• 9.3 Use Properties of Angles, Triangles, and the Pythagorean Theorem
  • Introduction
  • 1.1 Introduction to Whole Numbers
  • 1.2 Add Whole Numbers
  • 1.3 Subtract Whole Numbers
  • 1.4 Multiply Whole Numbers
  • 1.5 Divide Whole Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to the Language of Algebra
  • 2.1 Use the Language of Algebra
  • 2.2 Evaluate, Simplify, and Translate Expressions
  • 2.3 Solving Equations Using the Subtraction and Addition Properties of Equality
  • 2.4 Find Multiples and Factors
  • 2.5 Prime Factorization and the Least Common Multiple
  • Introduction to Integers
  • 3.1 Introduction to Integers
  • 3.2 Add Integers
  • 3.3 Subtract Integers
  • 3.4 Multiply and Divide Integers
  • 3.5 Solve Equations Using Integers; The Division Property of Equality
  • Introduction to Fractions
  • 4.1 Visualize Fractions
  • 4.2 Multiply and Divide Fractions
  • 4.3 Multiply and Divide Mixed Numbers and Complex Fractions
  • 4.4 Add and Subtract Fractions with Common Denominators
  • 4.5 Add and Subtract Fractions with Different Denominators
  • 4.6 Add and Subtract Mixed Numbers
  • 4.7 Solve Equations with Fractions
  • Introduction to Decimals
  • 5.1 Decimals
  • 5.2 Decimal Operations
  • 5.3 Decimals and Fractions
  • 5.4 Solve Equations with Decimals
  • 5.5 Averages and Probability
  • 5.6 Ratios and Rate
  • 5.7 Simplify and Use Square Roots
  • Introduction to Percents
  • 6.1 Understand Percent
  • 6.2 Solve General Applications of Percent
  • 6.3 Solve Sales Tax, Commission, and Discount Applications
  • 6.4 Solve Simple Interest Applications
  • 6.5 Solve Proportions and their Applications
  • Introduction to the Properties of Real Numbers
  • 7.1 Rational and Irrational Numbers
  • 7.2 Commutative and Associative Properties
  • 7.3 Distributive Property
  • 7.4 Properties of Identity, Inverses, and Zero
  • 7.5 Systems of Measurement
  • Introduction to Solving Linear Equations
  • 8.1 Solve Equations Using the Subtraction and Addition Properties of Equality
  • 8.2 Solve Equations Using the Division and Multiplication Properties of Equality
  • 8.3 Solve Equations with Variables and Constants on Both Sides
  • 8.4 Solve Equations with Fraction or Decimal Coefficients
  • 9.1 Use a Problem Solving Strategy
  • 9.2 Solve Money Applications
  • 9.4 Use Properties of Rectangles, Triangles, and Trapezoids
  • 9.5 Solve Geometry Applications: Circles and Irregular Figures
  • 9.6 Solve Geometry Applications: Volume and Surface Area
  • 9.7 Solve a Formula for a Specific Variable
  • Introduction to Polynomials
  • 10.1 Add and Subtract Polynomials
  • 10.2 Use Multiplication Properties of Exponents
  • 10.3 Multiply Polynomials
  • 10.4 Divide Monomials
  • 10.5 Integer Exponents and Scientific Notation
  • 10.6 Introduction to Factoring Polynomials
  • 11.1 Use the Rectangular Coordinate System
  • 11.2 Graphing Linear Equations
  • 11.3 Graphing with Intercepts
  • 11.4 Understand Slope of a Line
  • A | Cumulative Review
  • B | Powers and Roots Tables
  • C | Geometric Formulas

Learning Objectives

By the end of this section, you will be able to:

  • Use the properties of angles
  • Use the properties of triangles

Use the Pythagorean Theorem

Be Prepared 9.7

Before you get started, take this readiness quiz.

Solve: x + 3 + 6 = 11 . x + 3 + 6 = 11 . If you missed this problem, review Example 8.6 .

Be Prepared 9.8

Solve: a 45 = 4 3 . a 45 = 4 3 . If you missed this problem, review Example 6.42 .

Be Prepared 9.9

Simplify: 36 + 64 . 36 + 64 . If you missed this problem, review Example 5.72 .

So far in this chapter, we have focused on solving word problems, which are similar to many real-world applications of algebra. In the next few sections, we will apply our problem-solving strategies to some common geometry problems.

Use the Properties of Angles

Are you familiar with the phrase ‘do a 180 ’? 180 ’? It means to turn so that you face the opposite direction. It comes from the fact that the measure of an angle that makes a straight line is 180 180 degrees. See Figure 9.5 .

An angle is formed by two rays that share a common endpoint. Each ray is called a side of the angle and the common endpoint is called the vertex . An angle is named by its vertex. In Figure 9.6 , ∠ A ∠ A is the angle with vertex at point A . A . The measure of ∠ A ∠ A is written m ∠ A . m ∠ A .

We measure angles in degrees, and use the symbol ° ° to represent degrees. We use the abbreviation m m for the measure of an angle. So if ∠ A ∠ A is 27° , 27° , we would write m ∠ A = 27 . m ∠ A = 27 .

If the sum of the measures of two angles is 180° , 180° , then they are called supplementary angles . In Figure 9.7 , each pair of angles is supplementary because their measures add to 180° . 180° . Each angle is the supplement of the other.

If the sum of the measures of two angles is 90° , 90° , then the angles are complementary angles . In Figure 9.8 , each pair of angles is complementary, because their measures add to 90° . 90° . Each angle is the complement of the other.

Supplementary and Complementary Angles

If the sum of the measures of two angles is 180° , 180° , then the angles are supplementary.

If ∠ A ∠ A and ∠ B ∠ B are supplementary, then m ∠ A + m ∠ B = 180°. m ∠ A + m ∠ B = 180°.

If the sum of the measures of two angles is 90° , 90° , then the angles are complementary.

If ∠ A ∠ A and ∠ B ∠ B are complementary, then m ∠ A + m ∠ B = 90°. m ∠ A + m ∠ B = 90°.

In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications.

Use a Problem Solving Strategy for Geometry Applications.

  • Step 1. Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
  • Step 2. Identify what you are looking for.
  • Step 3. Name what you are looking for and choose a variable to represent it.
  • Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  • Step 5. Solve the equation using good algebra techniques.
  • Step 6. Check the answer in the problem and make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

The next example will show how you can use the Problem Solving Strategy for Geometry Applications to answer questions about supplementary and complementary angles.

Example 9.16

An angle measures 40° . 40° . Find ⓐ its supplement, and ⓑ its complement.

Try It 9.31

An angle measures 25° . 25° . Find its: ⓐ supplement ⓑ complement.

Try It 9.32

An angle measures 77° . 77° . Find its: ⓐ supplement ⓑ complement.

Did you notice that the words complementary and supplementary are in alphabetical order just like 90 90 and 180 180 are in numerical order?

Example 9.17

Two angles are supplementary. The larger angle is 30° 30° more than the smaller angle. Find the measure of both angles.

Try It 9.33

Two angles are supplementary. The larger angle is 100° 100° more than the smaller angle. Find the measures of both angles.

Try It 9.34

Two angles are complementary. The larger angle is 40° 40° more than the smaller angle. Find the measures of both angles.

Use the Properties of Triangles

What do you already know about triangles? Triangle have three sides and three angles. Triangles are named by their vertices. The triangle in Figure 9.9 is called Δ A B C , Δ A B C , read ‘triangle ABC ABC ’. We label each side with a lower case letter to match the upper case letter of the opposite vertex.

The three angles of a triangle are related in a special way. The sum of their measures is 180° . 180° .

Sum of the Measures of the Angles of a Triangle

For any Δ A B C , Δ A B C , the sum of the measures of the angles is 180° . 180° .

Example 9.18

The measures of two angles of a triangle are 55° 55° and 82° . 82° . Find the measure of the third angle.

Try It 9.35

The measures of two angles of a triangle are 31° 31° and 128° . 128° . Find the measure of the third angle.

Try It 9.36

A triangle has angles of 49° 49° and 75° . 75° . Find the measure of the third angle.

Right Triangles

Some triangles have special names. We will look first at the right triangle . A right triangle has one 90° 90° angle, which is often marked with the symbol shown in Figure 9.10 .

If we know that a triangle is a right triangle, we know that one angle measures 90° 90° so we only need the measure of one of the other angles in order to determine the measure of the third angle.

Example 9.19

One angle of a right triangle measures 28° . 28° . What is the measure of the third angle?

Try It 9.37

One angle of a right triangle measures 56° . 56° . What is the measure of the other angle?

Try It 9.38

One angle of a right triangle measures 45° . 45° . What is the measure of the other angle?

In the examples so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. So we will wait to draw the figure until we write expressions for all the angles we are looking for.

Example 9.20

The measure of one angle of a right triangle is 20° 20° more than the measure of the smallest angle. Find the measures of all three angles.

Try It 9.39

The measure of one angle of a right triangle is 50° 50° more than the measure of the smallest angle. Find the measures of all three angles.

Try It 9.40

The measure of one angle of a right triangle is 30° 30° more than the measure of the smallest angle. Find the measures of all three angles.

Similar Triangles

When we use a map to plan a trip, a sketch to build a bookcase, or a pattern to sew a dress, we are working with similar figures. In geometry, if two figures have exactly the same shape but different sizes, we say they are similar figures . One is a scale model of the other. The corresponding sides of the two figures have the same ratio, and all their corresponding angles have the same measures.

The two triangles in Figure 9.11 are similar. Each side of Δ A B C Δ A B C is four times the length of the corresponding side of Δ X Y Z Δ X Y Z and their corresponding angles have equal measures.

Properties of Similar Triangles

If two triangles are similar, then their corresponding angle measures are equal and their corresponding side lengths are in the same ratio.

The length of a side of a triangle may be referred to by its endpoints, two vertices of the triangle. For example, in Δ A B C : Δ A B C :

the length a can also be written B C the length b can also be written A C the length c can also be written A B the length a can also be written B C the length b can also be written A C the length c can also be written A B

We will often use this notation when we solve similar triangles because it will help us match up the corresponding side lengths.

Example 9.21

Δ A B C Δ A B C and Δ X Y Z Δ X Y Z are similar triangles. The lengths of two sides of each triangle are shown. Find the lengths of the third side of each triangle.

Try It 9.41

Δ A B C Δ A B C is similar to Δ X Y Z . Δ X Y Z . Find a . a .

Try It 9.42

Δ A B C Δ A B C is similar to Δ X Y Z . Δ X Y Z . Find y . y .

The Pythagorean Theorem is a special property of right triangles that has been used since ancient times. It is named after the Greek philosopher and mathematician Pythagoras who lived around 500 500 BCE.

Remember that a right triangle has a 90° 90° angle, which we usually mark with a small square in the corner. The side of the triangle opposite the 90° 90° angle is called the hypotenuse , and the other two sides are called the legs . See Figure 9.12 .

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the two legs equals the square of the hypotenuse.

The Pythagorean Theorem

In any right triangle Δ A B C , Δ A B C ,

where c c is the length of the hypotenuse a a and b b are the lengths of the legs.

To solve problems that use the Pythagorean Theorem, we will need to find square roots. In Simplify and Use Square Roots we introduced the notation m m and defined it in this way:

For example, we found that 25 25 is 5 5 because 5 2 = 25 . 5 2 = 25 .

We will use this definition of square roots to solve for the length of a side in a right triangle.

Example 9.22

Use the Pythagorean Theorem to find the length of the hypotenuse.

Try It 9.43

Try it 9.44, example 9.23.

Use the Pythagorean Theorem to find the length of the longer leg.

Try It 9.45

Use the Pythagorean Theorem to find the length of the leg.

Try It 9.46

Example 9.24.

Kelvin is building a gazebo and wants to brace each corner by placing a 10-inch 10-inch wooden bracket diagonally as shown. How far below the corner should he fasten the bracket if he wants the distances from the corner to each end of the bracket to be equal? Approximate to the nearest tenth of an inch.

Try It 9.47

John puts the base of a 13-ft 13-ft ladder 5 5 feet from the wall of his house. How far up the wall does the ladder reach?

Try It 9.48

Randy wants to attach a 17-ft 17-ft string of lights to the top of the 15-ft 15-ft mast of his sailboat. How far from the base of the mast should he attach the end of the light string?

ACCESS ADDITIONAL ONLINE RESOURCES

  • Animation: The Sum of the Interior Angles of a Triangle
  • Similar Polygons
  • Example: Determine the Length of the Hypotenuse of a Right Triangle

Practice Makes Perfect

In the following exercises, find ⓐ the supplement and ⓑ the complement of the given angle.

In the following exercises, use the properties of angles to solve.

Find the supplement of a 135° 135° angle.

Find the complement of a 38° 38° angle.

Find the complement of a 27.5° 27.5° angle.

Find the supplement of a 109.5° 109.5° angle.

Two angles are supplementary. The larger angle is 56° 56° more than the smaller angle. Find the measures of both angles.

Two angles are supplementary. The smaller angle is 36° 36° less than the larger angle. Find the measures of both angles.

Two angles are complementary. The smaller angle is 34° 34° less than the larger angle. Find the measures of both angles.

Two angles are complementary. The larger angle is 52° 52° more than the smaller angle. Find the measures of both angles.

