two lines intersect each other, then the vertically opposite angles are equal. Let AB and CD be two lines intersecting at O. (i) ∠AOC = ∠BOD (ii) ∠AOD = ∠BOC A line which intersects two or more lines at distinct points is called a transversal. Theorem 6.2 If a transversal intersects two parallel lines, then each pair of alternate interior angles are equal. Here ∠BQR=∠QRS and ∠AQR=∠DRQ Theorem 6.3 If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel. Theorem 6.4 If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Here ∠BQR+∠QRD=180∘ ∠AQR+∠QRC=180∘ Theorem 6.5 If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Theorem 6.6 Lines which are parallel to the same line are parallel to each other.
If line l∣∣ line m and line m∣∣ line n then line l∣∣ line n. Theorem 6.7 The sum of the angles of a triangle is 180∘. Theorem 6.8 If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.
Here ∠4=∠1+∠2 | 677.169 | 1 |
Exploring the Art of Shape Combinations: Discover the Endless Possibilities of Creating New Shapes
Shapes are the building blocks of art and design. But did you know that you can create new shapes by combining different shapes together? This may seem like a simple concept, but the possibilities are endless. In this article, we will explore the art of shape combinations and discover the endless possibilities of creating new shapes. We will look at various techniques and examples of how to combine shapes to create unique designs. So, get ready to unleash your creativity and discover the endless possibilities of shape combinations!
Understanding Basic Shapes
Polygons: Triangles, Rectangles, and Circles
Polygons are two-dimensional shapes with three or more sides. They are a fundamental building block of geometry and are used extensively in various fields, including art, design, and engineering. The three most basic polygons are triangles, rectangles, and circles.
Defining Polygons
A polygon is a two-dimensional shape with three or more sides. It is a closed shape, meaning that all the sides meet at a common point called the vertex. The sum of the internal angles of a polygon is always equal to 360 degrees.
Types of Polygons
There are many different types of polygons, but the three most basic are triangles, rectangles, and circles.
Triangles are polygons with three sides and three vertices. They can be classified based on their sides and angles. For example, an equilateral triangle has all three sides of equal length, while an isosceles triangle has two sides of equal length.
Rectangles are polygons with four sides and four vertices. They have two pairs of parallel sides.
Circles are polygons with an infinite number of sides and vertices. They are defined as the set of all points in a plane that are equidistant from a given point called the center.
Characteristics of Polygons
Polygons have a number of characteristics that make them useful in various contexts. For example, they can be used to represent shapes in art and design, to define areas in engineering and architecture, and to model real-world objects in mathematics.
One important characteristic of polygons is their perimeter, which is the distance around the shape. The perimeter of a rectangle is equal to the sum of the lengths of its sides, while the perimeter of a triangle is equal to the sum of the lengths of its sides.
Another important characteristic of polygons is their area, which is the space inside the shape. The area of a rectangle can be calculated by multiplying the length and width, while the area of a triangle can be calculated using the formula (base x height) / 2.
Polygons can also be combined in various ways to create new shapes. For example, a rectangle can be combined with a triangle to create a pentagon, while two rectangles can be combined to create a parallelogram.
Geometric Transformations: Rotation, Scaling, and Reflection
Definition of Geometric Transformations
Geometric transformations refer to the process of modifying a shape or image by applying specific mathematical operations. These operations alter the original shape's position, size, orientation, or reflection, resulting in a new shape.
Types of Geometric Transformations
Rotation: This transformation involves rotating the shape around a specific point, known as the pivot point. The shape is rotated clockwise or counterclockwise, depending on the angle and direction of the rotation.
Scaling: Scaling changes the size of the shape, either by stretching or compressing it. This transformation can be uniform or non-uniform, meaning that the shape can be scaled equally in all directions or differently in each direction.
Reflection: Reflection creates a new shape by reflecting the original shape across a line, known as the axis of reflection. This operation can create a mirror image of the original shape, or it can produce a new shape that is a combination of the original shape and its reflection.
How Geometric Transformations Work
Geometric transformations are performed using mathematical equations that define the specific operation to be applied to the shape. For example, the equation for rotation involves calculating the new position of each point on the shape based on its distance from the pivot point and the angle of rotation. Similarly, the equation for scaling involves multiplying each point on the shape by a scaling factor to adjust its size. Reflection involves using a mirroring equation that maps each point on the shape to its corresponding point on the other side of the axis of reflection.
By understanding the basics of geometric transformations, artists and designers can explore the endless possibilities of creating new shapes and combinations.
Fractals: Intricate Patterns Created by Repetition
Fractals are intricate patterns that are created by repeating a simple process over and over again. They are found in many different areas of mathematics and science, and they can be created using a wide variety of methods.
Definition of Fractals
A fractal is a geometric shape that has self-similarity, meaning that it looks the same at different scales. This is created by repeating a process over and over again, with each iteration being slightly different from the previous one.
Types of Fractals
There are many different types of fractals, including:
Iterated Function Systems (IFS): These are created by iterating a function over and over again, with each iteration being slightly different from the previous one.
Fractal Dimension: This is a measure of the complexity of a fractal, and it is used to describe how much surface area is contained within the shape.
Self-Similarity: This is the property of a fractal that makes it look the same at different scales.
Examples of Fractals in Nature
Fractals can be found in many different areas of nature, including:
Clouds: The shape of clouds is often fractal-like, with each branch of the cloud being slightly smaller than the one before it.
Coastlines: The shape of a coastline is often fractal-like, with each bend in the coastline being slightly smaller than the one before it.
Plants: The branching patterns of some plants, such as trees and shrubs, are often fractal-like.
Fractals are an interesting area of mathematics and science, and they can be used to create a wide variety of shapes and patterns. By understanding the basic concepts of fractals, you can begin to explore the endless possibilities of creating new shapes.
Combining Shapes to Create New Ones
Key takeaway: Shapes play a crucial role in various fields, including art, design, engineering, and science. They can be combined in various ways to create new shapes, such as tessellations, fractals, and 3D shapes. Understanding the basics of geometry and mathematical operations such as rotation, scaling, and reflection can help artists and designers explore the endless possibilities of creating new shapes.
Tessellations: Patterns That Continue Without Limit
Tessellations are a type of geometric pattern that is created by repeating a shape, called a tessellating shape, without any gaps or overlaps. These patterns can be formed by a variety of different shapes, including squares, triangles, hexagons, and more.
One of the most unique characteristics of tessellations is that they can continue indefinitely in any direction, creating a seamless pattern that never repeats. This makes them a popular choice for decorative designs, such as floor tiles, wallpaper, and even clothing.
There are several different types of tessellations, each with their own unique characteristics and properties. Some of the most common types include:
Regular tessellations: These are created by repeating a single shape, such as a square or a triangle, to form a pattern.
Semi-regular tessellations: These are created by repeating two or more different shapes to form a pattern, such as a square and a hexagon.
Ambiguous tessellations: These are created by repeating a shape that can be seen as two or more different shapes, such as a rectangle that can also be seen as a square and a parallelogram.
Throughout history, many famous artists and mathematicians have created stunning tessellations that continue to inspire and captivate people today. Some of the most famous tessellations include:
The tessellated floor at the Alhambra palace in Granada, Spain
The intricate patterns found in Islamic art and architecture
The geometric designs of M.C. Escher, a Dutch graphic artist known for his mathematically-inspired artwork
By combining different shapes and experimenting with different patterns, artists and mathematicians continue to discover new and exciting possibilities for creating tessellations. Whether you're looking to create a decorative design or simply explore the beauty of geometry, tessellations offer endless possibilities for creativity and inspiration.
Symmetry: Balanced and Harmonious Shapes
Definition of Symmetry
Symmetry is the repetition of a shape, line, or pattern in a balanced and harmonious manner. It can be found in various forms of art, nature, and even in everyday objects. The concept of symmetry is closely related to the idea of balance, as it creates a sense of stability and harmony when applied to a design.
Types of Symmetry
There are several types of symmetry that can be observed in different contexts:
Reflection Symmetry: This type of symmetry is created by folding a shape along a central axis, producing an exact mirror image on the other side. Reflection symmetry can be found in objects like flowers, coins, and buildings with identical sides.
Rotational Symmetry: Rotational symmetry occurs when an object can be rotated around a central point, maintaining its overall shape and design. This type of symmetry is often seen in patterns on circular objects like circular motifs in textiles, or in the petals of a flower.
Translation Symmetry: Translation symmetry is the repetition of a shape or pattern in a design, which remains the same when the design is shifted in a specific direction. This type of symmetry is common in tile patterns, wallpaper designs, and other repetitive elements in architecture and design.
Examples of Symmetry in Nature
Symmetry can be observed in various aspects of nature, showcasing the beauty and order found in the natural world. Some examples include:
Flower petals: Many flowers exhibit reflection symmetry, with petals arranged around a central axis, creating a beautiful and harmonious design.
Insects: Some insects, such as bees and butterflies, have a distinctive pattern on their wings that exhibits rotational symmetry.
Seashells: Seashells often have a spiral shape that showcases translation symmetry, with the same pattern repeating along the shell's length.
In conclusion, symmetry plays a crucial role in creating balanced and harmonious shapes by repetition and the alignment of elements. This fundamental principle of design can be found in various contexts, from nature to art and architecture, showcasing the endless possibilities of combining shapes to create new and visually appealing designs.
3D Shapes: Extending Shapes into Three Dimensions
3D shapes, also known as three-dimensional shapes, are objects that have length, width, and height. These shapes extend beyond the two-dimensional plane and possess depth, making them appear more lifelike and dynamic. In comparison to 2D shapes, 3D shapes offer a more comprehensive representation of objects in the real world.
Types of 3D Shapes
There are numerous types of 3D shapes, each with its unique characteristics and properties. Some of the most common 3D shapes include:
Cubes: These are six-sided shapes with equal dimensions, where each face is a square.
Spheres: These are perfectly round shapes with no flat surfaces, similar to a ball.
Cylinders: These are three-dimensional shapes with a circular base and a curved surface, resembling a tube or can.
Cones: These are three-dimensional shapes with a curved surface, tapering from a circular base to a pointed apex.
Pyramids: These are three-dimensional shapes with a triangular base and a pointed apex, often associated with ancient Egyptian architecture.
Examples of 3D Shapes in Real Life
3D shapes are found everywhere in our daily lives. Here are some examples:
Buildings: Many modern buildings have a 3D shape, with various heights, widths, and depths.
Furniture: Furniture like chairs, tables, and beds also have 3D shapes, as they are not simply flat surfaces.
Technology: Many technological devices, such as smartphones, laptops, and televisions, have 3D shapes to accommodate their various components and functions.
By understanding and exploring the different types of 3D shapes, it becomes evident that combining them can lead to endless possibilities for creating new shapes. This can be seen in various fields, including art, design, and engineering, where the principles of 3D shapes are utilized to create innovative and visually appealing designs.
Creating New Shapes: The Art of Combining and Transforming
Creating new shapes by combining and transforming existing ones is an art form that has captivated artists and designers for centuries. By exploring the process of creating new shapes, tips for combining and transforming shapes, and examples of creative shape combinations, we can unlock the endless possibilities of this art form.
The Process of Creating New Shapes
The process of creating new shapes by combining and transforming existing ones involves a deep understanding of geometry, proportion, and perspective. By examining the fundamental properties of shapes, such as their size, position, and orientation, artists and designers can experiment with different combinations to create new and unique shapes. This process requires a keen eye for detail, as even small changes in the position or size of a shape can result in dramatic changes in the overall composition.
Tips for Combining and Transforming Shapes
To create new shapes through combination and transformation, it is important to consider the following tips:
Start with simple shapes: Begin by combining basic shapes, such as circles, squares, and triangles, to create more complex shapes.
Experiment with different perspectives: Play with the size and position of shapes to create new and interesting perspectives.
Consider the relationship between shapes: Think about how different shapes interact with each other and how they can be combined to create a cohesive composition.
Pay attention to balance and symmetry: Balance and symmetry are key components of effective shape combinations, so be sure to consider how different shapes contribute to the overall balance of the composition.
The Golden Rectangle: A shape composed of a square and a rectangle, where the ratio of the length of the rectangle to the width is equal to the golden ratio.
The Pentagon: A shape composed of five equilateral triangles, arranged in a star-like pattern.
The Mandala: A circular shape composed of repeating geometric patterns, often used in meditation and spiritual practices.
By exploring the art of combining and transforming shapes, we can unlock endless possibilities for creating new and unique shapes. Whether you are an artist, designer, or simply a curious observer, the process of creating new shapes is a fascinating and rewarding pursuit.
Applications of Shape Combinations
Architecture: Using Shapes to Create Buildings
The Role of Shapes in Architecture
Architecture is the art and science of designing and constructing buildings. Shapes play a crucial role in this field, as they provide the foundation for the visual aesthetics of a building. The use of shapes in architecture can be seen in various forms, from the geometric shapes used in ancient buildings to the complex shapes found in modern structures.
Examples of Buildings That Use Shape Combinations
One of the most famous examples of buildings that use shape combinations is the Guggenheim Museum in Bilbao, Spain. The museum's unique shape, designed by Frank Gehry, is a combination of curves and angles that create a dynamic and visually striking structure. Another example is the Sydney Opera House, which features a series of shell-shaped structures that come together to form the iconic building.
Future of Shape-Based Architecture
As technology continues to advance, the possibilities for shape-based architecture are endless. Architects and designers are experimenting with new materials and techniques to create structures that push the boundaries of traditional architecture. For example, 3D printing technology is being used to create complex shapes and structures that were previously impossible to build.
Overall, the use of shapes in architecture is a crucial aspect of the field, and their importance is only set to increase in the future. As technology advances and new materials are developed, architects and designers will have even more tools at their disposal to create unique and striking buildings that push the boundaries of what is possible.
Art: The Inspiration of Shapes in Creative Works
The Role of Shapes in Art
Art has been a form of expression for centuries, and shapes have played a significant role in it. From the abstract forms of Pablo Picasso to the geometric designs of Wassily Kandinsky, artists have used shapes to convey meaning, create depth, and evoke emotion in their work.
Examples of Art That Features Shape Combinations
Many artists have used shape combinations in their work to create unique and striking compositions. For example, in the works of Piet Mondrian, squares and rectangles are combined to create dynamic and balanced compositions. In the work of Wassily Kandinsky, circles and triangles are used to create a sense of movement and energy.
The Future of Shape-Based Art
As technology continues to advance, shape-based art is becoming more accessible to a wider audience. Digital art programs and applications are making it easier for artists to create and experiment with shape combinations, leading to new and innovative forms of art. Additionally, the rise of digital art platforms is allowing artists to showcase their work to a global audience, opening up new opportunities for shape-based art.
In conclusion, shapes have been a fundamental aspect of art for centuries, and their combination can lead to endless possibilities for creating new and unique works of art. As technology continues to advance, shape-based art is becoming more accessible and will continue to play a significant role in the future of art.
Technology: Advancements Made Possible by Shape Combinations
The Role of Shapes in Technology
Shapes as Building Blocks for Innovation
Shapes as a Foundation for Functionality
Examples of Technological Advancements Made Possible by Shape Combinations
Nature: The Beauty of Shapes in the Natural World
The Role of Shapes in Nature
In nature, shapes play a crucial role in the survival and growth of living organisms. They serve as building blocks for various structures, such as the branches of trees, the shells of sea creatures, and the wings of birds. Moreover, shapes help organisms blend into their surroundings, allowing them to camouflage and evade predators. The unique shapes of flowers attract pollinators, ensuring the continuation of their species. Thus, shapes are not only aesthetically pleasing but also essential for the functioning of the natural world.
Examples of Natural Phenomena That Feature Shape Combinations
Many natural phenomena showcase the beauty of shape combinations. For instance, the intricate patterns of a spider's web are formed by combining various shapes, such as circles and triangles. Snowflakes, with their unique six-sided symmetrical shape, demonstrate the intricate combination of different shapes. In addition, the spiral shape of shells, seashells, and the coiling of vines can be seen as a combination of circular and linear shapes. These examples illustrate the limitless possibilities of shape combinations in nature and inspire artists and designers to create new and innovative designs.
The Future of Shape-Based Science
Shape-based science is a rapidly growing field that explores the properties and behavior of shapes in various contexts. Scientists are using advanced technologies such as computer simulations and 3D printing to study the properties of shapes and their interactions. This research has led to the development of new materials with unique properties, such as shape-memory alloys that can change shape in response to temperature or pressure. Additionally, shape-based science has applications in various fields, including engineering, medicine, and architecture, where it can be used to create more efficient and sustainable designs.
In conclusion, the natural world demonstrates the endless possibilities of shape combinations, inspiring artists, designers, and scientists alike. As our understanding of shapes and their properties continues to grow, we can expect to see even more innovative applications and discoveries in the future.
The Importance of Shape Combinations in Everyday Life
The Role of Shapes in Everyday Life
In our daily lives, shapes play a significant role in the design of various objects and systems. From the simple shapes of buttons and icons to the complex geometry of buildings and bridges, shapes are used to create functional and aesthetically pleasing designs.
Examples of Shape Combinations in Daily Life
One example of shape combinations in everyday life is the design of household appliances. Many appliances, such as refrigerators and washing machines, have shapes that are a combination of rectangles, circles, and triangles. These shapes are used to maximize space and create efficient designs.
Another example is the design of transportation vehicles. Cars, buses, and trains all have shapes that are a combination of rectangles, circles, and triangles. These shapes are used to optimize performance and create designs that are both functional and visually appealing.
The Future of Shape-Based Solutions
As technology continues to advance, shape combinations will play an increasingly important role in solving complex problems. Shape-based solutions will be used to design new materials, create innovative products, and solve challenging engineering problems. The future of shape combinations in everyday life is bright, and we can expect to see even more creative and innovative designs in the years to come.
FAQs
1. What is the art of shape combinations?
The art of shape combinations is the creative process of using different shapes to form new and unique designs. This can involve combining different geometric shapes, such as circles, squares, triangles, and rectangles, to create intricate patterns and images. It can also involve using more organic shapes, such as leaves, flowers, and animals, to create naturalistic designs.
2. What are some common shapes used in shape combinations?
Some common shapes used in shape combinations include circles, squares, triangles, and rectangles. These basic shapes can be combined in various ways to create more complex designs. For example, a circle can be combined with a square to create a circular frame around an image, or triangles can be combined to create a pyramid shape.
3. What are some tips for creating new shapes using combinations?
One tip for creating new shapes using combinations is to experiment with different arrangements of shapes. This can involve trying out different shapes, sizes, and orientations to see how they fit together. It can also involve using different colors and patterns to add visual interest to the design. Another tip is to start with a simple shape and build upon it, adding more complex shapes and details as needed.
4. What are some benefits of using shape combinations in art?
Some benefits of using shape combinations in art include the ability to create intricate and detailed designs, as well as the ability to convey different moods and emotions through color and pattern. Shape combinations can also be used to create abstract images that evoke a sense of movement or change. Additionally, using shape combinations can help artists develop their skills in geometry and spatial reasoning.
5. Are there any limitations to the shapes that can be combined?
There are no hard and fast rules when it comes to the shapes that can be combined. However, some shapes may be more difficult to combine than others, depending on their size, shape, and orientation. Additionally, some shapes may not fit together seamlessly, resulting in a less cohesive design. However, with practice and experimentation, artists can learn to work with a wide range of shapes and create unique and compelling designs. | 677.169 | 1 |
The magnitude of the projection of the vector $$2 \hat{i}+\hat{j}+\hat{k}$$ on the vector perpendicular to the plane containing the vectors $$\hat{i}+\hat{j}+\hat{k}$$ and $$\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$$ is
A
$$\frac{2}{\sqrt{6}}$$
B
$$\frac{1}{\sqrt{6}}$$
C
$$\frac{5}{\sqrt{6}}$$
D
$$\frac{7}{\sqrt{6}}$$
3
MHT CET 2023 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $$\bar{a}, \bar{b}$$ and $$\bar{c}$$ are any three non-zero vectors, then $$(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$$
Vectors $$\overline{\mathrm{a}}$$ and $$\overline{\mathrm{b}}$$ are such that $$|\overline{\mathrm{a}}|=1 ;|\overline{\mathrm{b}}|=4$$ and $$\bar{a} \cdot \bar{b}=2$$. If $$\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$$, then the angle between $$\bar{b}$$ and $$\bar{c}$$ is | 677.169 | 1 |
Symmetric Property
The symmetric property is an essential property in algebra that is used in various math concepts such as equality, matrices, relations, congruence, etc. In general, the symmetric property on a set states that if one element is related to the second element, the second element is also related to the first element according to the relation defined on the set.