In the following exercises, solve using properties of triangles.

The measures of two angles of a triangle are 26° 26° and 98° . 98° . Find the measure of the third angle.

The measures of two angles of a triangle are 61° 61° and 84° . 84° . Find the measure of the third angle.

The measures of two angles of a triangle are 105° 105° and 31° . 31° . Find the measure of the third angle.

The measures of two angles of a triangle are 47° 47° and 72° . 72° . Find the measure of the third angle.

One angle of a right triangle measures 33° . 33° . What is the measure of the other angle?

One angle of a right triangle measures 51° . 51° . What is the measure of the other angle?

One angle of a right triangle measures 22.5 ° . 22.5 ° . What is the measure of the other angle?

One angle of a right triangle measures 36.5 ° . 36.5 ° . What is the measure of the other angle?

The two smaller angles of a right triangle have equal measures. Find the measures of all three angles.

The measure of the smallest angle of a right triangle is 20° 20° less than the measure of the other small angle. Find the measures of all three angles.

The angles in a triangle are such that the measure of one angle is twice the measure of the smallest angle, while the measure of the third angle is three times the measure of the smallest angle. Find the measures of all three angles.

The angles in a triangle are such that the measure of one angle is 20° 20° more than the measure of the smallest angle, while the measure of the third angle is three times the measure of the smallest angle. Find the measures of all three angles.

Find the Length of the Missing Side

In the following exercises, Δ A B C Δ A B C is similar to Δ X Y Z . Δ X Y Z . Find the length of the indicated side.

On a map, San Francisco, Las Vegas, and Los Angeles form a triangle whose sides are shown in the figure below. The actual distance from Los Angeles to Las Vegas is 270 270 miles.

Find the distance from Los Angeles to San Francisco.

Find the distance from San Francisco to Las Vegas.

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

In the following exercises, use the Pythagorean Theorem to find the length of the missing side. Round to the nearest tenth, if necessary.

In the following exercises, solve. Approximate to the nearest tenth, if necessary.

A 13-foot 13-foot string of lights will be attached to the top of a 12-foot 12-foot pole for a holiday display. How far from the base of the pole should the end of the string of lights be anchored?

Pam wants to put a banner across her garage door to congratulate her son on his college graduation. The garage door is 12 12 feet high and 16 16 feet wide. How long should the banner be to fit the garage door?

Chi is planning to put a path of paving stones through her flower garden. The flower garden is a square with sides of 10 10 feet. What will the length of the path be?

Brian borrowed a 20-foot 20-foot extension ladder to paint his house. If he sets the base of the ladder 6 6 feet from the house, how far up will the top of the ladder reach?

Everyday Math

Building a scale model Joe wants to build a doll house for his daughter. He wants the doll house to look just like his house. His house is 30 30 feet wide and 35 35 feet tall at the highest point of the roof. If the dollhouse will be 2.5 2.5 feet wide, how tall will its highest point be?

Measurement A city engineer plans to build a footbridge across a lake from point X X to point Y , Y , as shown in the picture below. To find the length of the footbridge, she draws a right triangle XYZ , XYZ , with right angle at X . X . She measures the distance from X X to Z , 800 Z , 800 feet, and from Y Y to Z , 1,000 Z , 1,000 feet. How long will the bridge be?

Writing Exercises

Write three of the properties of triangles from this section and then explain each in your own words.

Explain how the figure below illustrates the Pythagorean Theorem for a triangle with legs of length 3 3 and 4 . 4 .

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Access for free at https://openstax.org/books/prealgebra-2e/pages/1-introduction
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Angle Worksheets: Add Insight to Your Preparation

This far-from-exhaustive list of angle worksheets is pivotal in math curriculum. Whether it is basic concepts like naming angles, identifying the parts of an angle, classifying angles, measuring angles using a protractor, or be it advanced like complementary and supplementary angles, angles formed between intersecting lines, or angles formed in 2D shapes we have them all covered for students in grade 4 through high school. Application seals concepts in the minds of children, and hence adding a little challenge into the routine in the form of free worksheets is definitely not a bad idea.

List of Angle Worksheets

Parts of an Angle

Naming Angles

Acute, Right, and Obtuse Angles

Classifying Angles

Reading Protractors

Measuring Angles

Drawing Angles

Estimating Angles

Angles on a Straight Line

Angles Around a Point

  • Complementary & Supplementary Angles

Adjacent Angles

Vertical Angles

Linear Pairs of Angles

Pairs of Angles

Angles Formed by a Transversal

Angles in Shapes

Explore the angle worksheets in detail.

How about some practice in identifying the vertex and arms of an angle? Get ahead of the pack with these parts of an angle pdfs and practice identifying and naming the vertex and arms of an angle.

Are you aware of the four ways of naming angles? Buckle up with these printable worksheets, and watch how accurately and effortlessly children name angles using the three points.

Spark interest and encourage children to identify acute, right, and obtuse angles with a bunch of fun-filled exercises like recognizing angles in a clock, angle types in real-life objects, and a lot more!

Become twice as conversant with identifying, classifying, and drawing all six types of angles: acute, right, obtuse, straight, reflex, and complete angles with this collection of pdfs.

Use the protractor tool like a pro to measure and draw angles. Printable protractor templates, a chart illustrating the parts and use of the tool, and protractor reading exercises await students in elementary school.

Reading the correct scale of the protractor to measure angles: the inner or outer scale, measuring and classifying angles, and solving linear equations are the skills grade 4 and grade 5 students acquire with these exercises.

Show your students how to construct angles using a protractor with these drawing angle pdfs. The exercises include constructing angles with 1° increments or 5°, drawing reflex angles, and more.

Expert-level skills aren’t built in a day, to acquire superior skills in estimating angles 4th grade and 5th grade children need to bolster practice with our printable estimating angles worksheets.

Work your way through this compilation of worksheets and examine the angles on a straight line that add up to 180°. Grade 4 and grade 5 students find the measures of the unknown angles by subtracting the given angles from 180°.

Did you know that the angles around a point add up to 360°? Keep this fact in mind as you figure out the measures of the unknown angles by adding the given angles and subtracting the sum from 360°.

Complementary and Supplementary Angles

If it's a pair of angles you see and are trying to figure out if they make a complementary or supplementary pair, the trick is just adding them up and if their sum is 90° they are complementary and if it is 180° they are supplementary. These worksheets are a sure-shot hit with 6th grade and 7th grade learners.

Explore this bunch of printable adjacent angles worksheets to get a vivid picture of the angle addition property exhibited by angles that share the same vertex and are next to each other.

Linked here are exercises on angles formed by intersecting lines! Know the congruent properties of vertical angles or vertically opposite angles and apply them to determine unknown angle measures.

Two angles that are both adjacent and supplementary are a linear pair. The measure of such a pair sum up to 180°. Get to the heart of such angle pairs with these pdf worksheets and solve equations for the unknown angle measures.

Tap your grade 7, and grade 8 student’s potential in identifying the different pairs of angles such as complementary and supplementary angles, linear pair, vertical angles and much more with our engaging set of worksheets.

Construct additional and experiential knowledge with these 8th grade and high school handouts to comprehend the seven types of angle pairs formed by a transversal that include corresponding angles, alternate angles, and consecutive angles.

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Chapter 2: Trigonometric Ratios

Exercises: 2.3 Solving Right Triangles

Suggested Homework Problems

Exercises Homework 2.3

Exercise group..

For Problems 1–4, solve the triangle. Round answers to hundredths.

For Problems 5–10,

  • Sketch the right triangle described.
  • Solve the triangle.

[latex]A = 42{^o}, c = 26[/latex]

[latex]B = 28{^o}, c = 6.8[/latex]

[latex]B = 33{^o}, a = 300[/latex]

[latex]B = 79{^o}, a = 116[/latex]

[latex]A = 12{^o}, a = 4[/latex]

[latex]A = 50{^o}, a = 10[/latex]

For Problems 11–16,

  • Without doing the calculations, list the steps you would use to solve the triangle.

[latex]B = 53.7{^o}, b = 8.2[/latex]

[latex]B = 80{^o}, a = 250[/latex]

[latex]A = 25{^o}, b = 40[/latex]

[latex]A = 15{^o}, c = 62[/latex]

[latex]A = 64.5{^o}, c = 24[/latex]

[latex]B = 44{^o}, b = 0.6[/latex]

For Problems 17–22, find the labeled angle. Round your answer to tenths of a degree.

For Problems 23–28, evaluate the expression and sketch a right triangle to illustrate.

[latex]\sin^{-1} 0.2[/latex]

[latex]\cos^{-1} 0.8[/latex]

[latex]\tan^{-1} 1.5[/latex]

[latex]\tan^{-1} 2.5[/latex]

[latex]\cos^{-1} 0.2839[/latex]

[latex]\sin^{-1} 0.4127[/latex]

For Problems 29–32, write two different equations for the statement.

The cosine of [latex]15 {^o}[/latex] is [latex]0.9659\text{.}[/latex]

The sine of [latex]70 {^o}[/latex] is [latex]0.9397\text{.}[/latex]

The angle whose tangent is [latex]3.1445[/latex] is [latex]65 {^o}\text{.}[/latex]

The angle whose cosine is [latex]0.0872[/latex] is [latex]85 {^o}\text{.}[/latex]

Evaluate the expressions, and explain what each means. \begin{equation*} \sin^{-1} (0.6), (\sin 6{^o})^{-1} \end{equation*}

Evaluate the expressions, and explain what each means. \begin{equation*} \cos^{-1} (0.36), (\cos 36{^o})^{-1} \end{equation*}

For Problems 35–38,

  • Sketch a right triangle that illustrates the situation. Label your sketch with the given information.
  • Choose the appropriate trig ratio and write an equation, then solve the problem.

The gondola cable for the ski lift at Snowy Peak is [latex]2458[/latex] yards long and climbs [latex]1860[/latex] feet. What angle with the horizontal does the cable make?

The Leaning Tower of Pisa is [latex]55[/latex] meters in length. An object dropped from the top of the tower lands [latex]4.8[/latex] meters from the base of the tower. At what angle from the horizontal does the tower lean?

A mining company locates a vein of minerals at a depth of [latex]32[/latex] meters. However, there is a layer of granite directly above the minerals, so they decide to drill at an angle, starting [latex]10[/latex] meters from their original location. At what angle from the horizontal should they drill?

The birdhouse in Carolyn’s front yard is [latex]12[/latex] feet tall, and its shadow at [latex]4[/latex] pm is [latex]15[/latex] feet [latex]4[/latex] inches long. What is the angle of elevation of the sun at [latex]4[/latex]pm?

For Problems 39–42,

[latex]a = 18, b = 26[/latex]

[latex]a = 35, b = 27[/latex]

[latex]b = 10.6 , c = 19.2[/latex]

[latex]a = 88, c = 132[/latex]

For Problems 43–48,

  • Make a sketch that illustrates the situation. Label your sketch with the given information.
  • Write an equation and solve the problem.

The Mayan pyramid of El Castillo at Chichen Itza in Mexico has [latex]91[/latex] steps. Each step is 26 cm high and 30 cm deep.

  • What angle does the side of the pyramid make with the horizontal?
  • What is the distance up the face of the pyramid, from base to top platform?

An airplane begins its descent when its altitude is 10 kilometers. The angle of descent should be [latex]3{^o}[/latex] from horizontal.

  • How far from the airport (measured along the ground) should the airplane begin its descent?
  • How far will the airplane travel on its descent to the airport?

A communications satellite is in a low earth orbit (LOE) at an altitude of 400 km. From the satellite, the angle of depression to earth’s horizon is [latex]19.728{^o}\text{.}[/latex] Use this information to calculate the radius of the earth.

The first Ferris wheel was built for the [latex]1893[/latex] Chicago world’s fair. It had a diameter of [latex]250[/latex] feet, and the boarding platform, at the base of the wheel, was [latex]14[/latex] feet above the ground. If you boarded the wheel and rotated through an angle of [latex]50{^o}\text{,}[/latex] what would be your height above the ground?

To find the distance across a ravine, Delbert takes some measurements from a small airplane. When he is a short distance from the ravine at an altitude of [latex]500[/latex] feet, he finds that the angle of depression to the near side of the ravine is [latex]56{^o}\text{,}[/latex] and the angle of depression to the far side is [latex]32{^o}\text{.}[/latex] What is the width of the ravine? (Hint: First find the horizontal distance from Delbert to the near side of the ravine.)

The window in Francine’s office is [latex]4[/latex] feet wide and [latex]5[/latex] feet tall. The bottom of the window is 3 feet from the floor. When the sun is at an angle of elevation of [latex]64{^o}\text{,}[/latex] what is the area of the sunny spot on the floor?

Which of the following numbers are equal to [latex]\cos 45{^o}\text{?}[/latex]

  • [latex]\displaystyle \dfrac{\sqrt{2}}{2}[/latex]
  • [latex]\displaystyle \dfrac{1}{\sqrt{2}}[/latex]
  • [latex]\displaystyle \dfrac{2}{\sqrt{2}}[/latex]
  • [latex]\displaystyle \sqrt{2}[/latex]

Which of the following numbers are equal to [latex]\tan 30{^o}\text{?}[/latex]

  • [latex]\displaystyle \sqrt{3}[/latex]
  • [latex]\displaystyle \dfrac{1}{\sqrt{3}}[/latex]
  • [latex]\displaystyle \dfrac{\sqrt{3}}{3}[/latex]
  • [latex]\displaystyle \dfrac{3}{\sqrt{3}}[/latex]

Which of the following numbers are equal to [latex]\tan 60{^o}\text{?}[/latex]

Which of the following numbers are equal to [latex]\sin 60{^o}\text{?}[/latex]

  • [latex]\displaystyle \dfrac{3}{\sqrt{2}}[/latex]
  • [latex]\displaystyle \dfrac{\sqrt{3}}{2}[/latex]
  • [latex]\displaystyle \dfrac{\sqrt{2}}{3}[/latex]
  • [latex]\displaystyle \dfrac{2}{\sqrt{3}}[/latex]

For Problems 53–58, choose all values from the list below that are exactly equal to, or decimal approximations for, the given trig ratio. (Try not to use a calculator!)

[latex]\cos 30{^o}[/latex]

[latex]\sin 45{^o}[/latex]

[latex]\tan 30{^o}[/latex]

[latex]\cos 60{^o}[/latex]

[latex]\sin 90{^o}[/latex]

[latex]\cos 0{^o}[/latex]

Fill in the table from memory with exact values. Do you notice any patterns that might help you memorize the values?

Fill in the table from memory with decimal approximations to four places.

For Problems 61 and 62, compare the given value with the trig ratios of the special angles to answer the questions. Try not to use a calculator.

Is the acute angle larger or smaller than [latex]45{^o}\text{?}[/latex]

  • [latex]\displaystyle \sin \alpha = 0.7[/latex]
  • [latex]\displaystyle \tan \beta = 1.2[/latex]
  • [latex]\displaystyle \cos \gamma = 0.65[/latex]

Is the acute angle larger or smaller than [latex]60{^o}\text{?}[/latex]

  • [latex]\displaystyle \cos \theta = 0.75[/latex]
  • [latex]\displaystyle \tan \phi = 1.5[/latex]
  • [latex]\displaystyle \sin \psi = 0.72[/latex]

For Problems 63–72, solve the triangle. Give your answers as exact values.

  • Find the perimeter of a regular hexagon if the apothegm is [latex]8[/latex] cm long. (The apothegm is the segment from the center of the hexagon and perpendicular to one of its sides.)
  • Find the area of the hexagon.

Triangle [latex]ABC[/latex] is equilateral, and its angle bisectors meet at point [latex]P\text{.}[/latex] The sides of [latex]\triangle ABC[/latex] are 6 inches long. Find the length of [latex]AP\text{.}[/latex]

Find an exact value for the area of each triangle.

Find an exact value for the perimeter of each parallelogram.

  • Find the area of the outer square.
  • Find the dimensions and the area of the inner square.
  • What is the ratio of the area of the outer square to the area of the inner square?
  • Find the area of the inner square.
  • Find the dimensions and the area of the outer square.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Mathematics LibreTexts

2.2: Solving Right Triangles.

  • Last updated
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  • Page ID 112417

  • Katherine Yoshiwara
  • Los Angeles Pierce College

Introduction

A triangle has six parts: three sides and three angles. In a right triangle, we know that one of the angles is \(90^{\circ}\). If we know three parts of a right triangle, including one of the sides, we can use trigonometry to find all the other unknown parts. This is called solving the triangle.

Example 2.31

The hypotenuse of a right triangle is 150 feet long, and one of the angles is \(35^{\circ}\), as shown in the figure. Solve the triangle.

Screen Shot 2022-09-16 at 7.58.47 AM.png

We can find the side opposite the \(35°\) angle by using the sine ratio.

\begin{aligned} \sin 35^{\circ} &=\dfrac{\text { opposite }}{\text { hypotenuse }} \\ 0.5736 &=\dfrac{a}{150} \\ a &=150(0.5736)=86.04 \end{aligned}

The opposite side is about 86 feet long. To find side \(b\) we could use the Pythagorean theorem now, but it is better to use given information, rather than values we have calculated, to find the other unknown parts. We will use the cosine ratio.

\begin{aligned} \cos 35^{\circ} &=\dfrac{\text { adjacent }}{\text { hypotenuse }} \\ 0.8192 &=\dfrac{b}{150} \\ b &=150(0.8192)=122.89 \end{aligned}

The adjacent side is about 123 feet long. Finally, the unknown angle is the complement of \(35^{\circ}\), or \(55^{\circ}\).

Checkpoint 2.32

Sketch a right triangle with

  • one angle of \(37^{\circ}\),
  • the side adjacent to that angle of length 5 centimeters.

Without doing the calculations, list the steps you would use to solve the triangle.

Use \(\tan 37^{\circ}\) to find the opposite side. Use \(\cos 37^{\circ}\) to find the hypotenuse. Subtract \(37^{\circ}\) from \(90^{\circ}\) to find the third angle.

Finding an Angle

While watching her niece at the playground, Francine wonders how steep the slide is. She happens to have a tape measure and her calculator with her, and finds that the slide is 77 inches high and covers a horizontal distance of 136 inches, as shown below.

Screen Shot 2022-09-16 at 8.08.47 AM.png

Francine knows that one way to describe the steepness of an incline is to calculate its slope, which in this case is

\(\Delta y / \Delta x = \dfrac{77}{136} = 0.5662\)

However, Francine would really like to know what angle the slide makes with the horizontal. She realizes that the slope she has just calculated is also the tangent of the angle she wants.

If we know the tangent of an angle, can we find the angle? Yes, we can: locate the key labeled \(\mathbf{TAN}^{−1}\) on your calculator; it is probably the second function above the \(\mathbf{TAN}\) key. Enter

2nd TAN 0.5662

and you should find that

\(\tan ^{-1} 0.5662=29.52^{\circ}\)

This means that \(29.52^{\circ}\) is the angle whose tangent is \(0.5662\). We read the notation as " inverse tangent of \(0.5662\) is \(29.52\) degrees."

When we find \(\tan ^{-1}\) of a number, we are finding an angle whose tangent is that number. Similarly, \(\sin ^{-1}\) and \(\cos ^{-1}\) are read as "inverse sine" and "inverse cosine." They find an angle with the given sine or cosine.

Example 2.33

Find the angle whose sine is \(0.6834\).

Enter 2nd \(\operatorname{SIN} 0.6834\) into your calculator to find

\(\sin ^{-1} 0.6834=43.11^{\circ}\)

So \(43.11^{\circ}\) is the angle whose sine is \(0.6834\). Or we can say that

\(\sin 43.11^{\circ}=0.6834\)

You can check the last equation on your calculator.

Note 2.34 In the last example, the two equations

\(\sin 43.11^{\circ}=0.6834 \quad \text { and } \quad \sin ^{-1} 0.6834=43.11^{\circ}\)

say the same thing in different ways.

Caution 2.35

The notation \(\sin ^{-1} x\) does not mean \(\frac{1}{\sin x}\). It is true that we use negative exponents to indicate reciprocals of numbers, for example \(a^{-1}=\frac{1}{a}\) and \(3^{-1}=\frac{1}{3}\). But "sin" by itself is not a variable.

  • \(\sin ^{-1} x\) means "the angle whose sine is \(x\)"
  • \(\frac{1}{\sin x}\) means "the reciprocal of the sine of angle \(x\)"

(You may recall that \(f^{-1}(x)\) denotes the inverse function for \(f(x)\). We will study trigonometric functions in Chapter 4.)

Checkpoint 2.36

Write the following fact in two different ways: \(68^{\circ}\) is the angle whose cosine is \(0.3746\).

\(\cos 68^{\circ}=0.3746\) or \(\cos ^{-1}(0.3746)=68^{\circ}\)

Example 2.37

Find the angle of inclination of a hill if you gain 400 feet in elevation while traveling half a mile.

A sketch of the hill is shown at right. (Recall that 1 mile = 5280 feet.)

\begin{aligned} \sin \theta &=\dfrac{400}{2640}=0 . \overline{15} \\ \theta &=\sin ^{-1} 0 . \overline{15}=8.71^{\circ} \end{aligned}

The angle of inclination of the hill is about \(8.7^{\circ}\).

Checkpoint 2.38

The tallest living tree is a coast redwood named Hyperion, at 378.1 feet tall. If you stand 100 feet from the base of the tree, what is the angle of elevation of your line of sight to the top of the tree? Round your answer to the nearest degree.

\(75^{\circ}\)

The Special Angles

The trigonometric ratios for most angles are irrational numbers, but there are a few angles whose trig ratios are "nice" values. You already know one of these values: the sine of \(30^{\circ}\). Because the sides of a right triangle are related by the Pythagorean theorem, if we know any one of the trig ratios for an angle, we can find the others.

Recall that the side opposite a \(30^{\circ}\) angle is half the length of the hypotenuse, so \(\sin 30^{\circ}=\frac{1}{2}\).

The figure at right shows a 30-60-90 triangle with hypotenuse of length 2. The opposite side has length 1, and we can calculate the length of the adjacent side.

Screen Shot 2022-09-16 at 8.28.45 AM.png

\begin{aligned} 1^2+b^2 &=2^2 \\ b^2 &=2^2-1^2=3 \\ b &=\sqrt{3} \end{aligned}

Now we know the cosine and tangent of \(30^{\circ}\).

\(\cos 30^{\circ}=\frac{\text { adjacent }}{\text { hypotenuse }}=\frac{\sqrt{3}}{2} \quad \quad \quad \tan 30^{\circ}=\frac{\text { opposite }}{\text { adjacent }}=\frac{1}{\sqrt{3}}\)

These are exact values for the trig ratios, but we can also find decimal approximations. Use your calculator to verify the following approximate values.

\begin{aligned} &\text{exact value} &&\text{approximation}\\ &\cos 30^{\circ} = \dfrac{\sqrt{3}}{2} \quad && \approx 0.8660 \\ &tan 30^{\circ} = \dfrac{1}{\sqrt{3}} \quad &&\approx 0.5774 \end{aligned}

Caution 2.39

It is important for you to understand the difference between exact and approximate values. These decimal approximations, like nearly all the other trig values your calculator gives you, are rounded off. Even if your calculator shows you ten or twelve digits, the values are not exactly correct -- although they are quite adequate for most practical calculations!

The angles \(30^{\circ}, 60^{\circ}\), and \(45^{\circ}\) are "special" because we can easily find exact values for their trig ratios, and use those exact values to find exact lengths for the sides of triangles with those angles.

Example 2.40

The sides of an equilateral triangle are 8 centimeters long. Find the exact length of the triangle’s altitude, \(h\).

The altitude divides the triangle into two \(30-60-90\) right triangles as shown in the figure. The altitude is adjacent to the \(30^{\circ}\) angle, and the hypotenuse of the right triangle is 8 centimeters. Thus,

Screen Shot 2022-09-16 at 3.33.18 PM.png

\begin{aligned} \cos 35^{\circ} &=\dfrac{\text { adjacent }}{\text { hypotenuse }} \quad &&\text{Fill in the values.} \\ \dfrac{\sqrt{3}}{2} &=\dfrac{h}{8} \quad &&\text{Multiply both sides by 8}\\ h &=8\left(\dfrac{\sqrt{3}}{2}\right)=4 \sqrt{3} \end{aligned}

The altitude is exactly \(4 \sqrt{3}\) centimeters long.

From this exact answer, we can find approximations to any degree of accuracy we like. You can check that \(4 \sqrt{3} \approx 6.9282\), so the altitude is approximately \(6.9\) centimeters long.

Checkpoint 2.41

Use the figure in the previous example to find exact values for the sine, cosine, and tangent of \(60^{\circ}\).

\(\sin 60^{\circ}=\frac{\sqrt{3}}{2}, \quad \cos 60^{\circ}=\frac{1}{2}, \quad \tan 60^{\circ}=\sqrt{3}\)

There is one more special angle: \(45^{\circ}\). We find the trig ratios for this angle using an isosceles right triangle. Because the base angles of an isosceles triangle are equal, they must both be \(45^{\circ}\).

The figure shows an isosceles right triangle with equal sides of length 1. You can use the Pythagorean theorem to show that the hypotenuse has length \(\sqrt{2}\), so the trig ratios for \(45^{\circ}\) are

Screen Shot 2022-09-16 at 3.37.47 PM.png

\begin{aligned} \sin 45^{\circ} &=\dfrac{\text { opposite }}{\text { hypotenuse }}=\dfrac{1}{\sqrt{2}} \approx 0.7071 \\ \cos 45^{\circ} &=\dfrac{\text { adjacent }}{\text { hypotenuse }}=\dfrac{1}{\sqrt{2}} \approx 0.7071 \\ \tan 45^{\circ} &=\dfrac{\text { opposite }}{\text { adjacent }}=1 \end{aligned}

The Trigonometric Ratios for the Special Angles

Here is a summary of the trig ratios for the special angles.

\begin{aligned} &\begin{array}{|c|c|c|c|} \hline \text { Trigonometric Ratios for the Special Angles: }\\ \hline \text { Angle } & \text { Sine } & \text { Cosine } & \text { Tangent } \\ \hline 30^{\circ} & \frac{1}{2}=0.5 & \frac{\sqrt{3}}{2} \approx 0.8660 & \frac{1}{\sqrt{3}} \approx 0.5774 \\ \hline 45^{\circ} & \frac{1}{\sqrt{2}} \approx 0.7071 & \frac{1}{\sqrt{2}} \approx 0.7071 & 1 \\ \hline 60^{\circ} & \frac{\sqrt{3}}{2} \approx 0.8660 & \frac{1}{2}=0.5 & \sqrt{3} \approx 1.732 \\ \hline \end{array} \end{aligned}

You should memorize the exact values for these trig ratios. A good way to remember them is to know the two special triangles shown below. From these triangles, you can always write down the three trig ratios for the special angles.