In this article, we will describe the symmetric property of equality, symmetric property of congruence, symmetric property of relations, and the symmetric property of matrices. We will solve various examples related to the symmetric property to better understand the concept.
What is Symmetric Property?
The symmetric property in algebra is defined as a property that implies if one element in a set is related to the other, then we can say that the second element is also related to the first element. We study different forms of symmetric properties as given below:
All symmetric properties are specific cases of the symmetric property of relations. For example, if we define a relation on the set of numbers as 'is equal to', then we get the symmetric property of equality. If the relation is defined on the set of geometric figures/triangles, then we get the symmetric property of congruence. Let us discuss the symmetric property of equality in the next section.
Symmetric Property of Equality
Now that we have understood the basic meaning of the symmetric property, let us describe the symmetric property of equality. It states that if a real number x is equal to a real number y, then we can say that y is equal to x. This property helps in proving the relation defined on the set of numbers as 'aRb if and only if a = b' to be symmetric relation. Mathematically, we can express the symmetric property of equality as 'If x = y, then y = x'. This property also helps in finding the value of variables in a system of equations.
Symmetric Property of Congruence
The symmetric property of congruence states that if a geometric figure is congruent to another, then we can say that the second figure is congruent to the first figure. For example, if triangle ABC is congruent to the triangle PQR, then we can also say that triangle PQR is congruent to the triangle ABC. Another example of the symmetric property of congruence is that if a line segment AB is congruent to another line segment CD, then we can say that CD is congruent to AB. We can use this property to angles, and other geometric figures.
Symmetric Property of Matrix
The symmetric property of a matrix states that if matrix A is symmetric, then matrix A is equal to its transpose, that is, A = AT. Some of the important properties of a symmetric matrix are:
Symmetric matrices are commutative, that is, if A and B are symmetric matrices, then AB = BA.
The sum and difference of two symmetric matrices give the resultant a symmetric matrix.
Symmetric Property of Relations
A relation R defined on a set A is said to be symmetric if for all a, b in A, if aRb then we must have bRa, that is, if (a, b) is in R, then (b, a) is in R. This symmetric property of relations is used to prove if a relation R is symmetric or an equivalence relation. The number of symmetric relations for a set having 'n' number of elements is given as N = 2n(n+1)/2, where N is the number of symmetric relations and n is the number of elements in the set.
Important Notes on Symmetric Property
Symmetric property of equality, symmetric property of congruence, symmetric property of relations, and the symmetric property of matrices are the different symmetric properties that we study in algebra.
The symmetric property of equality states that if a real number x is equal to a real number y, then we can say that y is equal to x.
Symmetric Property Questions
FAQs on Symmetric Property
What is Symmetric Property in Geometry?
The symmetric property in geometry states that if one figure is congruent to another, then we can say that the second figure is congruent to the first figure. For example, if angle A is congruent to angle B, then we can say that angle B is congruent to angle B.
What is Symmetric Property of Equality?
The symmetric property of equality states that if a real number x is equal to a real number y, then we can say that y is equal to x. Mathematically, we can express the symmetric property of equality as 'If x = y, then y = x'.
What is Symmetric Property of Congruence?
The symmetric property of congruence states that if a geometric figure is congruent to another, then we can say that the second figure is congruent to the first figure.
How Do You Prove Symmetric Property?
We can prove the symmetric property of relation by assuming an element is related to another and then prove that the latter is also related to the former.
What is Symmetric Property of an Angle?
The symmetric property of an angle states that if angle A is congruent to angle B, then we can say that angle B is also congruent to angle A.
What is the Difference Between the Reflexive Property and Symmetric Property?
The reflexive property states that every element is related to itself and the symmetric property states that if one element is related to the second element, the second element is also related to the first element according to the relation defined on the set.
What is an Example of Symmetric Property?
Some of the important examples of symmetric properties are:
If x + y = 3, then 3 = x + y
If triangle ABC is congruent to triangle XYZ, then triangle XYZ is congruent to triangle ABC. | 677.169 | 1 |
Flat|Definition & Meaning
Definition
In mathematics, the term flat refers to a surface that is uniform, smooth, or plane. A flat surface represents a horizontal plane and does not have any depth. Examples include planar and two-dimensional shapes, etc.
A significant portion of geometry is concerned with two-dimensional figures, like squares, circles, and triangles, which can be drawn on a sheet of paper. The study of spheres, cones, and cubes, along with other three-dimensional solid objects that are all around us, is what solid geometry is all about.
The part of an object that is visible to us and is located on its exterior is referred to as the surface of the thing. It does not have any thickness, but it does have an area. Smoothness can be seen, for instance, on the surface of a mirror. The surfaces of these solids might be flat or curved, depending on their shape.
A few good examples of flat surfaces include the surface of the walls, the floor, the tabletop of a table, and paper. In addition, the exterior of spheres, eggs, and even lemons have a curved shape. Examine the figure below for several illustrations of surfaces that are both flat. In the figure below, the square has 4 flat surfaces while the other figure has 0 flat surface, which is a circle.
Figure 1 – Square and circle figure
What Are Flat Shapes in Math?
In mathematics, there are two different kinds of shapes: two-dimensional and three-dimensional. The two-dimensional flat shapes are solely two-dimensional, that is, they have length and width but no thickness. Any flat surface or piece of paper can be used to draw flat forms.
Some examples of flat shapes in mathematics include the square, rectangle, circle, diamond, and triangle.
Square
A square is a shape that only exists in two dimensions and is flat. It has four sides that are all the same length and four vertices. The level land area that is completely contained within a square is referred to as a square region.
Example
A good illustration of a square is a wall or table that is completely square and has four sides that are all the same length.
Figure 2 – Square figure showing flat surfaces
Rectangle
A rectangle is a form that is flat and only has two dimensions. Its opposite sides generally equal in length and parallel to one another. The term "rectangular region" refers to the level of land that lies inside the boundaries of the rectangle.
Example
Some examples of objects that have a rectangle shape include the chalkboard, notes for the rupee and the dollar, and so on.
Figure 3 – Rectangle figure with 4 flat surfaces
Triangle
A triangle is a type of polygonal flat shape which has three vertices and three sides, making it a three-sided polygon. The term "triangular region" refers to the level area that is contained inside the confines of the triangle.
Example
One of the most well-known and impressive examples of a triangular shape is a pyramid.
Figure 4 – Triangle with 3 flat surfaces
Circle
The circle is indeed a shape that only exists in two dimensions and has a flat perimeter that is curved. The circle lacks both sides and corners all the way around. The region that can be found inside a circle is referred to as the circular region.
Example
A few examples of objects that have circular shapes include the disc, pizza, and wall clock.
Octagon
The octagon is a flat, flat shape with eight sides. It is a shape with eight sides. The area inside the octagon that is flat is called the octagonal region.
Example
The roadside stop sign is a great example of an octagon shape.
Definitions of Common Solid and Three-Dimensional Shapes
A shape that occupies space is referred to as a solid or a three-dimensional shape. They have three dimensions: the length, the width, and the height of the object. They are everything within our grasp at this moment. They are present in the activities that we do on a regular basis. Solid shapes include things like your books and a football, for instance.
There are several characteristics that are consistent throughout all the 3D shapes. Faces, edges, as well as vertices or corners, are all examples of these types of characteristics.
The flat surface of such a solid is also known as the face of a solid. A solid's face can be any flat shape, including squares, circles, triangles, etc. The edge is indeed the line segment that connects two faces of a solid. The place where two or even more edges meet is known as the vertex.
A solid shape may have a single surface or several surfaces. Let's investigate the various surface kinds that solid shapes can have.
In Geometry, What Is a "Flat Surface"?
In the field of geometry, a plane is another name for a level surface. Plane geometry focuses on shapes that are flat and two-dimensional, like squares, circles, and triangles, which can be drawn on a piece of paper. Length and width are the two dimensions that make up a plane or flat shape.
A shape that occupies space is referred to as a solid or a three-dimensional shape. The term "surface" refers to the layer of such solid that is located at its outside. It is referred to as a flat surface whenever the surface of the solid is indeed a plane surface without any depths, which is another way of saying that it is flat.
Every day, we encounter a variety of flat objects in our environment. For example, the top of a book, a table, or a dresser can have a flat surface.
How to Recognize Flat Surfaces in 3D Forms
According to the previous section, a solid's surface can be flat or curved. The surface of a three-dimensional shape is a planar figure, including such squares, circles, and triangles.
Observing cubes, cuboids, pyramids, and prisms in three dimensions reveals that they have only flat or level surfaces. Three-dimensional figures and shapes, such as a cylinder or cones, consist both of flat as well as curved surfaces. In addition, particular three-dimensional structures, such as spheres, have no flat surfaces and only one curved surface.
An object with a flat surface can be slid. If such an object seems to have a curved surface, it is possible to roll it along that surface. | 677.169 | 1 |
Construction of a triangleConstruction of a triangle ΔABC where BC=7cm, ∠B=75° and AB+AC=13cm.
Explore how a triangle can be constructed
with given base-length, one base angle and
the sum of other two sides. You can also see how, with a compass,
60°, 120°, 90°, and 75° angles are drawn in the animation.
Construction of a triangle given base-length, base-angle and added value of two sides DCBAPQY60°120°90°∠B = 75°BD = 13cm7cm75°Perpendicular bisector to CDBC = 7cmBegin animationDraw BC so that BC = 7cm.Draw BY so that ∠B = 75°.First draw a 60° angle.Then an angle of 120° is drawn.Bisecting 60° and 120° arcs,we get a right angle(90°).Again bisecting 60° and 90° arcs,we finally get the 75° angle.We cut 13cm BD from BY at D,and join CD.On line CD, drawperpendicular bisector PQ.Let PQ intersect BD at point A.Now, ▲ABC is the required triangle,where BC=7cm, ∠B=75°, AB+AC=13cm.
How to Construct a ΔABC where BC=7cm, ∠B=75° and AB+AC=13cm.
* Draw a base BC = 7cm.
* Draw segment BY so that ∠CBY = 75°.
How to draw a 75° angle»
* To draw a 75° angle, we need to draw a 60° and a 90° angle.
* To draw a 90° angle, we need to draw a 60° and a 120° angle.
* First we draw the 60° and 120° angle.
* In between 60° and 120° angles, another 60° angle is created.
* Bisecting this angle we get the 90° angle(60 + 60 / 2).
* In between 60° and 90° angles, a 30° angle is created.
* Bisecting this angle we get the 75° angle(60 + 30 / 2).
* Put point D on line BY so that BD=13 cm
* Join CD
* Draw a perpendicular bisector PQ on CD
* Let PQ intersect BD at point A.
* Join AC to get the required triangle ABC | 677.169 | 1 |
Platonic solid with 12 edges crossword
Written by Afbburppmc Nwnqabki
Any attempt to build a Platonic solid with S>6 would fail because of overcrowding. We have arrived at an important theorem, usually attributed to Plato: Plato's Theorem: There are exactly five Platonic solids: the tetrahedron, cube, octahedron, dodecahedron and icosahedron. Show more.GOAL: Investigate properties of the Platonic solids. ANDGOAL: Determine how the number of faces, edges, and vertices of a polyhedron are related.We found 3 answers for the crossword clue Platonic. A further 18 clues may be related. If you haven't solved the crossword clue Platonic yet try to search our Crossword Dictionary by entering the letters you already know! (Enter a dot for each missing letters, e.g. "P.ZZ.." will find "PUZZLE".)A few solid earnings reports have been posted but they may not be enough to turn this market, writes James "Rev Shark" DePorre, who says Tesla (TSLA) reports afte...Company launches comprehensive edge platform to integrate operational and information technology into a cloud operating model with an entry-point ... Company launches comprehensive...Conclusion. The icosahedron is one of the five Platonic solids, which are 3D geometric shapes with identical faces and angles. It has 20 faces, 30 edges, and 12 vertices. It is also one of the polyhedra, which are 3D shapes that are made up of flat surfaces. The icosahedron is a popular choice for use in mathematics, as it is a symmetrical ...Below are possible answers for the crossword clue platonic solid with 12 edges. Clue. Length. Answer. platonic solid with 12 edges. 4 letters. cube. Definition: 1. raise to the third power. View more information about cubeThe five Platonic Solids are the tetrahedron, cube, octahedron, icosahedron and dodecahedron (Figure 1). Figure 1: The Platonic Solids. Click to see a 3D Model that you can zoom and rotate. The following table describes the main properties of the Platonic Solids. The Dual of a solid is the polyhedron obtained joining the centers of adjacent ...The cube is a Platonic solid, which has square faces. The cube is also known as a regular hexahedron since it has six identical square faces. A cube consists of 6 faces, 12 edges, and 8 vertices. The opposite faces of a cube are parallel to each other. Each of the faces of the cube meets 4 other faces, one on each of its edges.12. 12. 30. 30. Vertices. 4. 8. 6. 20. 12. Edges from vertex. 3. 3. 4. 3. 5. Number of diagonals. 0. 4. 3. 100. 36. ... Inradiu. 6 a 12. a 2. 6 a 6. 1 2 25 + 11 5 10 a. 42 + 18 5 12 a. Midradius. 2 a 4. 2 a 2. a 2 (5 + 3) a 4 (1 + 5) a 4. Keywords: Platonic solids, also called the regular solids or regular polyhedra. Trigonometry Law of Sines ...The Platonic solids formula is the key to understanding these symmetrical 3D shapes. Learn how to calculate their properties, There are five distinct types of Platonic solids. ... It possesses 12 edges. There are 8 vertices (corners). Equal-Sided Faces: All the faces of a cube are square-shaped, which means that the length, breadth, and height ...The (general) icosahedron is a 20-faced polyhedron (where icos- derives from the Greek word for "twenty" and -hedron comes from the Indo-European word for "seat"). Examples illustrated above include the decagonal dipyramid, elongated triangular gyrobicupola (Johnson solid J_(36)), elongated triangular orthobicupola (J_(35)), …Find the latest crossword clues from New York Times Crosswords, LA Times Crosswords and many more. ... Platonic solid with 12 edges 2% 4 ORAL: Edges away from heartA polyhedron ( plural polyhedra) is a three-dimensional solid with flat polygon faces joined at their edges. The word polyhedron is derived from the Greek poly meaning "many", and the Indo-European hedron meaning "seat or face". A polyhedron's faces are bounding surfaces consisting of portions of intersecting planesAn overview of Platonic solids. Each of the Platonic solids has faces, edges, and vertices. When finding the surface area or volume of a Platonic solid, you will need to know the measurement of the edge. Luckily, all of the edges of a Platonic solid are the same. Let's take a look at the different Platonic solids and how to find the surface ...Platonic solid: Tetrahedron A tetrahedron has 4 faces which are equilateral triangles. It has 4 vertices (each touching 3 faces). It has 6 edges.The Crossword Solver found 30 answers to "solid with 12 faces platonic solid is a regular convex polyhedron.The term polyhedron means that it is a three-dimensional shape that has flat faces and straight edges. The term convex means that none of its internal angles is greater than one hundred and eighty degrees (180°).The term regular means that all of its faces are congruent regular polygons, i.e. the sides of all faces are of the same length, and ...A Platonic solid is a regular, convex polyhedron with congruent faces of regular polygons and the same number of faces meeting at each vertex. ... We can inscribe a cube in dodecahedron (see this), where $12$ faces of dodecahedron give the $12$ edges of the cube. Can we inscribe cube in icosahedron? geometry; polyhedra; platonic-solids; Groups ...Dec 17, 2023 · Clue: Platonic solid with 12 edges. Platonic solid with 12 edges is a crossword puzzle clue that we have spotted 1 time. There are related clues (shown belowIn geometry, a Platonic solid is a convex polyhedron that is regular, in the sense of a regular polygon. Specifically, the faces of a Platonic solid are congruent regular polygons, with the same number of faces meeting at each vertex. They have the unique property that the faces, edges and angles of each solid are all congruent. There are precisely five …Details on the five platonic solids, with graphs. Lauren K. Williams, PhD Applets; Resources; Teaching; CV; The Platonic Solids Tetrahedron Face: Equilateral Triangle Faces ... Vertices: 4 Dihedral Angle: 70.53° Dual: Self Hexahedron (Cube) Face: Square Faces: 6 Edges: 12Platonic solids rolling through their edge MN withdifferent rotation angles shown in Table 2. A body frame (O − e 1 e 2 e 3 ) is fixed at the center of each solid (left).Platonic solids. The name given to five convex regular polyhedra: the tetrahedron, the cube, the octahedron, the dodecahedron, and the icosahedron. The names of the polyhedra are Plato's names, who in his Timei (4th century B.C.) assigned them a mystical significance; they were known before Plato3 Coordinates and other statistics of the 5 Platonic Solids. They are the tetrahedron, cube (or hexahedron), octahedron, dodecahedron and icosahedron. Their names come from the number of faces (hedron=face in Greek and its plural is hedra). tetra=4, hexa=6, octa=8, dodeca=12 and icosa=20.Find the latest crossword clues from New York Times Crosswords, LA Times Crosswords and many more. Enter Given Clue. Number of Letters ... Platonic Solid With 12 Edges Crossword Clue; Perhaps Bluffers Got Involved In Robberies, Wiping Out Hotel Crossword Clue; Pound, For One Crossword Clue;Do you want to learn how to edge your lawn? Click here for a step-by-step guide explaining how to effectively and efficiently edge a lawn. Expert Advice On Improving Your Home Vide...The Crossword Solver found 30 answers to "The Platonic solid with the most faces12. What is the measure of each interior angle of a regular pentagon? (Use the formula S = 180(n - 2), where S is the sum of the interior angles and n is the number of sides) _____ 13. How many regular pentagons can be put together at a vertex to form a solid? _____ 14. Briefly explain why there cannot be more than five Platonic solids.Expert-verified. 23. [Euler's Theorem on Polyhedra] A polyhedron is a solid in three dimensions with polygonal faces and straight edges; the three-dimensional version of a polygon. A polyhedron is called a platonic solid if all of the faces are identical regular polygons. There are only five platonic solids: tetrahedron, with four triangularOrigami of Platonic Solids: Octahedron: There are many ways to make models of the Platonic Solids. This tutorial is using equilateral triangles with pockets in each edges to create a tetrahedron. This is ideal for math centers for your Geometry or Mathematics class and for home decors. ... Step 2: 12 Origami Connectors. This will be used to ...The Platonic Solids as Edge-Models Rudolf Hrach . 1 Introduction . The ve Platonic solids are attractive subjects in space geometry since Euklid's . ... Number of vertices 20 8 4 6 12. Link. 136 R. Hrach. Fig. 4 . The 5 vertex connectors . 