Screen Shot 2022-09-16 at 3.40.56 PM.png

You should also be able to recognize their decimal approximations.

Note 2.42 We can use the special angles as benchmarks for estimating and mental calculation. For example, we know that \(60^{\circ} \approx 0.8660\), so if \(sin \theta = 0.95\) for some unknown angle \(\theta\), we know that \(\theta > 60^{\circ}\), because as \(\theta\) increases from the sine of \(\theta\) increases also.

Example 2.43

If \(\cos \alpha > \dfrac{\sqrt{3}}{2}\), what can we say about \(\alpha\)?

As an angle increases from \(0^{\circ}\) to \(90^{\circ}\), its cosine decreases. Now, \(\cos 30^{circ} = \dfrac{\sqrt{3}}{2}\), so if \(\cos \alpha > \dfrac{\sqrt{3}}{2}\), then \(\alpha\) must be less than \(30^{\circ}\).

Checkpoint 2.44

If \(1 < \tan \beta < \sqrt{3}\), what can we say about \(\beta\)?

\(45^{\circ} < \beta < 60^{\circ}\)

Review the following skills you will need for this section.

Algebra Refresher 2.4

1. \(\sqrt{2} \sqrt{2}\) 2. \(\frac{3}{\sqrt{3}}\) 3. \(\sqrt{8}\) 4. \(\sqrt{\frac{3}{4}}\)

Rationalize the denominator.

5. \(\dfrac{1}{\sqrt{2}}\) 6. \(\dfrac{2}{\sqrt{3}}\) 7. \(\dfrac{6}{\sqrt{3}}\) 8. \(\dfrac{4}{\sqrt{8}}\)

2 \(\sqrt{3}\)

3 \(2 \sqrt{2}\)

4 \(\dfrac{\sqrt{3}}{2}\)

5 \(\dfrac{\sqrt{2}}{2}\)

6 \(\dfrac{2\sqrt{3}}{3}\)

7 \(2 \sqrt{3}\)

8 \(\sqrt{2}\)

Section 2.3 Summary

• Solve a triangle

• Inverse sine

• Inverse cosine

• Inverse tangent

• Special angles

• Exact value

• Decimal approximation

1 If we know one of the sides of a right triangle and any one of the other four parts, we can use trigonometry to find all the other unknown parts.

2 If we know one of the trigonometric ratios of an acute angle, we can find the angle using the inverse trig key on a calculator.

3 The exact values of trigonometric ratios of the special angles should be memorized.

\begin{array}{|c|c|c|c|} \hline \text { Angle } & \text { Sine } & \text { Cosine } & \text { Tangent } \\ \hline 30^{\circ} & \frac{1}{2}=0.5 & \frac{\sqrt{3}}{2} \approx 0.8660 & \frac{1}{\sqrt{3}} \approx 0.5774 \\ \hline 45^{\circ} & \frac{1}{\sqrt{2}} \approx 0.7071 & \frac{1}{\sqrt{2}} \approx 0.7071 & 1 \\ \hline 60^{\circ} & \frac{\sqrt{3}}{2} \approx 0.8660 & \frac{1}{2}=0.5 & \sqrt{3} \approx 1.732 \\ \hline \end{array}

4 You can remember the trig values for the special angles if you memorize two triangles:

Screen Shot 2022-09-16 at 3.53.21 PM.png

5 For the trigonometric ratios of most angles, your calculator gives approximations, not exact values.

Study Questions

1 How many parts of a right triangle (including the right angle) do you need to know in order to solve the triangle?

2 Why is it better to use the given values when solving a triangle, rather than values you have calculated?

3 What is the \(\sin^{-1}\) (or \(\cos^{-1}\) or \(\tan^{-1}\)) button on the calculator used for?

4 Which are the "special" angles, and why are they special?

Practice each skill in the Homework Problems listed.

1 Solve a right triangle #1-16, 63-74

2 Use inverse trig ratio notation #17-34

3 Use trig ratios to find an angle #17-22, 35-38

4 Solve problems involving right triangles #35-48

5 Know the trig ratios for the special angles #49-62, 75-78

Homework 2.3

In these Homework Problems, we use the following standard notation for a right triangle: in \(\triangle A B C, \angle C\) is a right angle. The side opposite \(\angle C\) has length \(c\), and so on. (See the figure at below.)

Screen Shot 2022-09-16 at 4.00.28 PM.png

For Problems 1–4, solve the triangle. Round answers to hundredths.

Screen Shot 2022-09-16 at 4.01.15 PM.png

For Problems 5–10,

a Sketch the right triangle described.

b Solve the triangle.

5. \(A=42^{\circ}, c=26\)

6. \(B=28^{\circ}, c=6.8\)

7. \(B=33^{\circ}, a=300\)

8. \(B=79^{\circ}, a=116\)

9. \(A=12^{\circ}, a=4\)

10. \(A=50^{\circ}, a=10\)

For Problems 11–16,

b Without doing the calculations, list the steps you would use to solve the triangle.

11. \(B=53.7^{\circ}, b=8.2\)

12. \(B=80^{\circ}, a=250\)

13. \(A=25^{\circ}, b=40\)

14. \(A=15^{\circ}, c=62\)

15. \(A=64.5^{\circ}, c=24\)

16. \(B=44^{\circ}, b=0.6\)

For Problems 17–22, find the labeled angle. Round your answer to tenths of a degree.

Screen Shot 2022-09-26 at 11.11.57 PM.png

For Problems 23–28, evaluate the expression and sketch a right triangle to illustrate.

23. \(\sin ^{-1} 0.2\)

24. \(\cos ^{-1} 0.8\)

25. \(\tan ^{-1} 1.5\)

26. \(\tan ^{-1} 2.5\)

27. \(\cos ^{-1} 0.2839\)

28. \(\sin ^{-1} 0.4127\)

For Problems 29–32, write two different equations for the statement.

29. The cosine of \(15^{\circ}\) is \(0.9659\).

30. The sine of \(70^{\circ}\) is \(0.9397\).

31. The angle whose tangent is \(3.1445\) is \(65^{\circ}\).

32. The angle whose cosine is \(0.0872\) is \(85^{\circ}\).

33. Evaluate the expressions, and explain what each means.

\(\sin ^{-1}(0.6), \quad\left(\sin 6^{\circ}\right)^{-1}\)

34. Evaluate the expressions, and explain what each means.

\(\cos ^{-1}(0.36), \quad\left(\cos 36^{\circ}\right)^{-1}\)

For Problems 35–38,

a Sketch a right triangle that illustrates the situation. Label your sketch with the given information.

b Choose the appropriate trig ratio and write an equation, then solve the problem.

35. The gondola cable for the ski lift at Snowy Peak is 2458 yards long and climbs 1860 feet. What angle with the horizontal does the cable make?

36. The Leaning Tower of Pisa is 55 meters in length. An object dropped from the top of the tower lands 4.8 meters from the base of the tower. At what angle from the horizontal does the tower lean?

37. A mining company locates a vein of minerals at a depth of 32 meters. However, there is a layer of granite directly above the minerals, so they decide to drill at an angle, starting 10 meters from their original location. At what angle from the horizontal should they drill?

38. The birdhouse in Carolyn's front yard is 12 feet tall, and its shadow at \(4 \mathrm{pm}\) is 15 feet 4 inches long. What is the angle of elevation of the sun at \(4 \mathrm{pm}\)?

For Problems 39–42,

39. \(a=18, b=26\)

40. \(a=35, b=27\)

41. \(b=10.6, c=19.2\)

42. \(a=88, c=132\)

For Problems 43–48,

a Make a sketch that illustrates the situation. Label your sketch with the given information.

b Write an equation and solve the problem.

43. The Mayan pyramid of El Castillo at Chichen Itza in Mexico has 91 steps. Each step is 26 cm high and 30 cm deep.

i What angle does the side of the pyramid make with the horizontal?

ii What is the distance up the face of the pyramid, from base to top platform?

44. An airplane begins its descent when its altitude is 10 kilometers. The angle of descent should be \(3^{\circ}\) from horizontal.

i How far from the airport (measured along the ground) should the airplane begin its descent?

ii How far will the airplane travel on its descent to the airport?

45. A communications satellite is in a low earth orbit (LOE) at an altitude of 400 km. From the satellite, the angle of depression to earth’s horizon is \(19.728^{\circ}\). Use this information to calculate the radius of the earth.

46. The first Ferris wheel was built for the 1893 Chicago world’s fair. It had a diameter of 250 feet, and the boarding platform, at the base of the wheel, was 14 feet above the ground. If you boarded the wheel and rotated through an angle of \(50^{\circ}\), what would be your height above the ground?

47. To find the distance across a ravine, Delbert takes some measurements from a small airplane. When he is a short distance from the ravine at an altitude of 500 feet, he finds that the angle of depression to the near side of the ravine is \(56^{\circ}\), and the angle of depression to the far side is \(32^{\circ}\). What is the width of the ravine? (Hint: First find the horizontal distance from Delbert to the near side of the ravine.)

48. The window in Francine’s office is 4 feet wide and 5 feet tall. The bottom of the window is 3 feet from the floor. When the sun is at an angle of elevation of \(64^{\circ}\), what is the area of the sunny spot on the floor?

49. Which of the following numbers are equal to \(\cos 45^{\circ}\)?

a \(\dfrac{\sqrt{2}}{2}\)

b \(\dfrac{1}{\sqrt{2}}\)

c \(\dfrac{2}{\sqrt{2}}\)

d \(\sqrt{2}\)

50. Which of the following numbers are equal to \(\tan 20^{\circ}\)?

a \(\sqrt{3}\)

b \(\dfrac{1}{\sqrt{3}}\)

c \(\dfrac{\sqrt{3}}{3}\)

d \(\dfrac{3}{\sqrt{3}}\)

51. Which of the following numbers are equal to \(\tan 60^{\circ}\)?

52. Which of the following numbers are equal to \(\sin 60^{\circ}\)?

a \(\dfrac{3}{\sqrt{2}}\)

b \(\dfrac{\sqrt{3}}{2}\)

c \(\dfrac{\sqrt{2}}{3}\)

d \(\dfrac{2}{\sqrt{3}}\)

For Problems 53–58, choose all values from the list below that are exactly equal to, or decimal approximations for, the given trig ratio. (Try not to use a calculator!)

\(\begin{array}{|ccccc|} \hline \sin 30^{\circ} & \cos 45^{\circ} & \sin 60^{\circ} & \tan 45^{\circ} & \tan 60^{\circ} \\ 0.5000 & 0.5774 & 0.7071 & 0.8660 & 1.0000 \\ \frac{1}{\sqrt{2}} & \frac{2}{\sqrt{2}} & \frac{3}{\sqrt{2}} & \frac{1}{2} & \frac{\sqrt{2}}{2} \\ \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & \frac{\sqrt{3}}{2} & \sqrt{3} & \frac{\sqrt{3}}{3} \\ \hline \end{array}\)

53. \(\cos 30^{\circ}\)

54. \(\sin 45^{\circ}\)

55. \(\tan 30^{\circ}\)

56. \(\cos 60^{\circ}\)

57. \(\sin 90^{\circ}\)

58. \(\cos 0^{\circ}\)

59. Fill in the table from memory with exact values. Do you notice any patterns that might help you memorize the values?

60. Fill in the table from memory with decimal approximations to four places.

For Problems 61 and 62, compare the given value with the trig ratios of the special angles to answer the questions. Try not to use a calculator.

61. Is the acute angle larger or smaller than \(45^{\circ}\)?

a \(\sin \alpha=0.7\)

b \(\tan \beta=1.2\)

c \(\cos \gamma=0.65\)

62. Is the acute angle larger or smaller than \(60^{\circ}\)?

a \(\cos \theta=0.75\)

b \(\tan \phi=1.5\)

c \(\sin \psi=0.72\)

For Problems 63–72, solve the triangle. Give your answers as exact values.

Screen Shot 2022-09-26 at 11.35.03 PM.png

a Find the perimeter of a regular hexagon if the apothegm is 8 cm long. (The apothegm is the segment from the center of the hexagon and perpendicular to one of its sides.)

b Find the area of the hexagon.

Screen Shot 2022-09-26 at 11.43.24 PM.png

Triangle ABC is equilateral, and its angle bisectors meet at point P . The sides of \(△ABC\) are 6 inches long. Find the length of \(AP\).

75. Find an exact value for the area of each triangle.

Screen Shot 2022-09-26 at 11.43.59 PM.png

76. Find an exact value for the perimeter of each rhombus.

Screen Shot 2022-09-26 at 11.45.15 PM.png

a Find the area of the outer square.

b Find the dimensions and the area of the inner square.

c What is the ratio of the area of the outer square to the area of the inner square?

Screen Shot 2022-09-26 at 11.51.15 PM.png

a Find the area of the inner square.

b Find the dimensions and the area of the outer square.