3.2 Construction of the Vertex-connector .From 5 Platonic Solids another set of semi-regular polyhedra, called the 13 Archimedean Solids, can be derived. Aside from the Truncated Tetrahedron, the other 12 fall into two distinct categories. Some are based on the Octahedron and Cube with octahedral symmetry, and another six are derived from the Dodecahedron and Icosahedron, that exhibit ...Clue: One of the Platonic solids. One of the Platonic solids is a crossword puzzle clue that we have spotted 1 time. There are related clues (shown below). Referring crossword puzzle answers. CUBE; Likely related crossword puzzle clues. Sort A-Z. Block; Die; Cut up, as ...POLYHEDRA, GRAPHS AND SURFACES 3.2. Platonic Solids and Beyond Classifying the Platonic Solids ... edges and faces for each of the Platonic solids and, if you do so, you'll end up with a table like the following. ... cube 4 3 8 12 6 octahedron 3 4 6 12 8 dodecahedron 5 3 20 30 12 icosahedron 3 5 12 30 20 The following diagram shows the fiveFind the latest crossword clues from New York Times Crosswords, LA Times Crosswords and many more. ... Platonic solid with 12 edges 2% 4 HIHO: Old cracker brand 2% 6 ...We do it by providing Washington Post Sunday Crossword 12/17/2023 answers and all needed stuff. If the Washington Post Sunday Crossword is suddenly upgraded, you can always find new answers to this site. So do not forget to add our site to your favorites and tell your friends about it. ... Platonic solid with 12 edges. Retailer with the blogtions between these ve planets and the ve Platonic solids. His model had each planet's orbit associated with a sphere and the distance between the spheres was determined by a Platonic solid, as seen in gure 1.2. The spheres of orbits cir-cumscribed and inscribed each Platonic solid. The out-most sphere represented the orbit of Saturn.All Platonic Solids (and many other solids) are like a Sphere... we can reshape them so that they become a Sphere (move their corner points, then curve their faces a bit).. For this reason we know that F + V − E = 2 for a sphere (Be careful, we cannot simply say a sphere has 1 face, and 0 vertices and edges, for F+V−E=1). So, the result is 2 again. ...cube has eight vertices, twelve edges and six faces, and it is another Platonic solid. • When four squares meet at a vertex, the sum of the angles is 360 degrees. Hence, by the same argument as for six equilateral triangles, there are no Platonic solids with more than three squares meeting at every vertex. ⊆. 10. MTCircular · Autumn 2018 ·It is one of the five Platonic solids. Create an account ... from others. For example, a square has 4 sides and 4 corners, while a 3-D cube has 6 faces, 8 vertices (or corners) and 12 edges ...respectively called edges and vertices of the given polytope. As for graphs, the degree of a vertex v of a polytope is the number of edges incident to v. Let P be a polytope. We make the following geometric observations. Remark 2. The boundary of every face of P consists of at least 3 edges. The degree of every vertex of P is at least 3.As we saw in earlier articles, the sum of the angles of the four Platonic solids that represent Fire, Air, Earth & Water (the 4 Earthly Elements) equals the diameter of the Earth in miles (99.97% accuracy). Earth's polar diameter in 2013 (NASA) = 7899.86 miles. The equatorial diameter = 7926.33 miles.Regular icosahedron (12 vertices, 30 edges, 20 equilateral triangles as faces) At the top right of this app's control panel, you can select one of the Platonic solids. The position in the space can be set with the big button; depending on the setting, a vertex, the center of an edge or the center of a face will lie on the upward pointing z-axis ...Study with Quizlet and memorize flashcards containing terms like what is a platonic solid ?, how many faces does a tetrahedron have?, how many vertices does a tetrahedron have ? and more.built on these platonic solids in his work "The Elements". He showed that there are exactly five regular convex polyhedra, known as the Platonic Solids. These are shown below. Each face of each Platonic solid is a convex regular polygon. Octahedron. 8 triangular faces 12 edges 8 vertices . Cube . 6 square facesStudy with Quizlet and memorize flashcards containing terms like Tetrahedron, Hexahedron, Octahedron and moreAnswers for Figure with 12 edges crossword clue, 4 letters. Search for crossword clues found in the Daily Celebrity, NY Times, Daily Mirror, ... Regular solid figures with twelve equal pentagonal faces (11) Advertisement. ENGLISH PATIENT: 1996 film with 12 Oscar nominations (with "The")What is the correct answer for a "Platonic solid with 12 edges" Washington Post Sunday Crossword Clue? The answer for a Platonic solid with 12 edges …Find the latest crossword clues from New York Times Crosswords, LA Times Crosswords and many more. ... Platonic solid with 12 edges 2% 4 HIHO: Old cracker brand 2% 6 ...Properties. The rhombic dodecahedron is a zonohedron. Its polyhedral dual is the cuboctahedron.The long face-diagonal length is exactly √ 2 times the short face-diagonal length; thus, the acute angles on each face measure arccos(1 / 3), or approximately 70.53°.. Being the dual of an Archimedean polyhedron, the rhombic dodecahedron is face-transitive, meaning the symmetry group of the solid ...The icosahedron's definition is derived from the ancient Greek words Icos (eíkosi) meaning 'twenty' and hedra (hédra) meaning 'seat'. It is one of the five platonic solids with equilateral triangular faces. Icosahedron has 20 faces, 30 edges, and 12 vertices. It is a shape with the largest volume among all platonic solids for its surface area.In geometry, a Platonic solid is a convex, ... The circumradius R and the inradius r of the solid {p, q} with edge length a are given by ... The orders of the proper (rotation) groups are 12, 24, and 60 respectively - precisely twice the number of edges in the respective polyhedra. The orders of the full symmetry groups are twice as much ...Platonic solids are regular polyhedrons, meaning all their faces, edges, and angles are congruent, regular polygons, and in which the same number of faces meet at each vertex. Platonic solids that we see in day-to-day life are dice. The five regular polyhedrons are: cube, tetrahedron, regular octahedron, regular dodecahedron, and regular ...10. We're going to take the 5 platonic solids ( tetrahedron, cube, octahedron, dodecahedron, and icosahedron) and suspend them in various ways (we'll assume that they are solid and of uniform density). Then we'll do a horizontal cut through the centre of gravity and describe the shape of the resulting cut face. The suspension methods will be:Regular polyhedra are also called Platonic solids (named for Plato). If you fix the number of sides and their length, there is one and only one regular polygon with that number of sides. That is, every regular quadrilateral is a square, but there can be different sized squares. Every regular octagon looks like a stop sign, but it may be scaled ...All five truncations of the Platonic solids are Archimedean solids. These are: 3. Truncated tetrahedron – creates triangular & hexagonal faces = 3600° It has: 4 triangular faces; 4 hexagonal faces; 8 total faces; 18 edges; 12 vertices . The net of the truncated tetrahedron: A shallow truncation of the tetrahedron: A full truncation ...For some reason, lots of people believe that the ability to solve crossword puzzles is a talent doled out at birth to a select few. This couldn't be farther from the truth. Crosswo...Crossword Clue. Here is the solution for the Platonic conceptsA synthesis of zoology and algebra Platonic Solids and Polyhedral Groups Symmetry in the face of congruence What is a platonic solid? A polyhedron is three dimensional analogue to a polygon A convex polyhedron all of whose faces are congruent Plato proposed ideal form of classical elements constructed from regular polyhedrons Examples of Platonic Solids Five such solids exist: Tetrahedron ...Platonic solids are particularly important polyhedra, but there are countless others. ... Truncated Tetrahedron 8 faces, 12 vertices, 18 edges. Cuboctahedron 14 faces, 12 vertices, 24 edges. Truncated Cube 14 faces, 24 vertices, 36 edges. Truncated Octahedron 14 faces, 24 vertices, 36 edges. Rhombicuboctahedron 26 faces, 24 vertices, 48 edges.The answer is yes. In other words, if we develop a Platonic solid by cutting along its edges, we always obtain a flat nonoverlapping simple polygon. We also give self-overlapping general ...A dodecahedron is a platonic solid that consists of 12 sides and 12 pentagonal faces. The properties of a dodecahedron are: A dodecahedron has 12 pentagonal sides, 30 edges, and 20 vertices and at each vertex 3 edges meet. The platonic solid has 160 diagonals.If you want to improve your finances take initiative and make a plan. Here are six elements of a solid personal financial plan to get you started. The College Investor Student Loan... Edges Crossword Clue. The Crossword Solver found 60 answers to "Edges", 6 letters crossword clue. The Crosswor
Close platonic relationship between men (informal) Crossword Clue Answers. Find the latest crossword clues from New York Times Crosswords, LA Times Crosswords and many more. Crossword Solver. Crossword Finders. Crossword Answers. Word Finders ... CUBE Platonic solid with 12 edges (4) 4% SISTER How to resist a …1 Discussion. This brief note describes the 5 Platonic solids and lists speci c vertex values and face connectivity indices. that allow you to build triangle or polygon meshes of the solids. In each of the sections the following notation. is used. v. number of vertices. A. dihedral angle between adjacent faces.The clue for your today's crossword puzzle is: "Platonic solid with 12 edges" ,published by The Washington Post Sunday. Please check our best answer below:Platonic solids as art pieces in a park. The Platonic solids are a group of five polyhedra, each having identical faces that meet at identical angles. Some of the earliest records of these objects ... Pythagoras (c.A three-dimensional shape that is made up of four triangles is called a tetrahedron. If it is a regular tetrahedron, then it contains four equilateral triangles as its faces. A reg...Platonic Relationships. Exercise: Get to know the five Platonic solids and the relationships between them. Start by counting the number of faces, edges, and vertices found in each of these five models. Make a table with the fifteen answers and notice that only six different numbers appear in the fifteen slots. faces edges verticesA three-dimensional figure with faces that are polygons that share a common side. flat surface formed by a polygon. point at which three or more edges intersect. A line segment where two faces intersect. many seated (sides) TEACHER. Start studying platonic solids. Learn vocabulary, terms, and more with flashcards, games, and other study tools.The Dodecahedron – 6480°. The dodecahedron is the most elusive Platonic solid. It has: 12 regular pentagonal faces. 30 edges. 20 corners. There are 160 diagonals of the dodecahedron. 60 of these are face diagonals. 100 are space diagonals (a line connecting two vertices that are not on the same face).12 edges, i.e. E = 12; Tetrahedron and cube are platonic solids in which three faces ( regular polygons ) meet at a point to form a vertex. Octahedron. Let us now move on to a new platonic solid in which four regular polygons meet at a point to form a vertex.Edges: 12 Vertices: 6 ... Dual: Dodecahedron Platonic Solids A Platonic solid is a three dimensional figure whose faces are identical regular, convex polygons. Only five such figures are possible: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. These polyhedra are named for Plato, ...Platonic solids, the 5 regular polyhedra, tetrahedron, hexahedron, octahedron, dodecahedron, icosahedron, polyhedron calculator and formulas. ... 12 edges 3 faces and 3 edges meet at each vertex Each face is a pentagon. 12 faces 20 vertices 30 edges 5 faces and 5 edges meet at each vertex ...either cyclic or dihedral or conjugate to Symm(X) for some Platonic solid X. The Tetrahedron The tetrahedron has 4 vertices, 6 edges and 4 faces, each of which is an equilateral triangle. There are 6 planes of reflectional symmetry, one of which is shown on the below. Each such plane contains one edge and bisects the opposite edge (this gives ...Find out the steps you need to take to polish a bullnose edge molding on a granite countertop from home improvement expert Danny Lipford. Expert Advice On Improving Your Home Video...Study with Quizlet and memorize flashcards containing terms like Tetrahedron D4, Cube D6, Octahedron D8 and more.Platonic solid. In geometry, a Platonic solid is a convex, regular polyhedron in three-dimensional Euclidean space. Being a regular polyhedron means that the faces are congruent (identical in shape and size) regular polygons (all angles congruent and all edges congruent), and the same number of faces meet at each vertex.There are exactly five Platonic solids: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. The above tape-and-cardboard discussion provides very strong evidence that this theorem is true, but we must acknowledge that more work would be required to achieve a completely airtight proof of this theorem.The regular dodecahedron is a Platonic solid having of 20 vertices, 30 edges, and 12 faces. Each face is a regular pentagon. The dodecahedron is the dual of the icosahedron which has 12 vertices, 30 edges and 20 faces. ... (All of the solids discussed here are Platonic Solids and all have both inscribed and circumscribed spheres.) In Figure 9.31. The radius of the sphere circumscribing the polyhedron; 2. The radius of the sphere inscribed in the polyhedron; 3. The surface area of the polyhedron; 4. The volume of the polyhedron. Tetrahedron: All four faces are equilateral triangles. canE = number of edges In this case, we are given that the Platonic solid has 8 vertices and 12 edges. Substituting these values into the formula, we have: F + 8 - 12 = 2 Simplifying the equation, we get: F - 4 = 2 Now, we can solve for F: F = 2 + 4 F = 6 Therefore, the Platonic solid with 8 vertices and 12 edges will have 6 faces.Advanced Math questions and answers. 3. (9 points) (a) For each of the five Platonic solids, give the rumber of vertices, edges and faces. (b) If V is the number of vertices, E is the number of exdges, and F is the number of faces, show that for every platonic solid, VE+F=2. (c) Compare the numbers for the cube against those for the octahedron.A vertex configuration is given as a sequence of numbers representing the number of sides of the faces going around the vertex. The notation "a.b.c" describes a vertex that has 3 faces around it, faces with a, b, and c sides. For example, "3.5.3.5" indicates a vertex belonging to 4 faces, alternating triangles and pentagons.Study with Quizlet and memorize flashcards containing terms like Tetrahedron faces, Tetrahedron Vertices, Tetrahedron edges and more. Scheduled maintenance: March 23, 2024 from 11:00 PM to 12:00 AM hello quizletlar polyhedra: (1) the same number of edges bound each face and (2) the same number of edges meet at every ver-tex. To illustrate, picture the cube (a regular polyhedron) at left. The cube has 8 verti-ces, 6 faces, and 12 edges where 4 edges bound each face and 3 edges meet at each vertex. Next, consider the tetrahedron (literally, "fourNov 11, 2021 · The crossword clue One of the Platonic solids with 4 letters was last seen on the November 11, 2021. We found 20 possible solutions for this clue. We think the likely answer to this clue is CUBE. You can easily improve your search by specifying the number of letters in the answer.Here is a picture of an octahedron, which is a regular (Platonic) solid with 8 triangular faces, 12 edges, and 6 vertices. You can imagine an octahedron as two pyramids with square bases, which are then glued together along their bases. octahedron We can turn a polyhedron into a graph by placing its vertices in the plane, and adding edges between …In geometry, a Platonic solid is a convex, regular polyhedron in three-dimensional Euclidean space. ... It has 12 faces, 20 vertices, 30 edges, and 160 diagonals. It is represented by the Schläfli symbol {5,3}. In geometry, a quasiregular polyhedron is a uniform polyhedron that has exactly two kinds of regular faces, which alternate around ...The Crossword Solver found 30 answers to "solid figure with twelve sidesThis item: Handmade Platonic Solid Set (SET OF 7, Clear Quartz) $2499. +. FemiaD 6 X 12 Novelty Funny Sign Sublime California Vintage Metal Tin Sign Wall Sign Plaque Poster for Home Bathroom and Cafe Bar Pub, Wall Decor Car Vehicle License Plate Souvenir. $1195.The Platonic solids are regular polyhedrons and consist of the tetra-, hexa-, octa-, dodeca- and the icosa-hedron. They can be built in a compact (face-model) and in an open (edge-model) form (see Fig. 1 ). The compact models are constructed in FUSION 360 and are practical for studying regular polygons. For completeness, the numbers of …It has: 8 triangular faces; 6 square faces; 14 total faces; 24 edges; 12 vertices. This is the first Archimedean solid we look at. It has square and equilateral triangle faces. Net of the cuboctahedron: 2. Icosidodecahedron - icosahedron and dodecahedron combined = 10080° ... All five truncations of the Platonic solids are Archimedean solids.A Platonic solid is a polyhedron, or 3 dimensional figure, in which all faces are congruent regular polygons such that the same number of faces meet at each vertex.There are five such solids: the cube (regular hexahedron), the regular tetrahedron, the regular octahedron, the regular dodecahedron, and the regular icosahedron.. The tetrahedron has four faces, all of which are triangles.A Platonic Solid is a 3D shape where: each face is the same regular polygon. the same number of polygons meet at each vertex (corner) Example: the Cube is a Platonic Solid. each face is the same-sized square. 3 squares meet at each corner. There are only five platonic solids.Crater edges Crossword Clue Answers. Find the latest crossword clues from New York Times Crosswords, LA Times Crosswords and many more. Crossword Solver Crossword ... CUBE Platonic solid with 12 edges (4) Show More Answers (29) To get better results - specify the word length & known letters in the search. 1) 2) Clues ...Platonic Solids and Tilings. Platonic solids and uniform tilings are closely related as shown below. Starting from the tetrahedron we have polyhedra with three triangles, squares and pentagons at each vertex. The next step is the plane tiling with three hexagons at each vertex.The Icosahedron - 3600°. The icosahedron is the shape that gives the most symmetrical distribution of points, edges, and surfaces on the sphere. It has: 20 Faces (20 equilateral triangles) 5 to a vertex. 30 edges. 12 corners. It's Dual is the dodecahedron.Some good ideas for science fair projects include recording the effects of different foods on the human heart rate, observing the influence of phrasing questions differently on the...The five platonic solids. tetrahedron, cube, octahedron, dodecahedron, icosahedron. Tetrahedron. A geometric solid with four sides that are all equilateral triangles. There are four faces and 4 vertices. At each vertex three triangles meet. Octahedron. A polyhedron having eight plane faces, each face being an equilateral triangleThe nested Platonic Solids can be elegantly represented in the Rhombic Triacontahedron, as shown in Rhombic Triacontahedron. ... Each cube has 12 edges, and each edge will be a diagonal of one of the 12 pentagonal faces of the dodecahedron. Since there are only 5 diagonals to a pentagon, there can only be 5 different cubes, each of which will ...Every Platonic Solid (and Archimedean Solid) is built out of regular polygons. This basically means that each edge is equal and each corner of the 2D shape is equal. The most basic regular polygon is a regular triangle. Add a corner more and you get a square, add another corner more and you get a pentagon.Platonic Solid Picture Number of Faces Shape of Faces Number of Faces at Each Vertex Number of Vertices Number of Edges Unfolded Polyhedron (Net) Dual (The Platonic Solid that can be inscribed inside it by connecting the mid-points of the faces) Tetrahedron: 4: Equilateral Triangle (3-sided) 3: 4: 6: Tetrahedron: Cube: 6: Square (4-sided) 3: 8: ... … solid that is a Platonic solid could be any one of the five shapes.. A Platonic solid is a three-dimensional shape with regular polygonal faces, all of which are congruent and have the same number of sides.. There are only five Platonic solids: tetrahedron, cube, octahedron, dodecahedron, and icosahedron. Each solid has its own …What are the 5 Platonic Solids? There are five total platonic solids: Tetrahedron: 4 faces, 4 points, 6 edges. Hexahedron: 6 faces, 8 points, 12 edges. Octahedron: 6 faces, 6 points, 12 edges. Icosahedron: 20 faces, 12 points, 30 edges. Dodecahedron: 12 faces, 20 points, 30 edges. The outlines of the five platonic solids. The Platonic solids, also called the regular solids or regular polyhedra, are convex polyhedra with equivalent | 677.169 | 1 |
Your two lines appear to be almost the same, almost straight, but measuring them, shows only ONE is straight
Pure BS. The simplest way to see this is to just cut out the sections between the middle and edges of the image, and get this:
This shows the left hand side, the middle and the right hand side are identical. You cannot measure the difference because the image is not accurate enough to do so. The curve you need to see here is much less than 1 pixel.
The same issue occurs in reality, that issue you continually fled from as you were unable to provide a device that is capable of measuring a variation of 20 n, over a distance of 1 m.
The simple fact is that if you are looking at a small enough portion of a curve you cannot tell if it is curved or straight. If you just measure the red line in the first image, you will not be able to measure the curve.
If you wish to spout such dishonest BS, then go and take that first image and show the measurement that allows you to determine the red line is curved.
You tried to show both lines overlapping each other, not as two separate lines - get serious!
Follow your own advice. I showed the 2 overlapping to show just how similar they look. Unless you had another image to compare with, you would not be able to tell which is straight and which is curved.
Now again, how about you stop with the pathetic deflection and explain how perspective magically stops and reverse to produce a horizon on a flat surface. Don't just repeatedly assert that it magically does. Clearly explain how.
Whenever we observe a surface which we can confirm to be flat, we see across the entire surface when above it. We do not get any magical horizon.
What confirms that a surface IS flat, specifically?
Since you believe we CAN confirm a surface is flat, how do we KNOW that it is flatNow again, how about you stop with the pathetic deflection and explain how perspective magically stops and reverse to produce a horizon on a flat surface. Don't just repeatedly assert that it magically does. Clearly explain howAnd surveyors on projects would measure an entire area, assuming it IS flat, measuring where it IS flat, and where it is NOT flat, right?The REAL surface, with REAL instruments, with as MUCH accuracy as possible.
Now, by this do you actually mean accuracy, or do you mean precision? The 2 are different.
I could say the curvature of Earth is 0 with a margin of error of 1 /m. That would technically be accurate, as I have included the range, but it is not precise.
Regardless, you are wrong. Only in science, where they care about discovering how the world works, do they go for as accurate and precise as possible. For engineering, they go for as accurate and precise as needed/practical.
They are quite happy to use tools which are imprecise, which cannot measure the curvature of Earth, because it isn't needed.
They aren't idiots. They aren't going to waste millions or billions of dollars on something that will be pointless due to the variance in materials.
No, I believe what is supported by the evidence, and makes logical sense.
Your delusional BS does not fit either criteria. You baselessly assert that they would try to measure it as accurately as possible, with no justification at all, nor any evidence to support that fantasy of yours. All for the implication that they should be measuring the curvature of Earth and accounting for it. All without even bothering with a simple bit of math to show how significant it should be.
Here is some simple math for you. Lets say we are building a skyscraper. It can be 100 m tall, and have a base of 100 m. That means from the centre out, there would be a drop due to a distance of 50 m. That is a drop of 0.196 mm. As a comparison, consider a 25 mm by 25 mm area of concrete from here: The average surface roughness was 0.1259 mm, and the total height variation was 2.047 mm.
That means the average roughness is comparable to the drop, and the total height variation was roughly 10 times the expected drop. Why would they bother with the drop due to the curvature of Earth when concrete would produce a larger variation?
As the edges of the building are 100 m apart, that means they are an angle of 3.2 arc seconds different. The 100 m of height, with this angle, will make the top of the building (which would typically be much smaller than the base, buts lets ignore that for now) 1.6 mm wider, if they construct the building entirely plumb. It could be 0 if the walls are angled just a tiny bit (1.6 arc seconds each, which is already an incredibly small angle). But again appealing to real materials, lets look at thermal expansion: Concrete has a coefficient of thermal expansion of 13-14 *10^-6 (or a structure is 9.8*10^-6). For simplicity lets use a value of 10*10^-6 = 10^-5. That means for each 1 degree celsius rise in temperature, our 100 m wide structure would expand (or contract) roughly 1 mm. So if the temperature fluctuates by 10 degrees, it would amount to a change of 10 mm. Already over 5 times that due to the curvature.
So with the curvature having such a minor effect, why should they account for it?
And unlike you, I don't need to lie and claim all the evidence is fake, or continually deflect by jumping topics.
Now again, how about you stop with the pathetic deflection and explain how perspective magically stops and reverse to produce a horizon on a flat surface. Don't just repeatedly assert that it magically does. Clearly explain how.
Or does your repeated refusal to do so amount to an admission that you cannot explain it and that the fact that objects appear to sink as they go beyond the horizon is a clear demonstration that Earth is curved?
And surveyors on projects would measure an entire area, assuming it IS flat, measuring where it IS flat, and where it is NOT flat, right?
Utah GEODETIC Surveying Geodetic surveying is similar to other types of land surveying, except that it takes into account the curvature of the Earth. Geodetic surveys generally take part over large areas, and provide great accuracy in both the linear and angular observations. a Geodetic survey takes two points located on the Earth's surface and treats them as arcs; the curvature of the Earth is then taken into account, and the angles between the lines help determine the distance between the two points. These surveys are often used to retrace the public land survey system, and replace or modify missing corners if necessaryIt's not practical to factor in curvature over small distances, but build a long tunnel or long bridge and you'll have to pull out your curvature measuring devicesIf you can't answer your own question, or make a point about it, nobody here is going to help you, including me. You should know that by now.
If you can't answer your own question, or make a point about it, nobody here is going to help you, including me. You should know that by now.
It has been answered, with the point made from it being repeatedly ignored by you, with you sometimes just outright lying by ignoring the answer.
The point is that due to how large Earth is, you cannot measure the curvature over a tiny portion of it. For Earth, with a radius of roughly 40 000 km, a 300 000 sided polygon would correspond to roughly 133 m. And the exterior angle of an n sided polygon is 360 degrees/n, which for this gives us 0.0012 degrees.
If you had an object 133 m away, standing upright at its location, then relative to you standing upright at your location, it would have tilted back 0.0012 degrees.
This shows just how tiny the curvature is, and why you don't need to care about it for most construction projects, and why you wont see obvious curvature just by looking at the ground at your feet.
You say Earth looks flat, because you cannot detect such a tiny change.
Also, have you considered following your own advice? How many times have you suggested we do something to try to prove your point, while refusing to do it yourself (likely because you know it will refute you).That doesn't mean we cannot MEASURE it as flat, OR as 'curved', within allowable tolerances, that's the point here.
If a surface 500 feet long, 400 feet wide, is measured as flat, within allowable tolerances, the same works with LONGER areas, and their relative tolerances, right?
The first area measures flat, within acceptable tolerances. An adjacent area measures flat, within tolerances. And a third area, adjacent to the second one, also measures flat, within tolerances.
Each of those 3 areas are measured flat, so to know if all 3 areas TOGETHER are flat, there's several ways to confirm it, and make sure all of them, separately, OR collectively, are flatWe can easily do this to measure for your supposed 'curvature' of Earth, too.
You believe that Earth has a 'curvature' of about 8 inches per mile, squared each additional mile. And you also claim we cannot MEASURE for it, within such 'small' areas, and we don't have instruments which can detect it, over those 'small' areas.That doesn't mean we cannot MEASURE it as flat, OR as 'curved', within allowable tolerances, that's the point here.
The point is that you are ignoring those tolerances. The point is that you are measuring it as "flat" with tolerances that cannot tell if it is a sphere of radius 6371 km, i.e. you cannot tell if Earth overall is flat or round.
Understand yet?
You trying to measure Earth's surface, over a distance of 1 m, to assert the RE model is wrong, is entirely pointless unless you are capable of measuring a variation of 20 nm over this distance of 1 m.
Yet you want to pretend that because you can't detect that tiny variation, that the RE model is wrong.
You want to pretend that a curve will always be easily and clearly visible as a curve, which is to completely ignore those facts about tolerances. Those tolerances mean that if the curve is large enough such that the curvature is small enough, you cannot tell if it is curved or flat.
If a surface 500 feet long, 400 feet wide, is measured as flat, within allowable tolerances, the same works with LONGER areas, and their relative tolerances, right?
If you actually measure it to those tolerances, or understand how to combine those tolerances/uncertainties to get the overall uncertainty/tolerance. You can't just say an area is flat to within 1 mm over a distance of 1 m so it must be flat to within 1 mm over 100 kmThis doesn't show what you think it does. Again, this is directly where that question of a 300 sided polygon comes in.
Lets say you have an area that is 100 m wide. You measure it and deem it to be "flat", but the uncertainty of your measurement is 0.1 degrees. That is your measurement is only capable of determining that one side of this area is within 0.1 degree, up or down, of the other side (in fact that would apply to the entire area). It doesn't matter how many of these 100 m wide areas you measure, nor how many times you measure it, nor how much overlap you have. All you will do with that is confirm that that span is within that tolerance.
To actually combine it, lets say you take 2 100 m wide areas, side by side. The expanded uncertainty becomes 0.2 degrees. After 10 such measurements you are up to an uncertainty of 1 degree. After 3600 such measurements you are up to an uncertainty of 360 degrees. That means after 360 km, you have no idea what the orientation of the surface is relative to the first point.
That means you have no idea if that 360 km span is entirely flat, or a curved into a circle. (Well technically you can figure out it isn't a circle by realising you aren't back where you started).Stop repeating the same refuted BS. You can't measure it over a tiny area like you want to. But over a large enough area you can. It is measured, it is seen, repeatedly.
Again, the very topic you continue to avoid shows it is round. The mere existence of the horizon should be a dead give away. A flat surface only has a horizon at the edge. But a curved surface will have a horizon where a line from your eye goes tangent to the curve. This means the horizon on a curved surface will move with you and be dependent upon your distance from it. This matches what is observed for Earth in reality, a horizon that moves with you and is dependent upon your altitude/height.
Likewise, as an object goes into the distance on a flat surface, it remains entirely visible, never appearing to sink. But on a curved surface, after enough distance the horizon and the curved surface start to obstruct the view. This causes the object to appear to sink and disappear from the bottom up. Again, this is observed.
You can directly measure the dip angle to the horizon, and with knowledge of your altitude, you can use that measurement to calculate the radius of Earth. You can measure the direction and distance along large roads or large maps, and try plotting them on a flat surface and seeing that it doesn't work, and then try on a round surface and adjusting the radius until it does fit.
You can also see its effects by looking at the stars or celestial objects. We can look at the moon and observe the same face (roughly) from all over Earth. Likewise, we can observe the constellations, looking the same (if visible) all over Earth. There is no apparent distortion which would be expected if you were looking at it from a different angle, nor do we see a significantly different face. This means they must be in the same direction for everyone. But their apparent direction (i.e. the direction relative to the surface of Earth at your location) varies. That means the orientation of Earth itself must vary. And if you measure how this varies, you see that Earth is round.
We can also use a ring laser gyroscope or Foucault's pendulum, to measure the rotation of Earth. A ring laser gyro is better because it can measure the angle of the axis, instead of needing to just observe that the apparent period changes as you do with Foucault's pendulum. You can then see how this axis or period varies across Earth, and see how well that fits a RE, and how it doesn't fit a flat Earth at all.
So again, your claim is pure BS. There is plenty of indication that Earth is curved. Plenty measures for it, and plenty is seen of it.
This not happening over a tiny area doesn't mean it doesn't happen over a much larger area.
Stop pretending not being able to measure the curve over 1 m means it isn't curved at all. Stop with all these pathetic deflections, and trying addressing the issue at hand.
Again, then why don't you be honest, admit you can't, and admit that this phenomenon is evidence for the curvature of Earth?
The point is that due to how large Earth is, you cannot measure the curvature over a tiny portion of it.
We certainly CAN measure for a 'curve', within a micron, by simply using the proper INSTRUMENTS!
I've heard over and over, that we cannot use a LASER level, to measure for 'curvature', which is complete BS.
It can easily be measured, using an accurate laser level, at it's minimum variance, which will be a certain maximum distance.
What your side does, is assume it must measure over a LONG distance, which is BS. It only has to measure what it CAN accurately measure, and nothing BEYOND what it can accurately measure, NOT because we COULDN'T measure for it, since we certainly can, and would be able to measure for it, and already WOULD have measured it, years ago, and would today, even MORE accurately than beforeThat is the ONLY distance which matters, to prove no 'curve' exists, and even though we CAN measure for it over such an area, with instruments of today, I'm going to explain how we can measure it with laser levels, easilyWhat they DIDN'T say, and DIDN'T do, is use their laser level, up to it's OPTIMUM distance for accuracy, to make it FAR MORE ACCURATE than any 'light beam' comes CLOSE to being!We could set up two or three lines with laser levels, at different points, to confirm our findings.
It could be done in much smaller sections, for more accurate measurements, too.
But don't tell me it's not possible to MEASURE for it, because that's complete BS, and you know it.
We certainly CAN measure for a 'curve', within a micron, by simply using the proper INSTRUMENTS!
Over a small distance, sure. But over that small distance you need to measure a variation of 10s of nm to measure the curve of Earth, so no, you can't.
Over a long enough distance, we can easily measure it with a theodolite, as people have. Even the FE high prophet Row Boat somewhat accepts that, but he then demands making the apparatus much worse by removing the lens to pretend it can't
Follow your own advice. As I have demonstrated, the curvature is insignificant at small distances, so there is no need to correct for it, and it almost certainly cannot be measured for over such small distances.
If you wish to disagree, then stop just asserting BS and start justifying your claims. Prove that such a variation can be measured for over a short distance, and stop ignoring all the measurements of curvature over a long distanceYour statement is meaningless. "measure a flat line" tells us nothing about just how flat it is. It does nothing to show if it can measure the curvature of Earth.
If you want to assert it can measure for the curvature of Earth, then provide the specifications with a reference. If you can't do that, then stop with the BS. And regardless of which path you take, stop just saying it can measure flat, without any specification of uncertainty
No it doesn't, but as already explained, if you want to try combining measurements, then you need to understand how that impacts uncertainty. Again, this is where a 300 000 sided polygon comes in. If you can only measure a variation of 0.1 degree, then you would measure each corner as "straight", and thus falsely conclude this 300 000 sided polygon is a straight line. Conversely, I recognise the limitation of the measuring device and how the uncertainty combines to realise that by the 3600th corner, we have no idea what the orientation of the next side is compared to the first side.
You can't even just use height like you want to. A laser level, if set up well, will level itself. That means if there is a curve, they will be offset at an angle. And so if you can't measure the curvature with one length, you wont be able to do so with 2But don't tell me it's not possible to MEASURE for it, because that's complete BS, and you know it.
Again, it is possible to measure for the curvature, it has been done. What I will continue to tell you is the truth, that it is impossible to measure it over a very small distance, because we do not have the tools that are accurate enough to measure such a variation over such a small distance.
Again, if you wish to disagree, then instead of spouting pure BS, justify it. Provide the specifications of a device to show it CAN measure the curvature, and provide a reference to those specs.
Using these 30 m, and the simple approximation of h=d^2/(2R) we get the drop due to the curvature over that distance of 30 m is 0.07 mm. Less than one 20th of the uncertainty. So it certainly can't measure it at that distance. Now sure, you can try going out more, but how accurate is it then, and can it even reach?
If we put in those 400 m we get an uncertainty of 19.4 mm using the 10 arcseconds, or 20 mm using the 1.5 mm per 30 m. But we get a drop due to curvature of 12.6 mm. Notice how the drop is less than the uncertainty.
So it appears this laser level, described as the most accurate, is unable to measure the curvature.
And again tell us, then why don't you be honest, admit you can't, and admit that this phenomenon is evidence for the curvature of Earth?It is a distance measurement device, not a level. That means it is entirely in appropriate for your suggested test, as it is not the correct device.
Do you even understand the difference between the 2?
Even if I were to accept your fantasy where it can magically switch to a laser level, it is accurate to 1.5 mm with a range of 100 m. But the drop due to curvature over that range is 0.78 mm. That is roughly half the accuracy of the device, so it can't measure the curveNo, it isn't. That isn't how it works at all. You are looking at the uncertainty for the measurement, while ignoring the uncertainty of the alignment.
That would be like suggesting I can take a 10 cm ruler, and measure from the start of a 1 m long object to the first 10 cm, put my finger down to mark that position and then move the ruler along, and repeat until I measure the entire object, then claiming that because the ruler is accurate to less than 1 mm, the overall result must be accurate to less than 1 cm. The problem is this ignores the issue of alignment. Each time the rule is moved, there is an uncertainty associated with the alignment, which is separate from the uncertainty in the measuring device.
That can make the overall uncertainty much larger.
So just how are you aligning your device? If we assume that it is a level, instead of a distance meter, and assume it will level itself, then if Earth is flat, such that level is the same everywhere, then it would be simple like you have done. But in reality, with a curved Earth, you are measuring different orientations, and you need to account for that change in orientation when combining your measurement.
Here is an example of an extreme case, drawn using different numbers, but drawn to scale: We see that the measurement has un uncertainty of 0.01 over the distance 0.1. But when we take our second point, Earth has curved, so it levels to a different orientation. This means our overall uncertainty is no longer simple addition, even though the second measurement also has an uncertainty of 0.01 over a distance of 0.1. Instead, the uncertainty is a combination of the individual measurement uncertainties and the alignment uncertainty. This results in an overall result of an error 0.038 over a distance of 0.197. Not the 0.02 error over a distance of 0.2 like you want to pretend.
So no, your method does not work, or at best, is entirely circular, assuming Earth is flat to assert the alignment is the same in each case to then claim you don't measure the curvature expected for a RE so Earth is flat. But it is only by already assuming your conclusion that Earth is flat that you reach the result that Earth is flat.
Without that baseless (and false) assumption, you have no idea what the alignment is, and thus no idea what the overall uncertainty is. Without that baseless (and false) assumption, Earth could be (and is) round, and thus the alignment can be different for each measurement, increasing the overall uncertainty allowing the curvature of Earth to sit entirely inside that uncertainty.
So unless you can use a laser level (not distance measurement tool), and describe the alignment procedure and the associated uncertainty with that procedure, your test is entirely useless.
And once all that is dealt with, you then get to deal with the surface of the lake. What is the uncertainty in the surface of the lake? If the lake's surface varies by a cm, then that will be an additional variation which needs to be accounted for which will expand the uncertainty in the overall result.
No, it is accurate to 1/4 inch over 300 feet, and 1/4 inch accurate over the next 300 feet, and so on, and to confirm those measurements are valid, over each section, as constant throughout them all, we use a second line of measurements, halfway up the first sections, to account for any error in each transition point, if any are found.
A third set of measurements halfway up from the second one, can further confirm those points of transition, too.
Our goal is to measure for the whole surface, by measuring a part of it, over and over again, which is confirmed by another set of measurements, staggered to the first, and so on, if necessary, to any degree of accuracyA laser level that is accurate to 1/4 inch over 300 feet, doesn't mean it WILL be 1/4 inch out all the time, it will usually be LESS than 1/4 inch off, the 1/4 inch is it's MAXIMUM variance.
These variances can always be accounted for, with other measurements checking for it's accuracy, and variances.
and to confirm those measurements are valid, over each section, as constant throughout them all, we use a second line of measurements, halfway up the first sections, to account for any error in each transition point, if any are found.
That has the exact same alignment issues. It doesn't matter how many extra points you use, you will still have these alignment issues which destroy your measurement.
Here is another simple diagram to demonstrate yet again that your claim is pure BS: A laser level mounted at the first point will record the height of both the second and third point to be level within uncertainty. A laser level mounted at the second point will record the height of the third point to be level within uncertainty. Each time they are accurate to 0.01 units over a distance of 0.1 units, or 0.005 over a distance of 0.05; but over the 0.2 units of distance (actually 0.197 when measured perpendicular to straight down at the first point), you have an uncertainty of 0.038. The uncertainty of the alignment increases the uncertainty of the measurement.
You have the uncertainty due to each of the 0.1 m long spans, as well as the uncertainty of the alignment which makes the overall uncertainty larger than simply adding up the 2 lots of 0.02.
It doesn't matter how much you want to ignore this fact, it wont magically change.
If you want the 0.02 as your uncertainty, you need to align them perfectly. It is that alignment which you need to focus on, not throwing in more pointsAs proven above, that is pure BS. If you wish to claim it can be done, then prove it. Either mathematically, or practically. Show that by using a measuring device incapable of measuring the curvature over a single span that you can measure it by combing multiple spans, including explaining how you are aligning these measurements.
No, it is accurate to 1/4 inch over 300 feet, and 1/4 inch accurate over the next 300 feet
Only if you perfectly align it. If there is any uncertainty/error in the alignment, then that will impact the uncertainty as well.
It WOULD be perfectly aligned, same as everything ELSE would be.
Laser levels would be almost useless, if they couldn't be perfectly aligned, or 'plumb', or 'square', before anything is measured with them! That's rather obvious, isn't it?
You seem to think we cannot measure any surface accurately, when it is 'too long', whatever that's supposed to mean!
I'll try to explain why you're wrong, that they CAN measure for 'curvature', which you say is so incredibly 'slight' of a curve, over the Earth, that we cannot measure it, over a 'small' distance, which is any distance we CAN, and HAVE, measured!Or we can simply use shorter distances, with the laser level, which would make it more accurate, over each measurement, that also would work.
I'm trying to explain to you, that we absolutely CAN measure a surface as being flat, or curved, over a one mile distance, or more, with absolute accuracy, that is a fact. It may not be spoken of, may not be claimed as a fact, and they never do, in public, at least. But they certainly KNOW it is a fact, and some have actually MEASURED it, proving that it IS a fact.