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CCSS Math Answers

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles

The students of middle school can get the Solution Key for Big Ideas Math Grade 8 Chapter 3 Angles and Triangles on this page. With the help of this Big Ideas Math Book 8th Grade Answer Key Chapter 3 Angles and Triangles you can finish your homework in time and also improve your performance in the exams. Get free step by step solutions for all the questions in Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles.

Big Ideas Math Book 8th Grade Answer Key Chapter 3 Angles and Triangles

Download Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles pdf for free of cost. The solutions for each and every question is prepared in an easy manner. Go through the table of contents shown in the below section to know the topics covered in Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles.

Performance

Angles and Triangles STEAM Video/Performance

Angles and triangles getting ready for chapter 3.

Lesson: 1 Parallel Lines and Transversals

Lesson 3.1 Parallel Lines and Transversals

Parallel lines and transversals homework & practice 3.1.

Lesson: 2 Angles of Triangles

Lesson 3.2 Angles of Triangles

Angles of triangles homework & practice 3.2.

Lesson: 3 Angles of Polygons

Lesson 3.3 Angles of Polygons

Angles of polygons homework & practice 3.3.

Lesson: 4 Using Similar Triangles

Lesson 3.4 Using Similar Triangles

Using similar triangles homework & practice 3.4.

Chapter 3 – Angles and Triangles

Angles and Triangles Connecting Concepts

Angles and triangles chapter review, angles and triangles practice test, angles and triangles cumulative practice.

STEAM Video

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 1

Watch the STEAM Video “Honeycombs.” Then answer the following questions.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 2

Answer: The sum of interior angles of the equilateral triangle = 180° x + x + x = 180° 3x° = 180° x = 180/3 x° = 60°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 3

Performance Task

Turtle Shells

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 4

Chapter Exploration

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 6

Answer: 8 angles are formed by the parallel lines and the transversal b. Which of these angles have equal measures? Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 9

EXPLORATION 1

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 10

Use the figure to find the measure of the angle. Explain your reasoning

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 12

Question 1. ∠1

Answer: 63°

Big Ideas Math Grade 8 Answers Chapter 3 img_1

Question 2. ∠2

Answer: 117°

BIM Grade 8 Chapter 3 Angles and triangles answer key img_2

Answer: ∠1 and 59° are the supplementary angles ∠1 + 59° = 180° ∠1 = 180° – 59° ∠1 = 121° ∠2 and 59° are vertical angles. They are congruent. So, the measure of ∠1 is 121° ∠3 and 59° are supplementary angles. ∠3 + 59° = 180° ∠3 = 180° – 59° ∠3 = 121° ∠4, ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel side. the measure of  ∠4 is 121° the measure of ∠5 is 59° the measure of  ∠6 is 121° the measure of ∠7 is 59°

In Example 3, the measure of ∠4 is 84°. Find the measure of the angle. Explain your reasoning.

Question 4. ∠3

Bigideas Math Answers Grade 8 Chapter 3 Angles and Triangles img_3

Question 5. ∠5

Answer: ∠4 and ∠5 are alternate interior angles formed by transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠5 is 84°

Question 6. ∠6

Answer: ∠3 and ∠6 are alternate exterior angles formed by transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠6 is 96°

Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING ANGLE MEASURES Use the figure to find the measures of the numbered angles.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 14

Answer: ∠1 and 120° are the supplementary angles. ∠1 + 120° = 180° ∠1 = 180 – 120 ∠1 = 60° Thus the measure of ∠1 is 60° ∠2 and 120° are the vertical angles. They are congruent. Thus the measure of ∠2 is 120° ∠3 and 120° are the supplementary angles. ∠3 + 120° = 180° ∠3 = 180 – 120 ∠3 = 60° ∠4, ∠5, ∠6, ∠7 are corresponding angles are formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠4 is 60° Thus the measure of ∠5 is 120° Thus the measure of ∠6 is 120° Thus the measure of ∠7 is 60°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 15

Answer: ∠1 and 35° are the supplementary angles. ∠1 + 35° = 180° ∠1 = 180 – 35 ∠1 = 145° Thus the measure of ∠1 is 145° ∠2 and 35° are the supplementary angles. ∠2 + 35° = 180° ∠2 = 180 – 35 ∠2 = 145° Thus the measure of ∠2 is 145° ∠3 and 35° are the vertical angles. They are congruent. Thus the measure of ∠3 is 35° ∠4, ∠5, ∠6, ∠7 are corresponding angles are formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠4 is 35° Thus the measure of ∠5 is 145° Thus the measure of ∠6 is 145° Thus the measure of ∠7 is 35°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 16

Answer: ∠2, ∠6 are corresponding angles are formed by transversal intersecting parallel lines. ∠6, ∠8 are vertical angles are formed by transversal intersecting parallel lines. ∠5 does not belong to the other three because all the other three measure are equal.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 17

Answer: The angle a and the angle of 70 degrees are complementary angles because they belong to a right triangle, where the third angle is the right angle. ∠a + 70 = 90 ∠a = 90 – 70 ∠a = 20°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 18

Answer: The lines AB and CD are parallel. ABC and BCD are the corresponding angles formed by transversal intersecting parallel lines. ∠BCD = 55° ∠BAC + ∠ABC + ∠ACB = 180° The sum of the angles in a triangle is 180° ∠BAC + 55°+ 52° = 180° ∠BAC + 107° = 180° ∠BAC = 180° – 107° ∠BAC = 73° So, the head tube angle of a bike is 73°

Review & Refresh

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 19

Answer: perimeter of red hexagon/perimeter of blue hexagon = \(\frac{3}{5}\) The values of the ratios of the perimeter is \(\frac{3}{5}\) Area of red hexagon/Area of blue hexagon = (\(\frac{3}{5}\))² = \(\frac{9}{25}\) The values of the ratios of the area is \(\frac{9}{25}\)

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 20

Answer: perimeter of red trapezium /perimeter of blue trapezium = \(\frac{7}{6}\) The values of the ratios of the perimeter is \(\frac{7}{6}\) Area of red hexagon/Area of blue hexagon = (\(\frac{7}{6}\))² = \(\frac{49}{36}\) The values of the ratios of the area is \(\frac{49}{36}\)

Evaluate the expression.

Question 3. 4 + 3 2

Answer: 4 + 9 = 13

Question 4. 5(2) 2 – 6

Answer: 5(4) – 6 20 – 6 = 14

Question 5. 11 + (-7) 2 – 9

Answer: 11 + 49 – 9 11 + 40 = 50

Concepts, Skills, & Problem Solving EXPLORING INTERSECTIONS OF LINES Use a protractor to determine whether lines a and b are parallel. ( See Exploration 1, p. 103.)

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 21

Answer: Use a protractor to measure ∠1 and ∠2 ∠1 ≈ 60° ∠2 ≈ 60° ∠1 and ∠2, it means the two angles are congruent. The angles are exterior alternate angles. According to the converse of the exterior alternate angles theorem, the two lines are parallel. a || b

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 22

Answer: Use a protractor to measure ∠1 and ∠2 ∠1 ≈ 50° ∠2 ≈ 60° ∠1 and ∠2, it means the two angles are not congruent. The angles are exterior alternate angles. According to the converse of the exterior alternate angles theorem, the two lines are not parallel.

FINDING ANGLE MEASURES Use the figure to find the measures of the numbered angles. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 23

Answer: ∠1 and 107° are corresponding angles. They are congruent. So, the measure of ∠1 is 107°. ∠1 and ∠2 are supplementary angles. ∠1 + ∠2 = 180° 107° + ∠2 = 180° ∠2 = 180° – 107° ∠2 = 73° So, the measure of ∠2 is 73°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 24

Answer: ∠3 and 95° are corresponding angles. They are congruent. Thus the measure of ∠3 is 95° ∠3 and ∠4 are supplementary angles. ∠3 + ∠4 = 180° 95° + ∠4 = 180° ∠4 = 180 – 95 ∠4 = 85° So the measure of ∠4 is 85°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 25

Answer: ∠5 and 49° are corresponding angles. They are congruent. So, the measure of ∠5 is 49° ∠5 and ∠6 are supplementary angles. ∠5 + ∠6 = 180° 49° + ∠6 = 180° ∠6 = 180° – 49° ∠6 = 131° So, the measure of ∠6 is 131°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 26

Answer: Since the two lines are not parallel. Hence ∠5 is not congruent to ∠6. By this, we can say that your friend is not correct.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 27

Answer: ∠1 and ∠2 are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. The measure of ∠1 is 60° so the measure of ∠2 is 60°

Question 13. OPEN-ENDED Describe two real-life situations that use parallel lines.

Answer: Example 1: The railroad tracks and the tram tracks are parallel lines. Example 2:  The shelves of a bookcase.

USING CORRESPONDING ANGLES Use the figure to find the measures of the numbered angles.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 28

Answer: ∠1 and 60° are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. ∠1 and ∠2 are supplementary angles. ∠1 + ∠2 = 180° 60° + ∠2 = 180° ∠2 = 180° – 60° ∠2 = 119° So, the measure of ∠2 is 119° ∠3 and ∠1 are vertical angles. They are congruent. So, the measure of ∠3 is 61° ∠4 and ∠2 are vertical angles. They are congruent. ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel lines. So, the measure of ∠5 is 119° So, the measure of ∠6 is 61° So, the measure of ∠7 is 119°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 29

Answer: ∠1 and 99° are supplementary angles. ∠1 + 99° = 180° ∠1 = 180° – 99° ∠1 = 81° Thus the measure of ∠1 is 81° ∠2 and 99° are vertical angles. They are congruent. The measure of ∠2 is 99° ∠3 and ∠1 are vertical angles. They are congruent. So, the measure of ∠3 is 81° ∠4, ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel lines. So, the measure of ∠4 is 99° So, the measure of ∠5 is 81° So, the measure of ∠6 is 99° So, the measure of ∠7 is 81°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 30

Answer: ∠1 and 90° are supplementary angles. ∠1 + 90° = 180° ∠1 = 180° – 90° ∠1 = 90° Thus the measure of ∠1 is 90° ∠2 and 90° are vertical angles. They are congruent. Thus the measure of ∠2 is 90° ∠3 and ∠1 are vertical angles. They are congruent. So, the measure of ∠3 is 90° ∠4, ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel lines. So, the measure of ∠4 is 90° So, the measure of ∠5 is 90° So, the measure of ∠6 is 90° So, the measure of ∠7 is 90°

USING CORRESPONDING ANGLES Complete the statement. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 30.1

Answer: ∠1 and ∠8 are corresponding angles. They are congruent. The measure of ∠1 = 124°, then the measure of ∠8 is 124° ∠8 and ∠4 are supplementary angles. ∠8 + ∠4 = 180° 124° + ∠4 = 180° ∠4 = 180° – 124° ∠4 = 56° So, the measure of ∠4 is 56°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 32

Answer: ∠2 and ∠7 are corresponding angles. They are congruent. The measure of ∠2 = 48°, then the measure of ∠7 is 48° ∠7 and ∠3 are supplementary angles. ∠7 + ∠3 = 180° 48° + ∠3 = 180° ∠3 = 180° – 48° ∠3 = 132° Thus the measure of ∠3 = 132°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 33

Answer: ∠4 and ∠2 are alternate interior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠2 is 55°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 34

Answer: ∠6 and ∠8 are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠8 is 120°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 35

Answer: ∠7 and ∠2 are corresponding angles. They are congruent. The measure of ∠7 is 50.5°, so the measure of ∠2 is 50.5° ∠2 and ∠6 are supplementary angle. ∠2 + ∠6 = 180° 50.5° + ∠6 = 180° ∠6 = 180° – 50.5° ∠6 = 129.5° So, the measure of ∠6 is 129.5°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 35.1

Answer: ∠3 and ∠6 are corresponding angles. They are congruent. The measure of ∠3 is 118.7° So, the measure of ∠6 is 118.7° ∠6 and ∠2 are supplementary angles. ∠6 + ∠2 = 180° 118.7° + ∠2 = 180° ∠2 = 180° – 118.7° ∠2 = 61.3° So, the measure of ∠2 is 61.3°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 31.1

Answer: ∠4 and ∠5 are alternate interior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠1 is 40°

Question 24. REASONING Is there a relationship between exterior angles that lie on the same side of a transversal? interior angles that lie on the same side of a transversal? Explain.

Big Ideas Math Answers grade 8 Chapter 3 Angles and Triangles img_4

Question 25. REASONING When a transversal is perpendicular to two parallel lines, all the angles formed measure 90°. Explain why.

Big Ideas Math Grade 8 Answers Chapter 3 img

Question 26. REASONING Two horizontal lines are cut by a transversal. What is the least number of angle measures you need to know to find the measure of every angle? Explain your reasoning.

BIM Grade 8 Chapter 3 Angles and triangles answers img_6

Answer: ∠1 and ∠7 are alternate exterior angles formed by a transeversal intersecting parallel lines. So, ∠1 and ∠7 are congruent. ∠1 and ∠5 are corresponding angles formed by a transeversal intersecting parallel lines. So, ∠1 and ∠5 are congruent. ∠5 and ∠7 are vertical angles so they are congruent. Hence ∠1 and ∠7 are congruent.