With laser levels of today, almost anyone can do it.
You're suggesting 'curvature' is too slight over 300 feet distance, because it has a 1/4 inch variance, which is less than 'curvature' is, over that same 300 foot distance.
So how did they know, when testing their laser level, that it actually WAS, at very most, inaccurate, over a distance of 300 feet, BY 1/4 inch?
Because that's the most important part you need to know about here.
They couldn't have known it was off by 1/4 inch at most, over 300 plus feet, unless they had MEASURED the CORRECT point, from 300 feet away!
Since you haven't disputed that figure, I assume you accept it as true, and we can move on...
Obviously, they couldn't know it was off by 1/4 inch, at most, unless they knew what WAS the correct point, from 300 feet away, or actually 324-5 feet away, or whatever it is, but they must have found out, it becomes less accurate at any LONGER distances, by measuring for the correct point there, as well.
And, if they have measured the correct point, where the laser level WOULD always hit, from 300 feet away, then obviously, we CAN measure it accurately, and HAVE measured it accurately, over a distance of 300 plus feet.
If THEY measured it, anyone else can measure it, too, the same way they did.
We know, they certainly knew, that a laser light is a perfectly straight line of light, so when they tested it for accuracy over 300 feet, or every 10 feet further out, or whatever, they had to set their targets at the exact same height, over all of the targets, at all distances outward from the laser level.
They had to set the targets at the exact same height, to a micron, or whatever is going to establish maximum accuracy.
When they measured a target at 300 or 400 feet away, they must have known the target point had to be at the EXACT SAME height as the laser light was, 300 feet way from it. They had to know it's exact direction, to set their target in the exact direction of the laser light, 300 feet away, at each side of it, too.
They set up targets at specific distances away from the laser level, with points on each target, set at the exact same height, exact same line of direction, as the laser level.
Why couldn't they have set the targets for 'curvature'? Because if they wanted to, each target would be set lower, to curvature from the laser level, to the targets, which would lower more and more, with each target further away, to match up with it.
To measure a point on a target, at 300 away from a laser level, means we can measure points on targets over any distance, because they never account for this completely made up 'curvature' of the Earth's surface.
Laser levels would be almost useless, if they couldn't be perfectly aligned, or 'plumb', or 'square', before anything is measured with them! That's rather obvious, isn't it?
So you are using the circular reasoning I think I pointed out before. You are assuming Earth is flat, to assume that when the levels level themselves they will all be levelling the same, to then claim they are all aligned. For your fantasy Earth that means they would all point the same way, but for the very real round Earth, that means each one will be pointing in a slightly different direction, just like I demonstrated in my image, where they are all pointing down towards the centre of Earth. That means you don't have the alignment you need.
This means your argument is pure garbage as it relies upon falsely assuming Earth is flat to try and demonstrate Earth is flat. But without that assumption it fails.
Each level in my diagram was perpendicular to the surface of Earth, that is level for a RE.
So you have been refuted once again. The errors cannot simply be added because it relies upon assuming the alignment which you cannot honestly assume.
Considering you started with such a false claim of being able to perfectly align them to demonstrate Earth is flat which relies upon you assuming Earth is flat, I highly doubt you will succeed, as I highly doubt I'm wrong. But go ahead and try.I have already demonstrated the math that shows that is not enough to detect the curvature. The best one I could find is 1.5 mm per 30 m, using the maximum range of 400 m, you have an error of 20 mm, while the curvature is only 12.6. This means the curvature is within the expected uncertainty. That means if you can't tell if it is flat or curved.
When you move it to another point, if Earth is curved, the alignment is out, so you can't just add the 2 uncertainties like you want toHow? By assuming Earth is flat and then measuring the height above the reference water level? If so, as already pointed out, that relies upon you assuming Earth is flat, and thus cannot be used to demonstrate Earth is flat. If Earth is round, you are measuring relative to a reference sphere (or geoid more generally), and that means you aren't ensuring the alignment is perfect and you aren't able to show Earth is or isn't round.
I'm trying to explain to you, that we absolutely CAN measure a surface as being flat, or curved, over a one mile distance, or more, with absolute accuracy, that is a fact.
No, you aren't explaining anything. You are just making more and more bold assertions with nothing to justify them.
The closest you have come to an explanation is by appealing to the laser's self-levelling. But while that doesn't give you the alignment it needs. So you failed with that. And that was before you claimed to try to explain it.
Your attempt at an "explanation" was nothing more than you just asserting yet again that you can measure it.
Try an actual explanation. One which doesn't rely upon you already assuming Earth is flat. And one which clearly explains how each step is done instead of just magically asserting that it happens.
But they certainly KNOW it is a fact, and some have actually MEASURED it, proving that it IS a fact.
Plenty of people have measured the curvature of Earth and know it is round. That is a fact. No one has been able to measure Earth to the required level of uncertainty to demonstrate it isn't curved as science shows us it is.
So how did they know, when testing their laser level, that it actually WAS, at very most, inaccurate, over a distance of 300 feet, BY 1/4 inch?
By understanding how the laser works. They don't need to actually measure at that distance to give an estimate of the uncertainty. The more accurately quoted uncertainty would be in arcseconds or another angular unit of measure. The stated accuracy comes from the beam divergence and the accuracy of its self-levelling.
That is the actual uncertainty of the instrument. The 1/4 of an archaic unit over 300 plus archaic units is based upon an angular uncertainty. The figure quoted on their website was 1.5 mm at 30 m. At that stage the curvature is just 0.07 mm. Tiny compared to the uncertainty of the laser.
You still don't seem to understand that you don't need to directly measure something to determine what the value is.
But no, they don't need to measure it. To show the insanity of that, consider all the other possible distances. Sure, you are going to roughly 100 m. But what about 50 m? Do they need to measure there as well? What about 1 m? 2m? 3? 4? 5? 6? and so on? Do they need to make measurements over 100 separate distances, performing multiple measurements at each to determine the accuracy? If so, what about at 50.5 m? or 50.1? or 50.01? and so on. They could be doing this forever, and according to you it still wouldn't be enough.
The entire rest of your argument seems to be based upon this fantasy of yours.
Back in reality, they measure the angular variance, which any rational, intelligent person (who understands basic trig) can use to calculate the uncertainty at any given range.
If THEY measured it, anyone else can measure it, too, the same way they did.
People can and have measured the curvature of Earth with a theodolite over long distances. A simple example of this is measuring the angle of dip to the horizon.
Again, you not liking reality wont change it. And your fantasy of how they set it up in no way helps you with your claim of magical perfect alignment, nor your claim that laser levels are accurate enough.
Summary All data and observations agree with the predictions of the Globe Model, which includes Terrestrial Refraction. The predictions for the Flat Earth Model, however, contradict the observations.
The Rainy Lake Experiment shows even better than the Bedford Level Experiment that the earth is a globe, since we also have GPS measurements that are not influenced by Refraction or Perspective, but are of a pure geometric nature. GPS measurements directly provide the radius of the earth.
Only one conclusion remains:
The earth cannot be flat, but is a globe with a mean radius of 6371 km!
With flat earther's debunking themselves
Quote
Flat-earthers tried to prove the Earth was flat with a videotaped experiment and it did not go well
When the experiment began, the light didn't appear on camera. A perplexed Jeran radioed Henrique to confirm the height of the light at 5.18 meters (17 feet) above sea level. On a flat Earth, he should be seeing the light. He then asked Henrique to lift the light above his head. Lo and behold, the light shined through.
"That's interesting," Jeran commented in the clip.
And you know what? It is interesting. This experimental set up has been a staple of flat-Earthers since 1836, when Samuel Birley Rowbotham first did it on the Old Bedford River. Time and time again, it has revealed the curvature of the Earth. Still, it is important to continue to repeat classic experiments as repetition is one of the cornerstones of science.
If you missed it
"when Samuel Birley Rowbotham first did it on the Old Bedford River. Time and time again, it has revealed the curvature of the Earth"
These are actual, solid, physical surfaces that are being measured with real, actual instruments, the only 'magic' is claiming it is measuring your made up 'curvature', without even MEASURING it, until they say they've measured it from 'space', or in 'orbit', where everything is 'proven', like a made up magical force called 'gravity', was also 'proven'.
'Space' is a magical place, which doesn't exist as they have claimed, not even close to it, in fact. The actual facts, real evidence, prove Earth IS flat, and more proof in future, it is inevitable.
Planes will measure for level flight, when they reach a specific altitude, while any adjustments during flight, are corrected for afterwards, and they return to level flight at altitude.
If planes measured for a level flight, but was for a curved flight path, that would require planes to make a physically curved path, to remain at altitude, while flying 'level' to a curved surface below it.
A level flight cannot be a curved flight, measured AS level to a curve. Because a curved path must be flown as a curve, as an arced path, as a PHYSICAL curve has to be flown by planes, to follow the same altitude of a curved surface below it, throughout the flight.
So planes can't measure for your 'curvature', when they only can measure around the plane itself, at very most, they cannot measure beyond that, they have to continually measure it, over and over again, during the whole flight.
If there WAS 'curvature', planes would measure it continually, over and over again, in SMALL, SEPARATE SEGMENTS, over the same, small distance, so even if there WERE 'curvature' on the surface of Earth, planes wouldn't be able to measure for it. Not with these instruments, anyway.
How could your magical made up force, 'gravity', make instruments measure level to a curved surface of a ball Earth in 'space'? Planes can only measure for level flight over that small distance, over and over again, of the same 0.0025 of 'curvature', each time, over and over again, throughout the flight, no matter how long it takes before we've developed an instrument to measure it, because that's how it would be, if Earth were a ball in 'space'.So, in short, you are incorrect.
No, surveyors that work on real projects say that they always 'assume' surfaces are FLAT, but it's far more than an assumption, because they actually WORK from this 'assumption', and then others will BUILD STRUCTURES on that same 'assumption', and they WORK PERFECTLY from that 'assumption', and they always DO work from this very same 'assumption', all the time!
They can call it anything they want to, but their actions show what it really is, and it is certainly NOT just an 'assumption', it is known to be flat, measured as flat, and built as flat, because it IS flat.
Surveying is about accurate measurements of the surface, and does NOT 'assume' anything, especially if they KNOW it is not true, and if they knew the surface was NOT flat, and also knew it was CURVED, they'd say it IS a curved surface, assume they're curved, work with a curved surface, make it flat for building on it, which would make perfect sense, and would be completely logical.
These are actual, solid, physical surfaces that are being measured with real, actual instruments
And none of that helps your claim. Again, in reality, laser levels level themselves. They don't magically align to some magical flat surface. They level themselves.
That makes your argument circular. You rely upon your false assumption that Earth is flat to falsely claim the levels are magically aligning themselves, to falsely claim they could measure the curvature, to falsely claim Earth is flat.
Then why are you entirely incapable of providing a single bit of it? Why do you instead resort to attacking the RE with pure BS? Why do you resort to repeating the same refuted nonsense again and again?
Yes, LEVEL! Not flat. We have been over planes quite a lot, with you fleeing from it because you couldn't defend your BS. Why come back to it just to get refuted all over again? Why flee from your laser levels? Does that mean you now fully accept your laser levels cannot measure the curvature of Earth because of issues with alignment?
If planes measured for a level flight, but was for a curved flight path, that would require planes to make a physically curved path, to remain at altitude, while flying 'level' to a curved surface below it.
And the amount they would need to curve would be hidden in the noise. They would not be able to tell, not without incredibly precise equipment.
If there WAS 'curvature', planes would measure it continually, over and over again ... planes wouldn't be able to measure for it. Not with these instruments
This shows just how pathetic your argument is. You can't even remain consistent for a single paragraph. On one hand you claim they should constantly measure it and need to adjust, only to turn around and say they wouldn't be able to
Again, pure BS. Again, when planes are flying, they maintain their altitude. The pilot or autopilot will constantly be adjusting the plane to do so. As they do that, there is no need for them to specifically account for the curvature.
Stop bringing up the same refuted BS as if it hasn't already been refuted.
they'd say it IS a curved surface, assume they're curved, work with a curved surface, make it flat for building on it, which would make perfect sense, and would be completely logical.
Again, stop repeating the same refuted BS. For the vast majority of projects, the curvature of Earth is less than the roughness of concrete. They are not going to need to care about that. It makes no sense to try to account for something that would be lost in the noise. | 677.169 | 1 |
properties of triangle and quadrilateral
Angle Properties Of Triangles And Quadrilaterals Worksheets – Triangles are among the most fundamental forms in geometry. Understanding triangles is important for getting more advanced concepts in geometry. In this blog post we will look at the different types of triangles with triangle angles. We will also discuss how to determine the size and perimeter of a triangle, and provide illustrations of all. Types of Triangles There are three types of triangles: equal, isosceles, and scalene. … Read more | 677.169 | 1 |
1.5 More Applications
Sometimes solving problems involving right triangles requires the use of a system of equations. A common method for determining the height of an object whose base is inaccessible is that of measuring the angle of elevation from two different places in front of the object. If you measure the angle of elevation to the top of of a radio antenna as 74 , then walk back 50 feet and measure the angle of elevation to the top of the antenna as 61 ', then we would have something like the diagram below:
One of the first things we can do is introduce some labels for the unknown distances:
Then, we can say that:
\(\tan 74^{\circ}=\frac{h}{x}\)
\(\tan 61^{\circ}=\frac{h}{x+50}\)
To solve this system of equations, we'll set the first one equal to \(h:\)
\[
\begin{array}{c}
\tan 74^{\circ}=\frac{h}{x} \\
x * \tan 74^{\circ}=h
\end{array}
\]
Then, substitute this into the second equation:
\[
\tan 61^{\circ}=\frac{h}{x+30}
\]
\[
\tan 61^{\circ}=\frac{x \tan 74^{\circ}}{x+50}
\]
Multiply on both sides by \(x+50:\)
\((x+50) \tan 61^{\circ}=\frac{x \tan 74^{\circ}}{x+30}(x+50)\)
\((x+50) \tan 61^{\circ}=x \tan 74^{\circ}\)
There are two options to solve this equation - we can hold on to the tangents
as they are and solve for \(x\) in terms \(\tan 74^{\circ}\) and \(\tan 61^{\circ},\) or we can approximate \(\tan 74^{\circ}\) and \(\tan 61^{\circ}\) and generate an approximate value for \(x\) and \(h .\) First we'll approximate:
\((x+50) \tan 61^{\circ}=x \tan 74^{\circ}\)
\[
(x+50) * 1.804 \approx 3.4874 x
\]
\(1.804 x+90.2024 \approx 3.4874 x\)
\(90.2024 \approx 1.6834 x\)
Exercises 1.5
1. Find the indicated height \(h\)
2. Find the indicated height \(h\)
3. A small airplane flying at an altitude of 5300 feet sights two cars in front of the plane traveling on a road directly beneath it. The angle of depression to the nearest car is \(62^{\circ}\) and the angle of depression to the more distant car is \(41^{\circ}\). How far apart are the cars?
4. A hot air balloon is flying above a straight road. In order to estimate their altitude, the people in the balloon measure the angles of depression to two consecutive mile markers on the same side of the balloon. The angle to the doser marker is \(17^{\circ}\) and the angle to the farther one is \(13^{\circ} .\) At what altitude is the balloon flying?
5. To estimate the height of a mountain, the angle of elevation from a spot
on level ground to the top of the mountain is measured to be \(32^{\circ} .\) From a point
1000 feet closer to the mountain, the angle of elevation is measured to be \(35^{\circ}\) How high is the mountain above the ground from which the measurements were taken?
6. The angle of elevation from a point on the ground to the top of a pyramid is \(35^{\circ} 30^{\prime} .\) The angle of elevation from a point 135 feet farther back to the top of the pyramid is \(21^{\circ} 10\). What is the height of the pyramid?
7. An observer in a lighthouse 70 feet above sea level sights the angle of depression of an approaching ship to be \(15^{\circ} 50^{\prime} .\) A few minutes later the angle of depression is sighted at \(35^{\circ} 40^{\prime} .\) Find the distance traveled by the ship during that time.
8. To estimate the height of a tree, one forester stands due west of the tree and another forester stands due north of the tree. The two foresters are the same distance from the base of the tree and they are 45 feet from each other. If the angle of elevation for each forester is \(40^{\circ},\) how tall is the tree?
9. A ship is anchored off of a long straight shoreline that runs east to west. From two observation points located 10 miles apart on the shoreline, the bearings of the ship from each observation point are \(S 35^{\circ} E\) and \(S 17^{\circ} W .\) How far from shore is the ship?
10. From fire lookout Station Alpha the bearing of a forest fire is \(N 52^{\circ} E .\) From lookout Station Beta, sited 6 miles due east of Station Alpha, the bearing is \(N 38^{\circ} \mathrm{W}\) How far is the fire from Station Alpha?
11. From a point 200 feet from the base of a church, the angle of elevation to the top of the steeple is \(28^{\circ},\) while the angle of elevation to the bottom of the steeple is \(20^{\circ} .\) How high off the ground is the top of the steeple?
12. A television tower 75 feet tall is installed on the top of a building. From a point on the ground in front of the building, the angle of elevation to the top of the tower is \(62^{\circ}\) and the the angle of elevation to the bottom of the tower is \(44^{\circ} .\) How tall is the building | 677.169 | 1 |
Items in this lesson
"Without mathematics, there's nothing you can do. Everything around you is mathematics. Everything around you is numbers." Do you agree with this quote? Why?
timer
1:30
Slide 1 - Open question
Circle
The parts of a circle are the radius, diameter, circumference, arc, chord, secant, tangent, sector and segment. A round plane figure whose boundary consists of points equidistant from a fixed point.
timer
4:00
Slide 2 - Slide
What is the numerical equivalent of Pi?
Slide 3 - Mind map
Pi
3.14
Slide 4 - Slide
What is the circumference 5 - Quiz
Finding the circumference of a circle using the radius
Slide 6 - Slide
What is the radius 7 - Quiz
Using the radius
Use the formula C = 2πr to find the circumference using the radius. In this formula, "r" represents the radius of the circle. Again, you can plug π into your calculator to get its numeral value, which is a closer approximation of 3.14.
A radius is any line segment that extends from the center of the circle and has its other endpoint on the edge of the circle.
You might notice this is similar to the C = πd formula. That's because the radius is half as long as the diameter, so the diameter can be thought of as 2r.
Slide 8 - Slide
Example
Slide 9 - Slide
Slide 10 - Video
What is the circumference of a circle that has a radius of 6cm?
timer
2:30
Slide 11 - Open question
Answer
37.7cm
Slide 12 - Slide
Finding the circumference of a circle using the diameter
Slide 13 - Slide
What is the diameter 14 - Quiz
Using the diameter
Use the formula C = πd to find the circumference if you know the diameter. In this equation, "C" represents the circumference of the circle, and "d" represents its diameter.
That is to say, you can find the circumference of a circle just by multiplying the diameter by pi. Plugging π into your calculator will give you its numerical value, which is a closer approximation of 3.14 or 22/7.
Diameter means a straight line segment that passes through the center of the circle and has its endpoints on the sides of the circle.
Slide 15 - Slide
Example
Slide 16 - Slide
What is the circumference of a circle that has a diameter of 9cm?
timer
2:30
Slide 17 - Open question
Answer
28.8cm
Slide 18 - Slide
Are people born with a specific personality, or is the character the result of their circumstances? | 677.169 | 1 |
Shapes And Angles
Shapes and AnglesLook around yourself. What do you see? Buildings , trees , books,tables, chairs, notebooks, sun, moon , stars, etc. Are they all same? Dothey have the same shape? No, not all of them are alike. The shape ofthe sun is different from that of a book. The notebooks are of the sameshape but different size. Today we will discuss the various shapes andthe properties of these shapes .ShapesWhat is shape? The literal meaning of shape is the external form of athing or its appearance. A figure can be made of different types ofshapes. Figures can be broadly classified as open or closed. Openfigures are those which do not end where they start. The figures thatstart and end at the same point are closed figures.Browse more Topics under Shapes And Angles Angles in Real LifeClassification of Shapes
Let us learn more about the closed figure. The shapes are made oflines or line segments arranged in an organized manner. Every objectaround us has a definite shape. Shapes can be classified into curves( circles ) and polygons .PolygonsA polygon is a simple closed figure with three or more than three linesegments. Poly means "many".Terminologies Associated with Polygon Sides: The lines segments that form the exterior of the polygonare the sides of the polygons. Adjacent Sides: Two sides of a polygon having a commonend-point are adjacent to each other. Vertex: The point where the sides of a polygon meet is calledthe vertex of the polygon. It is also referred to as end-point.The plural of vertex is vertices. Adjacent Vertices: The end-points of the same side of apolygon are adjacent vertices.
Polygons are classified into Regular and Irregular Polygons. APolygon in which all sides and all angles are equal is a regularpolygon. The sum of all interior angles of a regular polygon 180 (n 2), where n is the number of sides. A Polygon in which all sidesand all angles are not equal is an irregular polygon.Types of PolygonsLet us discuss the types of polygons on the basis of the number ofsides.TrianglesA polygon bounded by three line segments or sides is a triangle. Thesum of the interior angle of a triangle is 180 . On the basis of equalityof sides, triangles are of three types: Equilateral Triangle: A triangle with all sides equal is anequilateral triangle.
Isosceles Triangle: A triangle with two sides of equal length isan isosceles triangle. Scalene Triangle: A scalene triangle is the one with all unequalsides.
On the basis of the measure of angles, triangles are of following types: Acute-angled Triangle: A triangle in which each angle is acute(less than 90 ) is an acute-angled triangle. Right-angled Triangle: A right-angled triangle is the one inwhich one of the angles is a right angle (90 ).
Obtuse-angled Triangle: An obtuse-angled triangle is the one inwhich one of the angles is obtuse. Equiangular Triangle: If all the three angles are equal, thetriangle is an equiangular triangle.
Quadri lateralsA closed figure bounded by four line segments is known as aquadrilateral. A quadrilateral has four sides, four vertices, and fourangles. There are various types of the quadrilateral.ParallelogramA quadrilateral in which the pairs of opposite sides are parallel is aparallelogram. The sides of a quadrilateral that have no commonendpoint are opposite sides.Properties of a Parallelogram: Opposite sides are equal and parallel. Opposite angles are equal Diagonals bisect each other.
RectangleA rectangle is a special case of a parallelogram. In a rectangle, eachangle is of 90 .Properties of a Rectangle: Opposite sides are equal and parallel. All angles are right angles Both the diagonals are equal in length and bisect each other.SquareA parallelogram with all sides of the same length and all the angles of90 is a square.
Properties of a Square: All sides are equal. Opposite sides are parallel ( ). All angles are right angles. Both the diagonals are equal in length and bisect each other atright angles.RhombusA parallelogram in which all the four sides are equal in length is arhombus. A square is a special type of rhombus.
Properties of Rhombus: All sides are equal in length. Opposite sides are parallel. Opposite angles are equal. The diagonals bisect each other at right angles.TrapeziumA quadrilateral in which only one pair of opposite sides is parallel is atrapezium.
KiteA quadrilateral in which two pairs of adjacent sides are equal is a kite.Its diagonals bisect each other at right angle.PentagonIt is a five-sided polygon. The sum of the interior angles of a pentagonis 540 . ABCDE is a regular pentagon with all equal sides.HexagonA hexagon is a six-sided polygon. The sum of the angles is 720 .
A polygon with seven sides is a heptagon. With 8 sides it is anoctagon. A 9-sided polygon is a nonagon.CircleWhen a set of points is at the same distance from a fixed point thefigure obtained is a circle.Terminologies Associated with Circle Centre: The fixed point of the circle which is equidistance fromall the points on the circle is its centre. Point O is the centre ofthe circle. Radius: The line segment with one endpoint at the centre andthe other on the boundary of the circle is a radius of the circle.OA is the radius of the circle.
Diameter: A line segment with both of its endpoints on theboundary of the circle and passes through the centre is thediameter of the circle. AB is the diameter of the circle.Diameter 2 Radius. Chord: Any line segment with its endpoints on the boundary ofthe circle is a chord. The diameter is the longest chord whichpasses through the centre. AB is a chord of the circle.
Circumference: The length of the circle is the circumferenceof the circle. Arc: A part between any two points on a circle is an arc. Semicircle: Half of a circle is semicircle. Each part is asemicircle. A diameter divides a circle into two semicircles.Solved Examples for YouProblem: If the diameter of a circle is 40 cm, what is the value of itsradius?
Solution: Diameter 2 Radius.Solving we have, Radius 40 / 2 20 cm.Problem: What is the name of a polygon with the smallest number ofsides?Solution: Triangle (with 3 sides).Problem: Is a triangle possible with angles 54 , 63 , and 92 ?Solution: No, the sum of the angles 54 63 92 209 which isnot possible. The sum of angles of the triangle must be 180 .Angles in Real LifeDid you ever notice the two hands of a clock? The two hands togethermake different sets of lines from a common point. These sets of linesfrom a common point is called angle. The two hands form differentangles every minute of the time. A clock forms an example of anglesin real life. What are the other examples of angles in real life? Let's
study the various types of angles , say acute angle, in real life and theirexamples in detail.AnglesBefore studying angle, let us do one interesting thing. Take a piece ofpaper and drawn a dot anywhere on it. Name this dot O. This dot O iscalled point. From this point draw as many straight lines as you can.How many did you get?Draw another dot and name it P. Join OP. What did you get? A linesegment OP with endpoints O and P. A line segment is a part of a line.Do you know a line has no endpoints like line segment ? A line hasmany line segments in it.Have you ever thought why we say sun rays and not sun line? Is raydifferent from a line? Yes, it is. A ray has one fixed point . It cannot goin both the direction like that a line. For sun rays, the Sun is the fixedpoint and thus we say the light coming as sun rays.Draw one more point other than O and P and call it R. Now join ORand OP. What will you find? Two lines and a common point O. This is
an angle . Ask your friend to do this activity . Are the two angles sameevery time? Maybe it is not.Browse more Topics under Shapes And Angles Introduction to Shapes and AnglesTerminologies in AnglesTwo lines or line segments or rays with a common point form anangle. An angle is denoted by a symbol . An angle is measured inan anti-clockwise manner with the help of a protector in degree. Vertex: The common starting point in an angle is a vertex. Arms: The two rays forming an angle from a fixed point arearms. They are also called the sides of the angle.
Measuring Angle: Place the center of the protector at the fixedpoint, say O of the angle at one of the arm. The other armshows the degree of the angle when measured from the initialarm in an anti-clockwise manner.Types of AnglesThere are different types of angles formed by two lines or rays withone common starting point. Based on the degree of measurement , theangles are classified as:Zero AngleAn angle whose measure is 0 is zero angle. The initial and the finalarms are at the same position. AOB is zero angle.
Acute AngleAn acute angle is the one that is greater than 0 and less than 90 .Right AngleAny angle whose measure is equal to 90 is a right angle.Obtuse Angle
An angle is said to be an obtuse angle if it is greater than 90 but isless than 120 .Straight AngleAn angle whose measure is 180 is a straight angle. This angle got itsname as it forms a straight line. AOB is a straight angle.Reflex AngleA reflex angle is the one that is greater than 180 and less than 360 .
Complete AngleIf the measure of an angle is 360 , it is a complete angle. This angle isformed as the arm makes one completes turn and returns to its startingposition.Other Type of AnglesApart from the above-mentioned angles, there are other types ofangles too. Adjacent Angles: Two angles are adjacent if both of them haveone common arm and a common vertex. AOB and COBare the adjacent angles. Is there any other pair of adjacent
angles? AOC and COB, and AOB and AOC areanother sets of adjacent angles. Complementary Angles: Two angles are complementary toeach other if the sum of both the adjacent angles is 90 . Supplementary Angles: When the sum of a pair of adjacentangles is 180 , the angles are supplementary angles.
Vertically Opposite Angles: Suppose two straight lines AB andCD intersect each other at a point O. The pairs of the oppositeangles are vertically opposite angles (V.O.A). The V.O.A. areequal in measurement. Here, 1 2 and 3 4.Angles in Real LifeIn the beginning, we have talked about angles formed by the hands ofa clock. Consider an example, the clock shows the time is 3 o'clock inthe morning. What is the angle made by both hands of the clock? It is
a right angle. In direction, we can find various angles. Where else canwe find angles?Cloth-hangers, scissors, arrowhead, partly opened-doors, pyramids ,Set squares , an edge of a ruler, an edge of tables, cycle spokes, wheelsetc are examples of angles in real life. Different alphabets also formthe examples of angles. What is the angle in letter V? An acute angle.Even we make different angles in different postures while doing yogaand exercises.Solved Examples for YouProblem: Name the angle when the time in a clock is 3: 40 pm? Is itObtuse or Reflex or Acute Angle?Solution: The angle is measured in an anti-clockwise manner and thusthe angle formed is reflex and not obtuse or acute angle.Problem: What is the value of angles α, β?Solution: α 90 45 135 and β 180 .
An angle is said to be an obtuse angle if it is greater than 90 but is less than 120 . Straight Angle An angle whose measure is 180 is a straight angle. This angle got its name as it forms a straight line. AOB is a straight angle. Reflex Angle A reflex angle
Adjacent angles are two angles that share a common vertex and side, but have no common interior points. 5 and 6 are adjacent angles 7 and 8 are nonadjacent angles. Notes: Linear Pairs and Vertical Angles Two adjacent angles are a linear pair when Two angles are vertical angles when their noncommon sides are opposite rays.
vertical angles. Vertical angles have the same measure. Vertical angles are also called opposite angles. 1 and 2 are vertical angles. 3 and 4 are vertical angles. 14. In this triangle, name a pair of complementary angles. m T 30 m L 60 30 60 90 . So T and L are complementary angles. 15. In this parallelogram, name a pair of .
Vertical Angles Words Two angles are vertical angles if they are opposite angles formed by the intersection of two lines. Vertical angles are congruent. Examples 4 2 3 1 1 and 3 are vertical angles. 2 and 4 are vertical angles. EXAMPLE 2 Finding Angle Measures Find the value of x. a. 70 x The
Adjacent Angles: Two angles and with a common side ⃗⃗⃗⃗⃗ are adjacent angles if belongs to the interior of . Vertical Angles: Two angles are vertical angles (or vertically opposite angles) if their sides form two pairs of opposite rays. Angles on a Line: The sum of the me
two acute vertical angles 62/87,21 Vertical angles are two nonadjacent angles formed by two intersecting lines. You can use the corner of a piece of paper to see that Ø ZVY and Ø WVU are less than right angles. 7KHUHIRUH DQG DUHDFXWHYHUWLFDO angles. two obtuse adjacent angles 62/87,21 Adjacent angles are two angles that lie in the same
Microsoft Visio Workbook Page 7 Shapes e elements that can be dragged Visio shapes are pre-drawn pictur and dropped into a diagram to visually communicate information and processes. 1-D and 2-D Shapes There are two types of shapes in Visio: one-dimensional shapes (1-D shapes) and two-dimensional shapes (2-D shapes). | 677.169 | 1 |
Making use of Geometry to Visual Perceptual Relationships
A spatial relationship generally defines just how a subject is positioned in space general to a reference photograph. If the reference image is a lot larger than the object then the ex – is usually represented by an ellipse. The ellipse can be graphically represented using a corsa. The corsa has very similar aspects into a sphere introduced plotted on the map. Whenever we look carefully at an ellipse, we can see that it is shaped so that all of their vertices lie on the x-axis. Therefore a great ellipse may be thought of as a parabola with one concentration (its axis of rotation) and many points of orientation one the other side of the coin.
There are four main types of geometric diagrams that relate areas. These include: the area-to-area, line-to-line, geometrical construction, and Cartesian structure. The fourth type, geometrical construction is a little totally different from the other forms. In a geometrical structure of a pair of parallel straight lines is utilized to establish the areas within a model or construction.
The main difference among area-to-area and line-to-line is that an area-to-area relation relates just surface areas. This means that there are no spatial relationships involved. A point on a flat surface can be viewed as a point in an area-to-room, or an area-to-land, or a room to a bedroom or land. A point on a curved surface can also be deemed part of a living room to room or a part of a room to land regards. Geometries like the ring and the hyperbola can be considered element of area-to-room relationships.
Line-to-line is definitely not a space relationship but a mathematical one particular. It can be understood to be a tangent of geometries on a single set. The geometries in this relative are the location and the edge of the area of the two lines. The space relationship for these geometries is given by the food
Geometry performs an important role in video or graphic spatial associations. It enables the understanding of the three-dimensional (3D) world and it gives all of us a basis for comprehending the correspondence between the real world as well as the virtual universe (the online world can be described as subset belonging to the real world). A good example of a visual relationship is a relationship between (A, W, C). (A, B, C) implies that the distances (D, E) are equal when measured from (A, B), and that they maximize as the values on the distances lower (D, E). Visual spatial relations may also be used to infer the parameters of an model of real life.
Another program of visual spatial relationships may be the handwriting examination. Fingerprints kept by several people have been used to infer numerous aspects of ones personality. The accuracy of such fingerprint studies has much better a lot over the past few years. The accuracy of the analyses can be improved even more by using computerized methods, specifically for the large examples. | 677.169 | 1 |
The unit circle is fundamentally related to concepts in trigonometry. The trigonometric functions can be defined in terms of the unit circle, and in doing so, the domain of these functions is extended to all real numbers.
The unit circle is also related to complex numbers. A unit circle can be graphed in the complex plane, and all roots of unity will lie on this circle.
Contents
Relation to Right Triangles
Angles in the unit circle
Coordinates in the unit circle
Special angles on the unit circle
Relation to Right Triangles
Every point on the unit circle corresponds to a right triangle with vertices at the origin and the point on the unit circle. The right triangle has leg lengths that are equal to the absolute values of the \(x\) and \(y\) coordinates, respectively.
Since the hypotenuse of the right triangle is always 1 unit long, the values of the \(x\) and \(y\) coordinates of a point on the circle are always equal to the cosine and sine (respectively) of the angle \(\theta.\)
This angle is measured in a unit called radians, which corresponds to the distance around the unit circle from the point \((1,0).\) The circumference of the unit circle is \(2\pi,\) so \(2\pi\) radians is the same as 360\(^\circ.\) Any other angle less than 360\(^\circ\) can be represented as some fraction of \(2\pi\) radians. For example, A 90\(^\circ\) angle is the same as \(\frac{1}{4}\) of the way around the circle, which would be \(\frac{2\pi}{4}=\frac{\pi}{2}.\)
Some possible values of \(\theta\) are listed below, along with their corresponding values of sine and cosine.
The trigonometry we are familiar with so far is based on only right triangles and acute angles. However, with help of Unit Circle we can extend our understanding of trigonometric functions plus also become familiar with the use of non-acute angles.
More information about the circular system of angle measurement can be found on its wiki page.
Angles in the unit circle
An angle on Unit Circle is always measured from the positive \(x\)-axis, with its vertex at the origin. It is measured to a point on the unit circle. The ray that begins at the origin and contains the point on the unit circle is called the terminal side.
An angle is said to be positive if it is measured by going in anticlockwise direction from the positive \(x\)-axis and negative if it is measured by going in clockwise direction from the \(x\)-axis.
Since \(2\pi \text{rad}=360^\circ,\) any degree measurement can be converted to radians, and vice versa.
Let \(d\) be an angle's measurement in degrees, and let \(r\) be that same angle's measurement in radians.
Coordinates in the unit circle
A right triangle \(AOB\) with right angle at \(A\) lies on the Cartesian plane such that \(\overline{OA}\) lies on the \(x\)-axis , point \(O\) lies on the origin and point \(B\) lies anywhere on the Unit Circle. Note that \(OB=1\) units.
The sine and cosine trigonometric functions are given below. When defining these functions in terms of the unit circle, it is possible to have negative lengths. If \(\overline{OA}\) is along the negative \(x\)-axis, then \(OA\) is considered to be negative. Likewise, if \(\overline{AB}\) extends below the \(x\)-axis, then \(AB\) is considered to be negative.
By this convention, the sine of an angle is considered to be the \(y\)-coordinate of a point on the unit circle given by that angle. Likeways, the cosine of an angle is considered to be the \(x\)-coordinate of a point on the unit circle given by that angle. In general, to compute the sine or cosine of any angle \(\theta,\) look at the coordinates of the point on unit circle made by that angle.
Special angles on the unit circle
Main Article: Special Angles on Unit Circle
The special angles are angles on the unit circle for which the coordinates are well-known. These coordinates can be solved for with right-triangle relationships.
\[\text{The sixteen special angles (measured in radians) on the unit circle, each labeled at the terminal point.}\]
Given that a line passes through the unit circle with the angle \(\theta=\dfrac{\pi}{4}\) find the x,y. Refer to the given diagram.
We know that \(\begin{cases} \cos(\theta)= x \\ \sin(\theta)=y\end{cases}\). put the value of theta to find\[\begin{cases} x=\dfrac{\sqrt{2}}{2}\\y=\dfrac{\sqrt{2}}{2}\end{cases}\]We are done\(_\square\)
A line passes through the unit circle at the point at \(x=\dfrac{1}{2}\). find the value of \(\tan^2(\theta)\)
first, we know that \(x^2=\cos^2(\theta)=\dfrac{1}{4}\). we also know \(y^2=\sin^2(\theta)=1-cos^2(\theta)=\dfrac{3}{4}\). using another trig identity we have \(\tan^2(\theta)=\dfrac{\sin^2(\theta)}{\cos^2(\theta)}=\boxed{3}\) we are done\(_\square\)
\[R=0\]\[R≤0\]\[R<0\]\[R>0\]
If \(R = \sin 130^\circ+\cos 130^\circ\), then which of the following is true? | 677.169 | 1 |
The NCERT Solutions for Class 10 Maths Chapter 9, "Some Applications of Trigonometry," offers comprehensive solutions for every question presented in the NCERT Textbook. Chapter 9 consists of a discussion of basic trigonometry, heights, and distance, applications of trigonometry, Trigonometry Ratios, Angle of Elevation, and Angle of Depression. Prepared and reviewed by subject matter experts, these solutions are regularly updated to align with the latest CBSE Syllabus for the academic year 2024-25 and adhere to the guidelines set by the CBSE board.
Exercise 9.1: This exercisecovers four questions related to finding angles of elevation and depression. The questions involve finding the height, distance, or both of an object using trigonometric ratios. The solutions provided aim to help students understand the practical application of trigonometry in real-life situations.
1. A circus artist is climbing a \[20m\] long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is \[\mathbf{30}{}^\circ \].
Ans: By observing the figure, \[AB\] is the pole.
In\[\Delta ABC\],
$\frac{\text{AB}}{\text{AC}}=\sin {{30}^{{}^\circ }}$
$\Rightarrow \frac{\text{AB}}{20}=\frac{1}{2}$
$\Rightarrow \text{AB}=\frac{20}{2}=10$
Therefore, the height of the pole is$10m$.
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle \[\mathbf{30}{}^\circ \]with it. The distance between the foot of the tree to the point where the top touches the ground is \[\mathbf{8m}\]. Find the height of the tree.
Ans: Let \[AC\]was the original tree. Due to the storm, it was broken into two parts. The broken part \[AB\] is making 30° with the ground.
Let $\mathrm{AC}$ be the original tree. Due to the storm, it was broken into two parts. The broken part $\mathrm{A}^{\prime} \mathrm{B}$ is making ${{30}^{{}^\circ }}$ with the ground. In triangle${{\text{A}}^{\prime }}\text{BC}$,
3. A contractor plans to install two slides for the children to play in a park. For the children below the age of $5$ years, she prefers to have a slide whose top is at a height of\[\mathbf{1}.\mathbf{5m}\], and is inclined at an angle of \[\mathbf{30}{}^\circ \] to the ground, whereas for the elder children she wants to have a steep slide at a height of\[\mathbf{3m}\], and inclined at an angle of \[60{}^\circ \] to the ground. What should be the length of the slide in each case?