FINDING A VALUE Find the value of x.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 33.1

Answer: ∠1 and 50° are alternate interior angles. They are congruent. So, the measure of ∠1 is 50° ∠2 and ∠1 are corresponding angles. They are congruent. So, the measure of ∠2 is 50° ∠2 and x are supplementary angle. ∠2 + x = 180° 50° + x = 180° x = 180° – 50° x = 130° So, the measure of x is 130°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 34.1

Answer: ∠1 and 115° are corresponding angles. They are congruent. So, the measure of ∠1 is 115° ∠1 and x are alternate exterior angles. They are congruent. So, the measure of x is 115°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 35.2

Answer: 180° rotation, translation about line t.

Question 31. OPEN-ENDED Refer to the figure. a. Do the horizontal lines appear to be parallel? Explain.

Answer: The three horizontal lines seem to spread apart, even though in reality they are parallel.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 36

Answer: As the lines AB and CD are parallel and ∠BCD are alternate interior angles transversal BC, they are congruent. ∠ABC ≅ ∠BCD x = 64 b. How does the angle the puck hits the edge of the table relate to the angle it leaves the edge of the table?

Answer: m∠MBA + m∠ABC + m∠CBN = 180° 58 ° + 64° + m∠CBN = 180° 122 ° + m∠CBN = 180° m∠CBN = 180° – 122 ° m∠CBN = 58°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 38

Find the measures of the interior angles of the triangle.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 40

Answer: 81°, 25°, 74°

Explanation: Sum of all the angles in a triangle = 180° x° + 81° + 25° = 180° x° = 180° – 81° – 25° x = 74° Thus the measure of the interior angle is 74°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 41

Answer: 43°, 51°, 86°

Explanation: Sum of all the angles in a triangle = 180° x° + (x – 35)° + 43° = 180° x° + x° – 35° + 43° = 180° 2x° + 8° = 180° 2x° = 180° – 8° 2x° = 172° x° = 172°/2 x° = 86° The measure of the interior angle of the triangle (x – 35)° = 86 – 35 (x – 35)° = 51° x° = 51° + 35° x° = 86°

Self-Assessment for Concepts & Skills

Question 4. VOCABULARY How many exterior angles does a triangle have at each vertex? Explain.

BIM 8th Grade Answers Chapter 3 img_8

FINDING ANGLE MEASURES Find the value of x.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 42

Answer: Sum of all the angles in a triangle = 180° x° + 25° + 40° = 180° x° + 65° = 180° x° = 180° x° = 180° – 65° x° = 115° Thus the value of x is 115°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 43

Answer: x° = 50° + 55° x° = 105° Thus the value of x is 105°

Question 7. The Historic Triangle in Virginia connects Jamestown, Williamsburg, and Yorktown. The interior angle at Williamsburg is 120°. The interior angle at Jamestown is twice the measure of the interior angle at Yorktown. Find the measures of the interior angles at Jamestown and Yorktown. Explain your reasoning.

Big Ideas math Answers Grade 8 ch3 angles and triangle img_9

Answer: Given, A helicopter travels from point C to point A to perform a medical supply drop. The helicopter then needs to land at point B. A = 90° + 32° A = 122° Thus the helicopter should turn 122° at point A to travel towards point B.

Use the figure to find the measure of the angle. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 45

Question 1. ∠2

Answer: 82°

∠2 and 82° are alternate exterior angles formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠2 is 82°

Question 2. ∠6

∠6 and 82° are vertical angles formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠6 is 82°

Question 3. ∠4

∠4 and 82° are corresponding angles formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠4 is 82°

Question 4. ∠1

Answer: 98°

∠4 and 82° are corresponding angles formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠4 is 82° ∠4 and ∠1 are supplementary angles ∠4 + ∠1 = 180° 82° + ∠1 = 180° ∠1 = 180° – 82° ∠1 = 98°

You spin the spinner shown.

Question 5. What are the favorable outcomes of spinning a number less than 4?

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 46

Answer: 1, 2, 3

Explanation: The favorable outcome of spinning a number less than 4 is 1, 2, and 3.

Question 6. In how many ways can spinning an odd number occur?

Answer: two ways Odd numbers = 1 and 3 So, in two ways spinning an odd number can occur.

Concepts, Skills, & Problem Solving

USING PARALLEL LINES AND TRANSVERSALS Consider the figure below. (See Exploration 2, p. 111.)

Question 7. Use a protractor to find the measures of the labeled angles.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 47

Answer: Use a protractor to determine the measures of the angles A, B, C. m∠A = 30° m∠B = 105° m∠C = 45° m∠D = 150° m∠E = 75° m∠F = 105° m∠G = 30°

Question 8. Is ∠F an exterior angle of Triangle ABC ? Justify your answer.

Answer: An exterior angle is the angle between one side of a triangle and the extension of an adjacent side. ∠F is not an exterior angle of triangle ABC because it has a side of triangle ABC, but not the extension of the adjacent side DF.

USING INTERIOR ANGLE MEASURES Find the measures of the interior angles of the triangle.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 48

Answer: Sum of all the angles in a triangle = 180° x° + 90° + 30° = 180° x° + 120° = 180° x° = 180° – 120° x° = 60°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 49

Answer: Sum of all the angles in a triangle = 180° x° + 65° + 40° = 180° x° + 105° = 180° x° = 180° – 105° x° = 75°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 50

Answer: Sum of all the angles in a triangle = 180° x° + 35° + 45° = 180° x° + 80° = 180° x° = 180° – 80° x° = 100°

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 51

Answer: Sum of all the angles in a triangle = 180° x° + (x + 65)° + 25° = 180° x° + x° + 65° + 25° = 180° 2x° + 90° = 180° 2x° = 180° – 90° 2x° = 90° x° = 90°/2 x° = 45° (x + 65)° = 45 + 65 = 110 x° = 25°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 52

Answer: Sum of all the angles in a triangle = 180° x° + (x – 44)° + 48° = 180° x° + x° – 44° + 48° = 180° 2x° + 4° = 180° 2x° = 180° – 4° 2x° = 176° x° = 176°/2 x° = 88° (x – 44)° = 88 – 44 = 44 x° = 44°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 53

Answer: Sum of all the angles in a triangle = 180° x° + (x – 11)° + 73° = 180° x° + x° – 11° + 73° = 180° 2x° + 62° = 180° 2x° = 180° – 62° 2x° = 118° x° = 118°/2 x° = 59° (x – 11)° = 59 – 11 = 48 x° = 48°

FINDING EXTERIOR ANGLE MEASURES Find the measure of the exterior angle.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 54

Answer: x° = 38° + 90° x° = 128° The measure of exterior angle is 128°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 55

Answer: k° = 64° + 76° k° = 140° The measure of an exterior angle is 140°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 56

Answer: 2a° = (a + 10°) + 44° 2a° = a + 54° 2a° – a° = 54 a° = 54 The measure of the exterior angle = 2a = 2(54°) = 108°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 57

Answer: Sum of all the angles in a triangle = 180° x° + 75° + 75° = 180° x° + 150° – 150° = 180° – 150° x° = 30° Thus the angle that tornado direction needs to change is 30°.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 58

Answer: Your friend is not correct because the measure of the exterior angle is equal to the sum of two non-adjacent interior angles.

Question 20. REASONING The ratio of the interior angle measures of a triangle is 2 : 3 : 5. What are the angle measures?

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 59

Answer: Sum of all the angles in a triangle = 180° 2x° + 3x° + 5x° = 180° 10x° = 180° x = 180/10 x = 18° 2x° = 2(18°) = 36° 3x° = 3(18) = 54° 5x° = 5(18) = 90°

Question 21. PROBLEM SOLVING The support for a window air-conditioning unit forms a triangle and an exterior angle. What is the measure of the exterior angle?

Answer: The measure of the exterior angle DBC is m∠DBC = m∠ABC + m∠ACB m∠ABC + m∠ACB = 90° 5x – 6 + 3x = 90 8x – 6 = 90 8x = 90 + 6 8x = 96 x = 96/8 x = 12 m∠DBC = m∠BAC+ m∠ACB = 90° + 3(12)° = 126°

Question 22. REASONING A triangle has an exterior angle with a measure of 120°. Can you determine the measures of the interior angles? Explain.

Answer: A triangle has an exterior angle with a measure of 120° m∠ACB = m∠A + m∠B m∠A + m∠B = 120° According to the exterior angles We have m∠C + m∠ACD = 180° m∠C + 120° = 180° m∠C = 180° – 120° m∠C = 60°

ANGLES OF TRIANGLES

Determine whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 23. Given three angle measures, you can construct a triangle.

Answer: We can construct a triangle if the sum of the measure of the 3 angles is 180°. As a matter of fact, if the sum of the measures of the 3 angles is 180° We can build an infinity of triangles that are similar.

Question 24. The acute interior angles of a right triangle are complementary.

Answer: Let A, B, C be the angles of a right triangle with m∠A = 90° m∠A + m∠B + m∠C = 180° 90° + m∠B + m∠C = 180° m∠B + m∠C = 180° – 90° m∠B + m∠C = 90° This means ∠B and ∠C are complementary.

Question 25. A triangle has more than one vertex with an acute exterior angle.

Answer: An exterior angle of a triangle and the adjacent triangle’s angle are complementary. If an exterior angle is acute, it means the adjacent triangle’s angle is obtuse. Since we are given that more than one exterior angle is acute, it means the triangle would have more than one obtuse angle, which is impossible. The statement is never true.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 60

Answer: The angles z and w are supplementary z + w = 180° The sum of a triangle is 180° x + y + w = 180° z = 180° – w x + y = 180° – w z = x + y

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 61

Find the sum of the interior angle measures of the green polygon.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 63

Answer: S = (n – 2) . 180° S = (7 – 2) . 180° S = 5 . 180° S = 900° Thus the sum of the interior angle measure is 900°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 64

Answer: S = (n – 2) . 180° S = (6 – 2) . 180° S = 4 . 180° S = 720° Thus the sum of the interior angle measure is 720°

Question 5. WRITING Explain how to find the sum of the interior measures of a polygon.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 65

Answer: Steps to find the sum of the interior measurements of the polygon: 1. Count the number of sides of the polygon. 2. Subtract the number of sides by 2. 3. Multiply the result of the subtraction by 180°

Question 6. FINDING THE SUM OF INTERIOR ANGLE MEASURES Find the sum of the interior angle measures of the green polygon.

Answer: S = (n – 2) . 180° S = (4 – 2) . 180° S = 2 . 180° S = 360° Thus the sum of the interior angle measure is 360°

FINDING AN INTERIOR ANGLE MEASURE

Find the value of x.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 66

Answer: S = (n – 2) . 180° S = (5 – 2) . 180° S = 3 . 180° S = 540° Thus the sum of the interior angle measure is 540° x° + 160° + 110° + 105° + 95° = 540° x° + 470° = 540° x° = 540° – 470° x° = 70° Thus the value of x is 70°.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 67

Answer: S = (n – 2) . 180° S = (9 – 2) . 180° S = 7 . 180° S = 1260° Thus the sum of the interior angle measure is 1260° x° + 165° + 155° + 150° + 140° + 135° + 130° + 125° + 110° = 1260° x° + 1105° = 1260° x° = 1260° – 1105° x° = 155° Thus the value of x is 155°

Question 9. A company installs an octagonal swimming pool. a. Find the value of a for the pool shown at the left.

Answer: S = (n – 2) . 180° S = (8 – 2) . 180° S = 6 . 180° S = 1080° Thus the sum of the interior angle measure is 1080° a° + 120° + a° + 120° + a° + 120° + a° + 120° = 1080° 4a° + 480° = 1080° 4a° = 1080° – 480° 4a° = 600° a° = 600/4 a° = 150° Thus the value of x is 150°.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 67.1

Answer: No for any octagon the sum of the interior angles is 1080 degrees.

Question 10. DIG DEEPER! A Bronze Star Medal A is shown. a. How many interior angles are there?

Answer: 10 interior angles are there

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 68

Answer: S = (n – 2) . 180° S = (10 – 2) . 180° S = 8 . 180° S = 1440° Thus the sum of the interior angle measure is 1440°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 69

Answer: 60°

Explanation: Sum of all the angles = 180° x° + 60° + 60° = 180° x° + 120° = 180° x° = 180° – 120° x° = 60°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 70

Answer: 45°

Explanation: Sum of all the angles = 180° x° + x° + 90° = 180° 2x° + 90° = 180° 2x° = 180° – 90° 2x° = 90° x° = 45°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 71

Answer: 113°

Explanation: x° = 65° + 48° x° = 113° Thus the measure of an exterior angle is 113°

Solve the proportion.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 72

Explanation: \(\frac{x}{12}\) = \(\frac{3}{4}\) 12 . \(\frac{x}{12}\) = \(\frac{3}{4}\) . 12 x = 3 . 3 x = 9

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 73

Explanation: \(\frac{14}{21}\) = \(\frac{x}{3}\) 3 . \(\frac{14}{21}\) = \(\frac{x}{3}\) . 3 x = 2

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 74

Explanation: \(\frac{9}{x}\) = \(\frac{6}{2}\) 2. \(\frac{9}{x}\) = 6 18 = 6x x = 3

Concepts, Skills, & Problem Solving EXPLORING INTERIOR ANGLES OF POLYGONS Use triangles to find the sum of the interior angle measures of the polygon. (See Exploration 1, p. 117.)