Ans: It can be observed that $\text{AC}$ and $\text{PR}$ are the slides for younger and elder children respectively.
In $\vartriangle \text{ABC}$
$\Rightarrow \frac{\text{AB}}{\text{AC}}=\sin 30$
$\Rightarrow \frac{1.5}{\text{AC}}=\frac{1}{2}$
$\Rightarrow \text{AC}=3~\text{m}$
In $\vartriangle \text{PQR}$,
$\Rightarrow \frac{\text{PQ}}{\text{PR}}=\sin {{60}^{{}^\circ }}$
$\Rightarrow \frac{3}{\text{PR}}=\frac{\sqrt{3}}{2}$
$\Rightarrow \text{PR}=\frac{6}{\sqrt{3}}=2\sqrt{3}~\text{m}$
Therefore, the lengths of these slides are $3~\text{m}$ and $2\sqrt{3}~\text{m}$.
4. The angle of elevation of the top of a tower from a point on the ground, which is\[\mathbf{30m}\] away from the foot of the tower is \[\mathbf{30}{}^\circ .\] Find the height of the tower.
Ans: Let $\mathrm{AB}$ be the tower and the angle of elevation from point $\mathrm{C}$ (on ground) is $30^{\circ}$
In $\vartriangle \text{ABC}$ ,
$\Rightarrow \frac{\text{AB}}{\text{BC}}=\tan {{30}^{{}^\circ }}$
$\Rightarrow \frac{\text{AB}}{30}=\frac{1}{\sqrt{3}}$
$\Rightarrow \text{AB}=\frac{30}{\sqrt{3}}=10\sqrt{3}~\text{m}$
Therefore, the height of the tower is $10\sqrt{3}~\text{m}$.
5. A kite is flying at a height of \[\mathbf{60m}\] above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is \[\mathbf{60}{}^\circ .\] Find the length of the string, assuming that there is no slack in the string.
Ans: Let $\text{K}$ be the kite and the string is tied to point $\text{P}$ on the ground.
In $\vartriangle \text{KLP}$,
$\Rightarrow \frac{\text{KL}}{\text{KP}}=\sin {{60}^{{}^\circ }}$
$\Rightarrow \frac{60}{\text{KP}}=\frac{\sqrt{3}}{2}$
$\Rightarrow \text{KP}=\frac{120}{\sqrt{3}}=40\sqrt{3}~\text{m}$
Hence, the length of the string is $40\sqrt{3}~\text{m}$.
6. A \[\mathbf{1}.\mathbf{5m}\] tall boy is standing at some distance from a \[\mathbf{30m}\] tall building. The angle of elevation from his eyes to the top of the building increases from \[\mathbf{30}{}^\circ \] to \[\mathbf{60}{}^\circ \] as he walks towards the building. Find the distance he walked towards the building.
Ans : Let the boy was standing at point S initially. He walked towards the building and reached at point T.
7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a \[\mathbf{20m}\] high building are \[\mathbf{45}{}^\circ \] and \[\mathbf{60}{}^\circ \] respectively. Find the height of the tower.
Ans: Let \[AB\] be the statue, \[BC\] be the pedestal, and \[D\] be the point on the ground from where the elevation angles are to be measured.
8. A statue, \[\mathbf{1}.\mathbf{6m}\] tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is \[\mathbf{60}{}^\circ \] and from the same point the angle of elevation of the top of the pedestal is \[\mathbf{45}{}^\circ .\] Find the height of the pedestal.
Ans:
Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.
9. The angle of elevation of the top of a building from the foot of the tower is \[\mathbf{30}{}^\circ \] and the angle of elevation of the top of the tower from the foot of the building is \[\mathbf{60}{}^\circ .\] If the tower is \[\mathbf{50m}\] high, find the height of the building.
10. Two poles of equal heights are standing opposite each other on either side of the road, which is \[\mathbf{80m}\]wide. From a point between them on the road, the angles of elevation of the top of the poles are \[\mathbf{60}{}^\circ \]and \[\mathbf{30}{}^\circ \]respectively. Find the height of poles and the distance of the point from the poles.
Ans: Let \[AB\] and \[CD\] be the poles and \[O\] is the point from where the elevation angles are measured.
Therefore, the height of poles is $20\sqrt{3}$ and the point is $20m$ and $60m$ far from these poles.
11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is \[\mathbf{60}{}^\circ .\] From another point \[\mathbf{20m}\]away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is \[\mathbf{30}{}^\circ .\] Find the height of the tower and the width of the canal.
Therefore, the height of the tower is $10\sqrt{3}m$ and the width of the canal is $10m$.
12. From the top of a \[\mathbf{7m}\] high building, the angle of elevation of the top of a cable tower is \[\mathbf{60}{}^\circ \] and the angle of depression of its foot is \[\mathbf{45}{}^\circ .\] Determine the height of the tower.
Therefore, the height of the cable tower is $7\left( \sqrt{3}+1 \right)m$.
13. As observed from the top of a \[\mathbf{75m}\]high lighthouse from the sea-level, the angles of depression of two ships are \[\mathbf{30}{}^\circ \]and \[\mathbf{45}{}^\circ .\]If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Ans: Let $AB$be the lighthouse and the two ships be at point $C$ and \[D\]respectively.
In $\Delta \text{ABC}$,
$\Rightarrow \frac{\text{AB}}{\text{BC}}=\tan {{45}^{{}^\circ }}$
$\Rightarrow \frac{75}{\text{BC}}=1$
$\Rightarrow \text{BC=75m}$
In $\Delta \text{ABD}$,
$\Rightarrow \frac{AB}{BD}=\tan {{30}^{{}^\circ }}$
$\Rightarrow \frac{75}{BC+CD}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{75}{75+CD}=\frac{1}{\sqrt{3}}$
$\Rightarrow 75\sqrt{3}=75+CD$
$\Rightarrow 75\left( \sqrt{3}-1 \right)m=CD$
Therefore, the distance between the two ships is $75\left( \sqrt{3}-1 \right)m.$
14. A \[\mathbf{1}.\mathbf{2m}\] tall girl spots a balloon moving with the wind in a horizontal line at a height of \[\mathbf{88}.\mathbf{2m}\] from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is \[\mathbf{60}{}^\circ .\]After some time, the angle of elevation reduces to \[\mathbf{30}{}^\circ .\] Find the distance travelled by the balloon during the interval.
Ans: Let the initial position $A$ of balloon change to $B$ after some time and \[CD\] be the girl.
15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of \[\mathbf{30}{}^\circ ,\] which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be \[\mathbf{60}{}^\circ .\] Find the time taken by the car to reach the foot of the tower from this point.
Ans: Let \[AB\] be the tower. Initial position of the car is \[C\], which changes to \[D\] after six seconds.
What is Trigonometry?
Trigonometry is one of the most historical subjects studied by different scholars throughout the world. As you have read in Chapter 8 that trigonometry was introduced because its requirement arose to astronomy. Since then trigonometry is used to calculate the distance from the Earth to the stars and the planets. The most important use of trigonometry is to find out the height of the highest mountain in the world i.e. Mount Everest which is named after Sir George Everest. It is also widely used in Geography and navigation. The knowledge of trigonometry enables us to construct maps, evaluate the position of an island concerning the longitudes and latitudes.
Historical Facts. Let us Turn to the History of Trigonometry
The trigonometry was used by surveyors for centuries. One of the notable and the largest surveying projects of the nineteenth century was the " Great Trigonometric Survey" of British India for which the two largest theodolites were constructed. The highest mountain in the world was discovered during the survey in 1852. From a distance of over 160 km, this peak was seen from 6 distinct stations. This peak was named after Sir George Everest who had first used the theodolites. Theodolites are now exhibited in the museum of the surveys of Dehradun.
Height and Distance
In this topic, you will study about the line of sight, angle of elevation, horizontal level, and angle of depression. All these terms are explained in a detailed form along with some solved examples based on it. These solved examples based on the terms line of sight, angle of elevation and angle of depression will help you to understand the concepts thoroughly.
How to Calculate Height and Distance?
Trigonometric ratios are used to find out the height and the distance of the object. For example: In figure 1, you can see a boy looking at the top of the lampost. AB is considered as the horizontal level. This level is stated as the line parallel to the ground passing through the viewer's eyes. AC is considered as the line of sight. ∠A is known as the angle of elevation. Similarly, in figure 2, you can see PQ is the line of sight, PR is the horizontal level and ∠P is known as the angle of elevation.
An inclinometer or Clinometer is a device usually used for measuring the angle of elevation and the angle of depression.
Let us recall some trigonometric ratios which help to solve the questions based on class 10 maths Chapter 9.
Trigonometry Ratios
The ratio of the sides of a right-angle triangle in terms of any of its acute angle triangle is known as the trigonometric ratio of that specific angle.
In terms of ∠C, the ratio of trigonometry are given as:
Sine - The sine of an angle is stated as the ratio of the opposite side ( perpendicular side) to that angle to the hypotenuse side.
Hence, Sine C = Opposite side/Hypotenuse side
Cosine- The cosine of an angle is stated as the ratio of the adjacent side to that angle to the hypotenuse side.
Hence, Cosine C = Adjacent side /Hypotenuse side
Tangent - The tan of an angle is stated as the ratio of the opposite side (perpendicular side) to that angle to the side adjacent to that angle.
Hence, Tan C = Opposite side/Adjacent side
Cosecant- It is the reciprocal of sine.
Hence, Cosec C = Hypotenuse side/Opposite side
Secant- It is the reciprocal of cosine.
Hence, Sec C= Hypotenuse side/Adjacent side
Cotangent- It is the reciprocal of tangent.
Hence, Cot C = Adjacent side/Opposite side
The following trigonometry ratio table is used to calculate the questions based on applications of trigonometry class 10 NCERT solutions.
Trigonometric Ratio Table
∠C
0°
35°
45°
60°
90°
Sin C
0
1/2
1/2
3/2
1
Cos C
1
3/2
1/2
1/2
0
Tan C
0
1/3
1
3
Not defined
Cosec C
Not defined
2
2
2/3
1
Sec C
1
2 / 3
2
2
Not defined
Cot C
Not defined
3
1
1/3
0Important Terms to Remember in Height and Distance
Line of Sight - It is a line that is drawn from the eye of an observer to the point on the object viewed by the observer.
The Angle of Elevation - It is defined as an angle that is formed between the horizontal line and line of sight. If the line of sight lies upward from the horizontal line, then the angle formed will be termed as an angle of elevation.
Let us take another situation when a boy is standing on the ground and he is looking at the object from the top of the building. The line joining the eye of the man with the top of the building is known as the line of sight and the angle drawn by the line of sight with the horizontal line is known as angle of elevation.
The Angle of Depression - It is defined as an angle drawn between the horizontal line and line of sight. If the line of sight lies downward from the horizontal line, then the angle formed will be termed as an angle of depression.
Let us take a situation when a boy is standing at some height concerning the object he is looking at. In this case, the line joining the eye of the man with the bottom of the building is known as the line of sight and the angle drawn by the line of sight with the horizontal line is known as angle of depression.
Note: Angle of elevation is always equal to the angle of depression
The Important Point to Remember
The distance of the object is also considered as the base of the right angle triangle drawn through the height of the object and the line of sight.
The length of the horizontal level is also known as the distance of the object it forms the base of the triangle.
Line of sight is considered as a hypotenuse of the right-angle triangle. Hypotenuse side is calculated using Pythagorean Theorem if the height and distance of the object are given.The important topic " Height and Distance" covered in Some applications of trigonometry class 10 is followed by one exercise with 16 questions. The exercise aims to test your knowledge and how deeply you understood each formula and concept of the topic. The numerical questions given in this chapter are based on some applications of trigonometry.
To make you understand the topic and related concept, solved numerical problems are also given. Stepwise solutions are given for each of the solved examples. It will help you to understand which concept and formula will be used to solve the given questions accurately.
Overview of Deleted Syllabus for CBSE Chapter 9 Maths Class 10
Chapter
Dropped Topics
Some Applications of Trigonometry
9.1 Introduction
Class 10 Maths Chapter 9: Exercises Breakdown
Exercise
Number of Questions
Exercise 9.1
Exercise 9.1(15 Questions & Solutions)
Conclusion
Chapter 9 of Class 10 Maths, "Some Applications of Trigonometry," delves into practical applications of trigonometric concepts in real-life scenarios. Through this chapter, students explore various applications such as heights and distances, navigation, and more. By mastering these applications, students not only enhance their understanding of trigonometry but also recognize its significance in solving everyday problems. Therefore, practicing a variety of problems from NCERT Solutions and previous year papers can enhance preparation and confidence for exams. By mastering Introduction to Trigonometry, students not only excel in mathematics but also build skills that are useful in various real-world scenarios.
1. How downloading "NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry" PDF from Vedantu can help you to score better grades in exams?
Vedantu executed well-designed and reviewed applications of trigonometry class 10 ncert solutions which include the latest guidelines as recommended by CBSE board. The solutions are easily accessible from the Vedantiu online learning portal. It can be easily navigated as Vedantu retains a user-friendly network. The detailed solution of all the questions asked in the NCERT book will help you to understand the basic concepts of some applications of trigonometry class 10 thoroughly. Along with the NCERT solutions, the experts of Vedantu also provide various tips and techniques to solve the questions in the exam. Sign in to Vedantu and get access to the ncert solutions for class 10 maths chapter 9 some applications of trigonometry -Free pdf and you can download the updated solution of all the exercises given in the chapter.
2. What is the importance of trigonometry applications in real-life?
The application of trigonometry may not be directly used in solving practical issues but used in distinct fields. For example, trigonometry is used in developing computer music as you must be aware of the fact that sound travels in the form of waves and this wave pattern using sine and cosine functions helps to develop computer music. The following are some of the applications where the concepts of trigonometry and its functions are applicable.
It is used to measure the height and distance of a building or a mountain
It is used in the Aviation sector
It is used in criminology
It is used in Navigation
It is used in creating maps
It is used in satellite systems
The basic trigonometric functions such as sine and cosine are used to determine the sound and light waves.
It is used in oceanography to formulate the height of waves and tides in the ocean.
3. The distance from where the building can be viewed is 90ft from its base and the angle of elevation to the top of the building is 35°. Calculate the height of the building.
Given: Distance from where the building can be viewed is 90ft from its base and angle of elevation to the top of the building is stated as 35°
To calculate the height of the building, we will use the following trigonometry formula
Tan 35° = Opposite Side/ Adjacent Side
Tan 35° = H/90
H = 90 x Tan 35°
H = 90 x 0.7002
H = 63.018 feet
Hence, the height of the building is 63.018 feet.
4. From a 60 meter high tower, the angle of depression of the top and bottom of a house are 'a and b' respectively. Calculate the value of x, if the height of the house is [60 sin (β − α)]/ x.
H = d tan β and H- h = d tan α
60/60-h = tan β – tan α
h = [60 tan α – 60 tan β] / [tan β]
h= [60 sin (β – α)/ [(cos α cos β)] [(sin α sin β)]
x = cos α cos β
5. The top of the two different towers of height x and y, standing on the ground level subtended angle of 30° and 60° respectively at the center of the line joining their feet. Calculate x: y.
In ΔABE,
x/a = Tan 30°
x/a = 1/3
x = a/3
In ΔCDE
y/a = Tan 60°
y/a = 3 → y = a x 3
x/a ÷ y/a = a/3÷ a3
x/a × y/a = a/3 × 1/a×3
x/y = 1/3
6. What are the main points to study in Class 10 Maths NCERT Chapter 9?
The main points to study in Class 10 Maths NCERT Chapter 9 include:
Fundamental Basics Of Trigonometry Applications
The History Behind Trigonometry
Concepts On Height And Distance
The Study About The Line Of Sight
Angle Of Elevation, Horizontal Level
Angle Of Depression
The Calculation Methods For Heights And Distance
The Trigonometry Ratios And Angle Tables For The Ratios
A clear explanation for the solutions of the same can be found on the Vedantu website.
7. How many exercises are there in Chapter 9 of Class 10 Maths?
Chapter 9 of Class 10 Maths consists of one exercise that is Exercise 9.1 at the end of this chapter. This section of the chapter has 16 questions that are covered in the first two parts. The questions asked in the exercise are based on the basic concepts of trigonometry and its application, height distance etc. Following this, is a summary of the chapter.
8. How can I score full marks in Class 10 Chapter 9 Maths?
Maths is all about practice. If you practice your exercises regularly you will gradually lean towards perfection and accuracy. It will also increase your speed and hence you will be able to finish your paper on time, attempting all the questions, thereby scoring full marks in this subject. For this, you can refer to the Vedantu website or download the Vedantu app which will provide you with the best solutions at free of cost and help you fulfil your aim of scoring top grades in Maths.
9. Where can I get the best solutions for Maths chapter 9?
You can find the best solutions for NCERT Class 10 Maths Chapter 9 on Vedantu. Here, you will find the best possible trigonometry explanations and their solutions for respective exercises. These are provided in PDF format, which is totally free to download and you can study from it even in offline mode. For this:
Visit the page NCERT Solutions for Class 10 Maths Chapter 9.
Click on the download PDF option.
Once you're redirected to a page you will be able to download it.
10. What is the advantage of using Vedantu?
Vedantu outshines all other learning apps and websites because it has the best team of experts who analyse and form the best answers for the students to learn and practice from. This has a complete chapter-wise explanation and their respective solutions. This also has options for parents to take weekly homely tests from Vedantu so that they can keep a check on their wards. | 677.169 | 1 |
If two sides of a triangle are of lengths 5cm and 1.5cm, then the length of third side of the triangle cannot be
A
3.6cm
B
4.1cm
C
3.8cm
D
3.4cm
Views: 5,194 students
Updated on: Aug 13, 2023
Text solutionVerified
According to the triangle inequality, "the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side". Therefore, the length of the third side cannot be greater than 3.4 cm. Hence, the correct answer is 3.4 cm.
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If two sides of a triangle are of lengths 5cm and 1.5cm, then the length of third side of the triangle cannot be | 677.169 | 1 |
Let the position vectors of the vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a triangle be $2 \hat{\imath}+2 \hat{\jmath}+\hat{k}, \hat{\imath}+2 \hat{\jmath}+2 \hat{k}$ and $2 \hat{\imath}+\hat{\jmath}+2 \hat{k}$ respectively. Let $\ell_1, \ell_2$ and $\ell_3$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ respectively, then $\ell_1^2+\ell_2^2+\ell_3^2$ equals: | 677.169 | 1 |
Syntax
Description
angle = readRotationAngle(mygyrosensor) reads
the total amount of rotation since the creation of the connection
to the sensor, and returns the measurement in degrees. You can use
the resetRotationAngle function to reset this
value to zero. | 677.169 | 1 |
Math
Humanities
... and beyond
A pilot observes the measure of the angle of depression of a marker to be 36 degrees. the plane is 2000m above the ground. How Far from the marker is the point on the ground directly beneath the plane? | 677.169 | 1 |
A Side View of the Curvature of the Earth at Lake Pontchartrain
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Senior Member.ModeratorAgreed. It is side-to-side curvature, which occurs because the centre towers are closer to the viewer than the towers at the side, so of course they are less far over the curve (which is present in all directions). You can't really separate out the curvature into "front to back" and "left to right": a sphere curves equally in all directions.
Closed Account
SPOILER ALERT: This will seem totally mental to all but amateur geometry geeks who like pointless puzzles (and maybe them too).
So I've been playing some more with the idea of left-right curves, etc, and seeing what happens if we 'isolate' the 'different curves' by making the earth a cylinder. First off, there's the left-to-right cylinder, like this:
And then the head-on cylinder, like this:
In the attached spreadsheet you can generate what the curve of a line of towers would look like for these two cylinders, as well as the sphere and the flat earth, for a variable viewer height and variable height of towers. Like, for a viewer at 100 feet looking at a line of towers 100 feet high - nearest tower 4 miles away; furthest tower 8 miles from that, perpendicular to the viewer - the curves of the 4 different models look like this:
What's interesting about this is that for the first couple of miles the sphere curve and the head-on cylinder curve are similar, as are the left-to-right cylinder and the flat earth, whereas by the end they've flipped, as can be more clearly visualised here:
(Top lines tops of towers; bottom lines the bases.)