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 75

Answer: 360°

Explanation: Number of sides = 4 Number of interior triangles in the given figure = 2 The Sum of the measures of the interior angles using triangle = 2 . 180° = 360°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 76

Answer: 1260°

Explanation: Number of sides = 9 Number of interior triangles in the given figure = 7 The Sum of the measures of the interior angles using triangle = 7 . 180° = 1260°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 77

Answer: 540°

Explanation: Number of sides = 5 Number of interior triangles in the given figure = 3 The Sum of the measures of the interior angles using triangle = 3 . 180° = 540°

FINDING THE SUM OF INTERIOR ANGLE MEASURES Find the sum of the interior angle measures of the polygon.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 78

Explanation: S = (n – 2) . 180° S = (4- 2) . 180° S = 2 . 180° S = 360° Thus the sum of the interior angle measure is 360°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 79

Answer: 1080°

Explanation: S = (n – 2) . 180° S = (8- 2) . 180° S = 6 . 180° S = 1080° Thus the sum of the interior angle measure is 1080°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 80

Explanation: S = (n – 2) . 180° S = (9- 2) . 180° S = 7 . 180° S = 1260° Thus the sum of the interior angle measure is 1260°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 81

Answer: To find the sum of the interior angle measures he should subtract 2 from the number of sides of the polygon and then multiply by 180° S = (n – 2) . 180° By this, we can say that your friend is not correct.

FINDING AN INTERIOR ANGLE MEASURE Find the value of x.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 82

Answer: S = (n – 2) . 180° S = (4- 2) . 180° S = 2 . 180° S = 360° Thus the sum of the interior angle measure is 360° x° + 155° + 25° + 137° = 360° x° + 317° = 360° x° = 360° – 317° x° = 43°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 83

Answer: S = (n – 2) . 180° S = (6- 2) . 180° S = 4 . 180° S = 720° Thus the sum of the interior angle measure is 720° x° + x° + x° + x° + 90° + 90° = 720° 4x° + 180° = 720° 4x° = 720° – 180° 4x° = 540° x° = 540/4 x° = 135°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 84

Answer: S = (n – 2) . 180° S = (6- 2) . 180° S = 4 . 180° S = 720° Thus the sum of the interior angle measure is 720° 3x° + 45° + 135° + x° + 135° + 45° = 720° 4x° + 360° = 720° 4x° = 720° – 360° 4x° = 360° x° = 360/4 x° = 90°

FINDING A MEASURE Find the measure of each interior angle of the regular polygon.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 85

Answer: S = (n – 2) . 180° S = (3- 2) . 180° S = 1 . 180° S = 180° Thus the sum of the interior angle measure is 180° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 3. 180 ÷ 3 = 60°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 86

S = (n – 2) . 180° S = (9 – 2) . 180° S = 7 . 180° S = 1260° Thus the sum of the interior angle measure is 1260° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 9. 1260 ÷ 9 = 140°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 87

S = (n – 2) . 180° S = (12 – 2) . 180° S = 10 . 180° S = 1800° Thus the sum of the interior angle measure is 1800° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 12. 1800 ÷ 12 = 150°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 88

Answer: No, my friend is not correct because to find the measure of each interior angle of a regular 20-gon, he should divide the sum of the measured interior angles by the number of interior angles, in this case, 20 but your friend divide it by 18 so he is not correct.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 89

Answer: S = (n – 2) . 180° S = (5- 2) . 180° S = 3 . 180° S = 540° Thus the sum of the interior angle measure is 540° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 5. 540÷ 5 = 108°

b. RESEARCH Why are firehydrants made this way?

Question 22. PROBLEM SOLVING The interior angles of a regular polygon each measure 165°. How many sides does the polygon have?

Answer: (n – 2) . 180 = 165 . n 180n – 360 = 165n 180n – 360 + 360 – 165n = 165n + 360 – 165n 15n = 360 n = 360/15 n = 24 Therefore the polygon has 24 sides

Question 23. STRUCTURE A molecule can be represented by a polygon with interior angles that each measure 120°. What polygon represents the molecule? Does the polygon have to be regular? Justify your answers.

Answer: (n – 2) . 180 = 120 . n 180n – 360 = 120n 180n – 120n = 360 60n = 360 n = 6

Question 24. PROBLEM SOLVING The border of a Susan B. Anthony dollar is in the shape of a regular polygon. a. How many sides does the polygon have?

Answer: The polygon has 11 sides.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 91

Answer: S = (n – 2) . 180° S = (11 – 2) . 180° S = 9 . 180° S = 1620° Thus the sum of the interior angle measure is 1620° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 11. 1620 ÷ 11 = 147°

Question 25. REASONING The center of the stained glass window is in the shape of a regular polygon. What are the measures of the interior angles of the green triangle?

Answer: (n-2)180°/n = (8-2)180°/8 = 135° m∠OAB = m∠OBA = 135/2 = 67.5° m∠AOB + m∠OAB + m∠OBA = 180° m∠AOB + 67.5° + 67.5° = 180° m∠AOB + 135° = 180° m∠AOB = 180° – 135° m∠AOB = 45°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 92

Answer: The given polygon has 7 sides. S = (n – 2) . 180° S = (7 – 2) . 180° S = 5 . 180° S = 900° Thus the sum of the interior angle measure is 900° 4 . 135° + 3 . x° = 900° 540° + 3x° = 900° 3x° = 900° – 540° 3x° = 360° x° = 360/3 x° = 120°

Using Similar Triangles

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 93.1

Tell whether the triangles are similar. Explain.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 95

Answer: Yes

Explanation: x° + 66° + 90° = 180° x° + 156° = 180° x° = 180° – 156° x° = 24° y° + 24° + 90° = 180° y° + 114° = 180° y° = 180° – 114° y° = 66° The triangles have two pairs of congruent angles. Thus the triangles are similar.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 97

Explanation: We are not given any information about the lengths of the sides either, therefore with only a pair of congruent angles, we cannot tell whether the triangles are similar.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 98

Explanation: x° + 54° + 63° = 180° x° + 107° = 180° x° = 180° – 107° x° = 63°

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 98.1

Answer: Option B

Explanation: ΔPQR and ΔTSR are congruent as TS || PQ leads to two pairs of correspondent congruent angles. ΔPQR is a dilation of ΔTSR because their sides are proportional, the constant of proportionality being greater than 1. ΔPQR is a scale drawing of ΔTSR because their sides are proportional. The question that does not fit is “Are ΔPQR and ΔTSR the same size and shape?” because the triangles do not have the same size, but they have the same shape.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 100

Answer: Aqueduct/2.6 = 5/1 Aqueduct = 5 × 2.6 Aqueduct = 13 Thus the length of the Aqueduct is 13 km.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 102

Answer: a/10 = 3/6 6 × a = 3 × 10 6a = 30 a = 30/6 a = 5 The length from point Z to point Y is 5 miles. Time to travel from point Z to point Y = 5/3.5 = 1.56 hour

Find the measure of each interior angle of the regular polygon.

Question 1. octagon

Answer: The measure of each interior angle is 135°

Explanation: S = (n – 2) . 180° S = (8- 2) . 180° S = 6 . 180° S = 1080° Thus the sum of the interior angle measure is 1080° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 8. 1080÷ 8= 135°

Question 2. decagon

Answer: The measure of each exterior angle is 144°

Explanation: S = (n – 2) . 180° S = (10 – 2) . 180° S = 8 . 180° S = 1440° Thus the sum of the interior angle measure is 1440° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 10. 1440÷ 10= 144°

Question 3. 18-gon

Answer: The measure of each interior angle is 160°

Explanation: S = (n – 2) . 180° S = (18- 2) . 180° S = 16 . 180° S = 2880° Thus the sum of the interior angle measure is 2880° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 18. 2880 ÷ 18= 160°

Solve the equation. Check your solution.

Question 4. 3.5 + y = -1

Answer: Given the equation 3.5 + y = -1 y = -1 – 3.5 y = -4.5

Question 5. 9x = 54

Answer: Given the equation 9x = 54 x = 54/9 x = 6

Question 6. -4 = \(\frac{2}{7}\)p

Answer: Given the equation -4 = \(\frac{2}{7}\)p -4 × 7 = 2p 2p = -28 p = -28/2 p = -14

Concepts, Skills, & Problem Solving CREATING SIMILAR TRIANGLES Draw a triangle that is either larger or smaller than the one given and has two of the same angle measures. Explain why the new triangle is similar to the original triangle. (See Exploration 1, p. 123.)

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 103

IDENTIFYING SIMILAR TRIANGLES Tell whether the triangles are similar. Explain.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 105

Answer: The triangles have two pairs of congruent angles. So, the third angles are congruent, and the triangles are similar.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 106

Answer: x° + 36° + 72° = 180° x° + 108° = 180° x° = 180° – 108° x° = 72° y° + 33° + 72° = 180° y° + 105° = 180° y° = 180° – 105° y° = 72° The triangles do not have two pairs of congruent angles. Therefore the triangles are not similar.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 107

Answer: x° + 64° + 85° = 180° x° + 149° = 180° x° = 180° – 149° x° = 31° y° + 26° + 85° = 180° y° + 111° = 180° y° = 180° – 111° y° = 69° The triangles do not have two pairs of congruent angles. Therefore the triangles are not similar.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 108

Answer: x° + 48° + 81° = 180° x° + 129° = 180° x° = 180° – 129° x° = 51° y° + 48° + 51° = 180° y° + 99° = 180° y° = 180° – 99° y° = 81° The triangles have two pairs of congruent angles. Therefore the triangles are similar.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 109

Answer: 2x + 90 = 180° 2x = 180 – 90° 2x = 90° x = 90/2 x = 45° The ruler on the left and the ruler on the right both have the shape of a right triangle with 45° angles, therefore they are similar in shape, while the middle ruler has 60°, 30° angles.

STRUCTURE Tell whether the triangles are similar. Explain.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 110

Answer: m∠APB + m∠B = 90° m∠APB + 51° = 90° m∠APB = 90° – 51° m∠APB = 39° m∠APB + m∠BPD + m∠DPC = 180° 39° + 102° + m∠DPC = 180° m∠DPC + 141° = 180 m∠DPC = 180 – 141° m∠DPC = 39° m∠A = m∠C m∠APB = m∠DPC

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 111

Answer: ∠APB ≅ ∠CPD m∠APB = m∠CPD m∠APB = 29° m∠A + m∠B + m∠APB = 180° m∠A + 88° + 29° = 180° m∠A  + 117° = 180° m∠A = 180° – 117° m∠A = 63° m∠PDC + m∠PDE = 180° m∠PDC + 91° = 180° m∠PDC = 180° – 91° m∠PDC = 89°

IDENTIFYING SIMILAR TRIANGLES Can you determine whether the triangles are similar? Explain.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 112

Answer: PS || QR ∠PSQ and ∠SQR are interior angles using the transversal QS, thus they are congruent. ∠PSQ ≅ ∠SQR

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 113

Answer: As AB || DE there are two pairs of congruent alternate interior angles, using the transversals AE and BD. ∠A≅ ∠E ∠B≅ ∠D The two pairs of congruent angles are enough to prove that the triangles are similar. ΔABC ∼ ΔEDC

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 114

Answer: ΔAMN ∼ ΔABC MN/BC = AM/AB 1.5/d = 5/50 d = 1.5 × 10 d = 15 feet Therefore 15 feet is not an appropriate location.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 115

Answer: The two triangles are similar because they are right triangles and ∠AXB ≅ ∠PXQ because they are vertical angles. PQ/300 = 80/240 240PQ = 24000 PQ = 24000/240 PQ = 100 steps

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 116

Answer: Given, A person who is 6 feet tall casts a 3-foot-long shadow. A nearby pine tree casts a 15-foot-long shadow. ΔXAB ∼ ΔXPQ AB/PQ = XB/XQ 6/PQ = 3/15 PQ = 30 ft

Question 21. OPEN-ENDED You place a mirror on the ground 6 feet from the lamppost. You move back 3 feet and see the top of the lamppost in the mirror. What is the height of the lamppost?

Grade 8 BIM Chapter 3 Answer Key img_13

Question 22. DIG DEEPER! In each of two right triangles, one angle measure is two times another angle measure. Can you determine that the triangles are similar? Explain your reasoning.

Answer: We are given the right triangle ABC m∠A = 2m∠B Case 1: m∠A = 90° 90° = 2m∠B m∠B = 45° m∠C = 180° – 90° – 45° = 45° Case 2: m∠B = 90° m∠A = 2 × 90° = 180° Case 3: m∠C = 90° m∠A + m∠B = 180 – m∠C = 180° – 90° = 90° 2m∠B + m∠B = 90° 3m∠B = 90° m∠B = 30° m∠A = 2 . 30° = 60°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 117

Answer: ΔABG ∼ ΔACF ΔACF ∼ ΔADE ΔABG ∼ ΔADE AB = BC = CD = BD/2 = 6.32/2 = 3.16 AB/CD = BG/DE 3BG = 6 BG = 2 feet ΔACF ∼ ΔADE AC/AD = CF/DE 2/3 = CF/6 3CF = 2(6) CF = 4 feet

Using the Problem-Solving Plan

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 118

Understand the problem You know two dimensions of a dog park and the ratio of the perimeter of the small dog section to the perimeter of the entire park. You are asked to find the area of each section. Make a plan Verify that the small triangle and the large triangle are similar. Then use the ratio of the perimeters to find the base or the height of each triangle and calculate the areas. Solve and check. Use the plan to solve the problem. Then check your solution.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 119

Review Vocabulary

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 121

Choose and complete a graphic organizer to help you study the concept.