And making the line of towers 25 miles long changes the picture again:
This illustrates that the left-to-right cylinder is closest to the sphere earth, in the overall shape of the curve - though I'm not really sure what it illustrates about our ideas that we can see 'different curves' here on the actual earth.
Probably that we were wrong all along: there's just CURVE, and theNote: all this is assuming I've got the geometry more or less correct, which I'm not quite 100% sure on. For one thing, there's a weird kink in the curve between the 1st and 2nd points that I can't figure out. For another, I was expecting the flat earth tower bases to form a straight line, given that, as far as I can tell, it's only perspective playing a part in that one.
Seem pretty much right though. Perhaps some other puzzle-lovin' freak can iron out any creases in the 'quations.
Senior Member
Closed Account16 miles is arbitary. It's an 8-mile long line of towers and I chose to plot 50 points in total to give it a smooth curve.
Senior Member4 miles is arbitary. It's an 8-mile long line of towers and I chose 0.4 miles to give 20 points in total and a smooth curve.
Closed Account
You could say that. Points are calculated every 1/50th of the total length of the line of towers and the curve is generated from that. So, in essence, it's a line of 51 towers - but you'd also be able to see where towers in between those towers would be too. If that makes sense.
New Member
The way I see it, for once the true answer is perspective, when looking at an unmarked surface, it is really hard to tell if it is flat or curved, but add a visual cue (power towers for example) and combine it with the effect of perspective (compressing the image along the line of sight) and you can notice that the surface of the water is curved.
It is only through the effect of perspective (objects becoming smaller the further they are) do we notice the curve: the lines converging along a curve rather than a straight line, it is true on a sphere the curve is equal in all directions but because of the way our eyes or cameras work, convergence happens only along the line of sight.
check this out:
The rails are converging but the sides of the crossing are parallel.
As for the left-to-right direction, it is unrealistic to expect to discern curvature from low altitudes on this axis because perspective is not a factor here, Flat-earthers have the expectation that side-to-side curvature can be calculated with the 8 inches per mile2 formula, but this formula is for calculating the drop (in feet or meters) due to curvature in the z-axis where the curve is curving vertically down and away from the viewer.
We cannot use the same formula to calculate the curve of the horizon (in radians or degrees) on the x-axis as it curves around the viewer.
Closed Account
ThePerhaps that's technically true, to a point, but something I've learned recently is that some people differentiate between 'the curve of the earth' (aka, 'geodetic curve') and 'the curve of the horizon'.
This came up during a discussion about Neil deGrasse Tyson famously saying that Felix Baumgartner wouldn't have seen the earth's curvature from 128,000 feet, when obviously he would have seen a fairly pronounced 'curve', given that it can be seen from aeroplanes, and photographed from much lower elevations.
His statement was explained by saying he was referring to the geodetic curve. Though since the curve of the horizon exists because of the curve of the earth, I don't really see a reason to separate the two.
Funny old thread, this one. It made perfect sense at the time, but reading it back I can barely make head nor tail of it: a good lesson in realising that things that seem clear may not seem clear to everyone (e.g., clear to past me, and bemusingly perplexing to present me). | 677.169 | 1 |
Explanation:
Yes, there is a way to name \(\overline{J K}\) by switching the letters, \(\overline{K J}\) and when we are naming a line segment, either point can be named first.
Angles
You can name an angle by the vertex. When you name an angle using 3 points, the vertex is always the point in the middle.
Answer:
We cannot say ∠B here because that could be one of three angles. So we have to use either ∠ABD or ∠ABC, or ∠DBC. We need to remember when we name an angle using 3 points, the vertex is always the point in the middle.
Explanation:
Angles are classified by the size of the opening between the rays.
Activity 2 Classify an angle.
Materials
paper
To classify an angle, you can compare it to a right angle.
Make a right angle by using a sheet of paper. Fold the paper twice evenly to model a right angle. Use the right angle to classify the angles below. Write acute, obtuse, right, or straight.
a.
Answer:
Obtuse
Explanation:
An angle that measures greater than 90° degrees is called an obtuse angle.
b.
Answer:
Acute
Explanation:
An angle that measures less than 90° degrees is called an acute angle.
c.
Answer:
Acute
Explanation:
An angle that measures less than 90° degrees is called an acute angle.
Explanation:
By using the figure given above the name of the right angle is ∠TUW.
Problem Solving
Lise the picture of the bridge for 7 and 8.
Question 7.
Classify ∠A.
Answer:
Right
Question 8.
Which angle appears to be obtuse?
Answer:
∠C
Explanation:
The angle C appears to be obtuse.
Question 9. H.O.T. Use Diagrams How many different angles are in Figure X? List them.
Answer:
There are 10 different angles.
Explanation:
The different angles that are in Figure X are ∠ABD, ∠DBE, ∠EBF, ∠FBC, ∠ABE, ∠DBF, ∠EBC, ∠ABF, ∠DBC, ∠ABC.
Question 10. H.O.T. Multi-Step What's the Error? Vanessa drew the angle at the right and named it ∠ TRS. Explain why Vanessa's name for the angle is incorrect. Write a correct name for the angle.
Answer:
∠TSR or ∠RST
Explanation:
The vertex should be the middle letter in the angle's name. The correct name for the angle is ∠TSR or ∠RST. | 677.169 | 1 |
Elementary Geometry: Practical and Theoretical
Ex. 1697. On bases of 5 in. and 3 in. describe two similar triangles; calculate their areas, and find the ratio of their areas. Is it 5: 3?
▲ ABD = ||ogram ABCD, and ▲A'B'D' = |ogram A'B'C'D'.
The parallelograms ABCD, A'B'C'D' are divided up into congruent parallelograms; the squares are divided up into congruent
Ex. 1698. The ratio of corresponding altitudes of similar triangles
is equal to the ratio of corresponding sides.
THEOREM 6.
The ratio of the areas of similar triangles is equal to the ratio of the squares on corresponding sides.
Ex. 1699. What is the ratio of the areas of two similar triangles on bases of 3 in. and 4 in.?
Ex. 1700. The area of a triangle with a base of 12 cm. is 60 sq. cm. ; find the area of a similar triangle with a base of 9 cm.
What is the area of a similar triangle on a base of 9 in.?
Ex. 1701. The areas of two similar triangles are 100 sq. cm. and 64 sq. cm.; the base of the greater is 7 cm.; find the base of the smaller.
and
Ex. 1702. The areas of two similar triangles are 97.5 sq. cm. 75.3 sq. cm.; the base of the first is 17·2 cm.; find the base of the second.
Ex. 1703. The sides of a triangle ABC are 7·2 in., 3·5 in., 5·7 in.; the sides of a triangle DEF are 7·2 cm., 3·5 cm., 5·7 cm.; find the ratio of the area of the first triangle to that of the second.
Ex. 1704. Find the ratio of the bases of two similar triangles one of which has double the area of the other.
Draw two such triangles taking 1 in. as the base of the smaller.
Ex. 1705. Describe equilateral triangles on the side and diagonal of a square; find the ratio of their areas.
Ex. 1706. Draw a straight line parallel to the base of a triangle to bisect the triangle.
Ex. 1707. Describe equilateral triangles on the sides of a right-angled triangle whose sides are 1.5 in., 2 in., 2.5 in. What connection is there between the areas of the three equilateral triangles?
Ex. 1708. Prove that, if similar triangles are described on the three sides of a right-angled triangle, the area of the triangle described on the hypotenuse is equal to the sum of the other two triangles.
Ex. 1709. ABC, DEF are two triangles in which B=LE; prove that ▲ ABC: A DEF AB. BC: DE. EF.
[Draw AX 1 to BC, and DY to EF.]
Ex. 1710. What is the ratio of the areas of two circles whose radii are R, r? 3 in., 2 in.?
Ex. 1711. Draw two similar quadrilaterals ABCD, PQRS; calculate their areas (join AC, PR); find the ratio of their areas, and compare this with the ratio of corresponding sides.
RECTANGLE PROPERTIES.
Ex. 1712. XYZ is a triangle inscribed in a circle, XN is an altitude of
XY YD
=
XN NZ'
the triangle and XD a diameter of the circle; prove that rectangle property can be obtained from this by clearing of fractions?
Ex. 1713. With the same construction as in Ex. 1712, prove that
XZ. NY=XN.ZD.
What
[You will have to pick out two equal ratios from two equiangular triangles. If you colour XZ, NY red and XN, ZD blue you will see which are the triangles.]
Ex. 1714.
ABCD is a quadrilateral inscribed in a circle; its diagonals intersect at X. Prove that (i) AX. BC=AD. BX, (ii) AX. XC=BX. XD.
Ex. 1715. ABCD is a quadrilateral inscribed in a circle; AB, DC produced intersect at Y. Prove that
(i) YA. BD=YD. CA, (ii) YA. YB=YC. YD.
Ex. 1716. The rectangle contained by two sides of a triangle is equal to the rectangle contained by the diameter of the circumcircle and the altitude drawn to the base.
[Draw the diameter through the vertex at which the two sides intersect.]
Ex. 1717. The bisector of the angle A of ▲ ABC meets the base in P and the circumcircle in Q. Prove that the rectangle contained by the sides AB, AC rect. AP. AQ.
Ex. 1718. In Ex. 1680, prove that PQ. SR=PR.TQ.
Ex. 1719. The sum of the rectangles contained by opposite sides of a cyclic quadrilateral is equal to the rectangle contained by its diagonals. (Ptolemy's theorem.)
[Use the construction of Ex. 1680.]
....
Take
Ex. 1720. Draw a circle of radius 7 cm.; mark a point P 3 cm. from the centre O; through P draw five or six chords APB, CPD, Measure their segments and calculate the products PA. PB; PC. PD; . the mean of your results and estimate by how much per cent. each result differs from the mean. (Make a table.)
Ex. 1721. Draw a circle of radius 7 cm. and mark a point P 10 cm. from the centre O; through P draw a number of chords of the circle, and proceed as in Ex. 1720.
[Remember that if P is in the chord AB produced, PA, PB are still regarded as the segments into which P divides AB; you must calculate PA. PB, not PA. AB.]
Ex. 1722. What will be the position of the chord in Ex. 1721 when the two segments are equal? | 677.169 | 1 |
Identify Right Triangles
Drag and drop the triangle or triangles with a right angle to the box.
Don't forget to click the blue 'check' button to check your answer before moving onto the next slide.
Use the right arrow to navigate to Remember that a right angle is an angle that measures ninety degrees. So to be a right triangle, it must have a right angle. Can a triangle have more than one right angle? Why or why not? | 677.169 | 1 |
...the angle f. if. i. DTY is equal f to the angle GTS. therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, aid one fide equal to one fide, oppolite to two of the equal angles-, viz. DY to GS ; for they are...
...and the angle l)TY is equal f to the angle G PS : Therefore in the triangles D l"Y, GTS tlv;re are two angles in the one equal to two angles in the other, and one lide equal to one fide, oppofite to two of the equal angles, viz.. DY to GS ; for they are...
...equal c ; and the angle DTY is equal t to the angle GTS : Therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, and one fide equal to one^fide, oppofite to two of the equal angles, viz. DY to GSj for they arc the...
...vertical angles AED, CEB are equal a, and alfo the alternate angles EAD, ECB b, the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each : but the fides AD arid BC, which are opppfite tp equal angles in thefe trier 34. i. angles,...
...equal to GDC, QpM is equal to GDB but GDB is equal to MDq; in the two triangles QJD M, MD q, there are two angles in the one equal to two angles in the other, and one fide MD common to both, therefore QJM will be equal to My. The fame may be proved alfo of the...
...equal c ; and the angle DTY is equal f to the angle GTS. therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, and one fide equal to one fide, oppofite to two of the equal angles, viz. DY to GS; for they are the l,th. 17), namely, the angle A equal to the angle c, which 1, th. 17), namely, the angle A equal to the angle c, which...
...ADB and С В D are equal. Therefore the triangles BD С and DBA are identical, (Theo. 2.) ; having two angles in the one equal to two angles in the other, and the side BD between the equal angles, common to both; the triangle ABD is therefore equal to the...
...and the angle DTY, is equal (IS. 1.) to the angle GTS : Therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, and one side equal to one side, opposite to two of the equal angles, viz. DY to GS ; for they are the... | 677.169 | 1 |
Geometry Lesson 7.3.5 Circumference and Wheels
Instructor Materials
Learning goal: Compare wheels of different sizes and explain (orally) why a larger wheel needs fewer rotations to travel the same distance. Generalize that the distance a wheel rolls in one rotation is equal to the circumference of the wheel. Write an equation to represent the proportional relationship between the number of rotations and the distance a wheel travels | 677.169 | 1 |
Trigonometry in Astronomy
Feeling:
Dumb
Language:
Arabic
Prompt:
Trigonometry in Astronomy
Trigonometry in astronomy is used to calculate angles and distances between celestial objects. By using trigonometric functions such as sine, cosine, and tangent, astronomers can determine the positions of stars, planets, and other objects in the sky.
One example of how trigonometry is used in astronomy is in calculating the distance between Earth and a distant star. By measuring the angle between the star and the horizon from two different locations on Earth, astronomers can use trigonometry to calculate the distance to the star.
Another example is the calculation of the size and distance of planets in our solar system. By measuring the angle between the planet, the Sun, and Earth, astronomers can use trigonometry to determine the planet's size and distance from the Sun.
One famous application of trigonometry in astronomy is the calculation of the distance to the Moon using the parallax method. By measuring the angle between the Moon and a distant star from two different locations on Earth, astronomers can use trigonometry to calculate the distance to the Moon.
One verifiable fact is that trigonometry has been used in astronomy for centuries, dating back to ancient civilizations such as the Greeks and Babylonians. These early astronomers used trigonometry to make accurate predictions about the movements of celestial objects. Today, trigonometry continues to play a crucial role in modern astronomy, helping astronomers to better understand the universe and its many mysteries | 677.169 | 1 |
Elementary Geometry for College Students (6th Edition)
by
Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter 7 - Section 7.1 - Locus of Points - Exercises - Page 317: 23
Answer
We can see a sketch of the parabola below.
Work Step by Step
We can find these five points such that the distance between each point and the focus is equal to the horizontal distance of the point to the left of the directrix.
We can see a sketch of the parabola below. | 677.169 | 1 |
Related Angles – Interior, Exterior, Corresponding Angles
Related angles are nothing but the pairs of angles and assigned with specific names that we come across. Related angles have some conditions to mention. Learn the detailed concept of Related angles with images and examples in this article. Improve your preparation level by reading the entire concept without missing any subtopic. We have given complete information about Lines and Angles on our website for free of cost.
Different Types of Related Angles
Check different types of Related Angles along with examples to clearly understand the concept. We have explained each of them with definitions, solved examples, etc. Refer to the following modules and get a grip on it.
Complementary Angles
If the sum of the measures of two angles is about 90°, those angles are called complementary angles.
Facts of Complementary Angles
The two right angles never complement each other.
Also, the two obtuse angles never complement each other.
Two complementary angles are always acute but there is no possibility of vice versa.
Example:
Let us take two angles which are complementary angles. If one angle is a, then the other angle is 90° – a.
An angle of 40° and another angle of 50° are complementary angles of each other.
The ∠XOY is 40° and ∠MON is 50°. By adding two angles ∠XOY and ∠MON, we get 90° as they are the Complementary Angles.
Therefore, the sum of the ∠XOY and ∠MON is 90°
∠XOY + ∠MON = 90°.
Supplementary Angles
Supplementary Angles are the angles when the sum of the measures of two angles is 180°. If the sum of the two angles forms a straight angle, then those angles are called Supplementary Angles. If one angle is a, then the other angle is 180° – a.
Example:
Let us take two angles which are Supplementary angles. If one angle is x, then the other angle is 180° – x.
An angle of 110° and another angle of 70° are supplementary angles of each other. Also, a supplement of 110° is 180° – 110° = 70°.
And the supplement of 60° is 180° – 70° = 110°.
From the above figure, the ∠XOY is 110° and ∠MON is 70°. By adding two angles ∠XOY and ∠MON, we get 180° as they are the Supplementary Angles.
Therefore, the sum of the ∠XOY and ∠MON is 180°
∠XOY + ∠MON = 180°.
Adjacent Angles
The two angles are called to be Adjacent angles when they have a common arm, a common vertex, and also the non-common arms present on the opposite side of the common arm.
From the above figure, ∠ABD and ∠CBD are adjacent angles with the common arm BD. The B is the common vertex and BA, BC is opposite sides of BD.
Linear Pair
When two adjacent angles form a linear pair of angles with the non-common arms are two opposite rays. In other words, the sum of two adjacent angles is 180°.
From the above figure, the ∠XOY and ∠XOZ are two adjacent angles. By adding two angles ∠XOY and ∠XOZ, we get 180°.
Therefore, the sum of the ∠XOY and ∠XOZ is 180°
∠XOY + ∠XOZ = 180°.
Vertically Opposite Angles
The arms of the lines are opposite in direction and both lines are interesting to each other in Vertically opposite angles. The pair of vertically opposite angles are equal.
From the above figure, the ∠MOR and ∠SON and ∠MOS and ∠RON are pairs of vertically opposite angles.
Theorems on Related Angles
1. If a ray stands on a line, then the sum of adjacent angles formed is 180°.
Given: A ray BE standing on (AC) ⃡ such that ∠ABE and ∠CBE are formed. | 677.169 | 1 |
Why is the circumference of a circle 360 degrees instead of 100 degrees or 200 degrees?
There are many views on the origin of 360 degrees, including two main views.
1. Related to the ephemeris
In ancient times, people used the Sun, Moon, stars and other natural phenomena to measure time and create calendars.
The Sumerians observed the Sun, Moon and five visible planets (Mercury, Venus, Mars, Jupiter and Saturn), mainly for omens.
The number 360 was probably chosen because it was the number of days in the year for ancient people.
2. Related to base 60
In the early days, people often used their fingers to count, so the number base was 10, called the decimal system.
But in ancient times, some civilizations used 60 as the base, called the hexadecimal system (Base 60 numbering system).
The reason why 60 is used instead of other numbers may be because 60 is a relatively small number and can be divided into 10 real factors of 2, 3, 4, 5, 6, 10, 12, 15
In addition, using base 60 is also convenient for drawing with rulers and rulers.
As Greek geometry developed, it created the concept of angles and degrees.
The base 60 counting method was later spread to Europe by the Arabs, eventually becoming the mainstream method, currently widely used in time, angles, geographical coordinates and other fields.
Around 300 to 100 BC, the Babylonians broke down hours into base-60 fractions: 60 minutes in an hour and 60 seconds in a minute.
The reason why the circle's circumference is 360 degrees instead of 60 degrees is probably because the ancients discovered that there are exactly 6 equilateral triangles with side lengths being the radius of the circle and the angles of the equilateral triangle are also 60 degrees.
One degree is divided into 60 minutes, and one minute into 60 seconds.
Therefore, using 360 degrees to represent an angle is more convenient than 100, because it can be divided into many different angles and these angles are all integers, making calculations more convenient and reducing the possibility of errors.
In short, the origin of 360 degrees as an angular unit can be traced back to ancient civilizations, it has advantages and meanings in many aspects such as history, culture, technology and mathematics.
Although the number 100 is simpler, it does not meet these requirements well, so 360 degrees has become a commonly used unit of angle measurement. | 677.169 | 1 |
Triangles: Solving Right The learner will be able to
find the measure of each unknown side (to the nearest tenth or hundredth) or angle (to the nearest degree), solve real world trigonometry problems, and include the angle of depression or elevation in calculations.
Trig. Ratios: Identify Specific Values The learner will be able to
identify the trigonometric values of sine, cosine, and tangent for the following angles and their multiples: 0, 30, 45, 60, 90, and 180.
Question: What is the angle of depression or elevation in a real world trigonometry problem? Answer: The angle between the horizontal and a line of sight to an object. | 677.169 | 1 |
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