  • interior angles formed by parallel lines and a transversal
  • exterior angles formed by parallel lines and a transversal
  • interior angles of a triangle
  • exterior angles of a triangle
  • similar triangles

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 123

Chapter Self-Assessment

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 124

3.1 Parallel Lines and Transversals (pp. 103–110)

Question 1. ∠8

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 125

Answer: 140°

Explanation: ∠8 and 140 degrees angle are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠8 is 140°

Question 2. ∠5

Explanation: ∠5 and 140 degrees angle are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠5 is 140°

Question 3. ∠7

Answer: 40°

Explanation: ∠5 and 140 degrees angle are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠5 is 140° ∠5 and ∠7 are supplementary angle. ∠5 + ∠7 = 180° 140° + ∠7 = 180° ∠7 = 180° – 140° ∠7 = 40° So, the measure of ∠7 is 40°

Question 4. ∠2

Explanation: 140 and ∠2 are supplementary angle. 140° + ∠2 = 180° ∠2 = 180° – 140° ∠2 = 40° So, the measure of ∠2 is 40°

Question 5. ∠6

Explanation: ∠5 and 140 degrees angle are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠5 is 140° ∠5 and ∠6 are supplementary angle. 140° + ∠6 = 180° ∠6 = 180° – 140° ∠6 = 40° So, the measure of ∠6 is 40°

Complete the statement. Explain your reasoning.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 126

Answer: 123°

Explanation: ∠1 and ∠7 are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠7 is 123°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 127.1

Answer: 122°

Explanation: ∠2 and ∠6 are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠6 is 58° ∠5 and ∠6 are supplementary angle. ∠5 + ∠6 = 180° 58° + ∠5 = 180° ∠5 = 180° – 58° ∠5 = 122° So, the measure of ∠5 is 122°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 128

Answer: 119°

Explanation: ∠3 and ∠5 are alternate interior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠3 is 119°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 129

Explanation: ∠4 and ∠6 are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent. So, the measure of ∠4 is 60°

Question 10. In Exercises 6–9, describe the relationship between ∠2 and ∠8.

Answer: ∠2 ≅ ∠8

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 129.1

Answer: ∠1 = 108°, ∠2 = 108°

Explanation: ∠3 and 72° are alternate interior angles. They are congruent. So, the measure of ∠3 is 72° ∠3 + ∠1 = 180° 72° + ∠1 = 180° ∠1 = 180° – 72° ∠1 = 108° So, the measure of ∠1 is 108° ∠1 and ∠2 are alternating interior angles. They are congruent.

3.2 Angles of Triangles (pp. 111 – 116)

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 129.2

Answer: Sum of all the angles in a triangle = 180° x° + 50° + 55° = 180° x° + 105° = 180° x° = 180° – 105° x° = 75°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 130

Answer: Sum of all the angles in a triangle = 180° x° + (x + 8)° + 90° = 180° 2x° + 8° + 90° = 180° 2x° + 98° = 180° 2x° = 180° – 98° 2x° = 82 x° = 82/2 x° = 41° (x + 8)° = (41 + 8)° = 49°

Find the measure of the exterior angle.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 131

Answer: s° = 50° + 75° s° = 125° Thus the measure of the exterior angle is 125°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 132

Answer: Sum of all the angles in a triangle = 180° t° + (t + 10)° + (t + 20)° = 180° 3t° + 10° + 20° = 180° 3t° + 30° = 180° 3t° = 180° – 30° 3t° = 150° t° = 150/3 t° = 50° Exterior angle: t° + (t + 10)° t° + t° + 10° 2t° + 10° 2(50)° + 10° = 100° + 10° = 110° Thus the measure of the exterior angle is 110°.

Question 16. What is the measure of each interior angle of an equilateral triangle? Explain.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 133

Answer: Sum of all the angles in a triangle = 180° x° + 30° + 56° = 180° x° + 86° = 180° x° = 180° – 86° x° = 94° Thus the measure of the interior angle of the triangle at Chertan = 94°

3.3 Angles of Polygons (pp. 117–122)

Find the sum of the interior angle measures of the polygon.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 134

Answer: The polygon has 13 sides S = (n – 2) . 180° S = (13- 2) . 180° S = 11 . 180° S = 1980° Thus the sum of the interior angle measure is 1980°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 135

Answer: The polygon has 9 sides S = (n – 2) . 180° S = (9- 2) . 180° S = 7 . 180° S = 1260° Thus the sum of the interior angle measure is 1260°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 136

Answer: S = (n – 2) . 180° S = (4 – 2) . 180° S = 2 . 180° S = 360° Thus the sum of the interior angle measure is 360° x° + 60° + 128° + 95° = 360° x° + 283° = 360° x° = 360° – 283° x° = 77° Thus the value of x is 77°.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 137

Answer: S = (n – 2) . 180° S = (7 – 2) . 180° S = 5 . 180° S = 900° Thus the sum of the interior angle measure is 900° x° + 135° + 125° + 135° + 105° + 150° + 140° = 900° x° + 790° = 900° x° = 900° – 790° x° = 110°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 138

Answer: S = (n – 2) . 180° S = (6 – 2) . 180° S = 4 . 180° S = 720° Thus the sum of the interior angle measure is 720° x° + 120° + 140° + 92° + 125° + 130° = 720° x° + 607° = 720° x° = 720° – 607° x° = 113° The value of x° is 113°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 139

Answer: The given polygon is an octagon. It has 8 sides. S = (n – 2) . 180° S = (8 – 2) . 180° S = 6 . 180° S = 1080° Thus the sum of the interior angle measure is 1080° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 8. 1080 ÷ 3 = 135°

3.4 Using Similar Triangles (pp. 123–128)

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 140

Answer: x° + 68° + 90° = 180° x° = 180° – 158° x° = 22° y° + 22° + 90° = 180° y° + 112° = 180° y° = 180° – 112° y° = 68° The triangles have two pairs of congruent angles. So, the triangles are similar.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 141

Answer: x° + 100° + 30° = 180° x° + 130° = 180° x° = 180° – 130° x° = 50° y° + 100° + 50° = 180° y° + 150° = 180° y° = 180° – 150° y° = 30° The triangles have two pairs of congruent angles. So, the triangles are similar.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 142

Answer: x° + 50° + 85° = 180° x° + 135° = 180° x° = 180° – 135° x° = 45° y° + 85° + 35° = 180° y° + 120° = 180° y° = 180° – 120° y° = 60° The triangles do not have two pairs of congruent angles. So, the triangles are not similar.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 143

Answer: ∠B ≅ ∠D ∠A ≅ ∠C ∠AXB ≅ ∠CXD ∠AXB and ∠CXD are vertical angles. ΔAXB ∼ ΔCXD

Question 28. A person who is 5 feet tall casts a shadow that is 4 feet long. A nearby building casts a shadow that is 24 feet long. What is the height of the building?

Answer: Given, A person who is 5 feet tall casts a shadow that is 4 feet long. A nearby building casts a shadow that is 24 feet long. Let the height of the building = x ft x/24 = 5/4 24 . x/24 = 5/4 . 24 x = 30 Thus the height of the building is 30 ft.

Practice Test

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 144

Question 1. ∠7

Answer: 47°

Explanation: ∠7 and 47° angles are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠7 is 47°

Explanation: ∠6 and 47° angles are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠6 is 47°

Answer: 133°

Explanation: ∠4 and 47° are supplementary angles. 47° + ∠4 = 180° ∠4 = 180° – 47° ∠4 = 133° Thus the measure of ∠4 = 133°

Question 4. ∠5

Explanation: ∠6 and 47° angles are corresponding angles formed by transversal intersecting parallel lines. The angles are congruent. Thus the measure of ∠6 is 47° ∠6 + ∠5 = 180° 47° + ∠5 = 180° ∠5 = 180° – 47° ∠5 = 133° Thus the measure of ∠5 = 133°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 144.1

Answer: 28°

Explanation: Sum of all the angles in a triangle = 180° x° + 129° + 23° = 180° x° + 152° = 180° x° = 180° – 152° x° = 28° Thus the value of x° is 28°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 145

Answer: 68°

Explanation: Sum of all the angles in a triangle = 180° x° + (x – 24)° + 68° = 180° x° + x° – 24° + 68° = 180° 2x° + 44° = 180° 2x° = 180° – 44° 2x° = 136° x° = 68° (x – 24)° = (68 – 24)° = 44°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 146

Answer: j° = 40° + 90° j° = 130° The measure of an exterior angle is 130°.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 147

Answer: The coin has 7 sides. S = (n – 2) . 180° S = (7 – 2) . 180° S = 5 . 180° S = 900° Thus the sum of the interior angle measure is 900°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 148

Answer: S = (n – 2) . 180° S = (5 – 2) . 180° S = 3 . 180° S = 540° Thus the sum of the interior angle measure is 540° 2x° + 125° + 90° + 2x° + 125° = 540° 4x° + 340° = 540° 4x° = 540° – 340° 4x° = 200° x° = 200/4 x° = 50° The value of x° is 50°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 149

Answer: S = (n – 2) . 180° S = (6 – 2) . 180° S = 4 . 180° S = 720° Thus the sum of the interior angle measure is 720° In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 6. 720 ÷ 6 = 120°

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 150

Answer: To find x°: x° + 61° + 70° = 180° x° + 131° = 180° x° = 180° – 131° x° = 49° To find y°: x° + 39° + 70° = 180° x° + 109° = 180° x° = 180° – 109° x° = 71° The triangles do not have two pairs of congruent angles. So, the triangles are not similar.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 151

Answer: ∠A ≅ ∠QPB ∠C ≅ ∠PQB ΔBPQ ∼ ΔBAC

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 152

Answer: One way: ∠3 and 65° are supplementary angles. ∠5 and ∠3 are alternate interior angles. Another way: ∠8 and 65° are alternate exterior angles. ∠5 and ∠8 are supplementary angles.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 153

Answer: Given, You swim 3.6 kilometers per hour. d/105 = 80/140 105 . d/105 = 80/140 . 105 d = 60 The length of the pond is 60 m. Speed = 3.6 km per hour = 1 m sec Distance = d = 60m Time it will take to swim across the pond = distance/speed = 60/1 = 60 sec = 1 min

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 154

Answer: C = 11 + 1.6t C – 11 = 1.6t 1.6t = C – 11 t = (C – 11)/1.6 Thus the correct answer is option B.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 156

Answer: 5(x – 4) = 3x 5x – 20 = 3x 5x – 3x = 20 2x = 20 x = 20/2 x = 10 Thus the correct answer is option I.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 157

Answer: △PQR is similar to △STU PQ = 12 ST = 16 SU = 20 TU = 18 PQ/ST = QR/TU 12/16 = X/18 16X = 12 × 18 X = 216/16 X = 13.5 cm Thus the correct answer is option C.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 158

Answer: ∠y and 125° are supplementary angles. 125° + ∠y = 180° ∠y = 180° – 125° ∠y = 55° So, the measure of ∠y = 55° ∠x and ∠y are alternate interior angles. They are congruent. So, the measure of ∠x = 55°

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 159

Answer: My friend made the error by multiplying both sides by –\(\frac{2}{5}\). To correct the error she should multiply both sides by –\(\frac{5}{2}\) instead of –\(\frac{2}{5}\) Thus the correct answer is option F.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 160

Answer: Given, X(-6,-1) Y(-3,-5) X(-2,-3) Reflecting a point (x,y) in the y-axis. (x, y) = (-x, y) X(-6,-1) = X'(6, -1) Y(-3,-5) = Y'(3, -5) X(-2,-3) = Z'(2, -3) Thus the correct answer is option B.

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 162

Answer: S = (n – 2) . 180° Part B A quadrilateral has angles measuring 100°, 90°, and 90°. Find the measure of its fourth angle. Show your work and explain your reasoning.

Answer: The quadrilateral has 4 sides S = (n – 2) . 180° S = (4 – 2) . 180° S = 2 . 180° S = 360 ° Thus the sum of the interior angles is 360 ° x° + 100° + 90° + 90° = 360° x° + 280° = 360° x° = 360° – 280° x° = 80° Thus the value of x° is 80°

Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 163

Answer: Number of sides = 3 The number of interior triangles in the given figure = 3 Sum of the interior angles measure using triangle = 3 × 180° = 540

Conclusion:

I wish the details prevailed in the above article is beneficial for all the 8th grade students. Hope our Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles helped you a lot to overcome the difficulties in this chapter. Feel free to post your comments in the comment box. Stay tuned to our ccssmathanswers.com to get step by step explanation for all the Grade 8 chapters.

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Geometry (all content)

Unit 1: lines, unit 2: angles, unit 3: shapes, unit 4: triangles, unit 5: quadrilaterals, unit 6: coordinate plane, unit 7: area and perimeter, unit 8: volume and surface area, unit 9: pythagorean theorem, unit 10: transformations, unit 11: congruence, unit 12: similarity, unit 13: trigonometry, unit 14: circles, unit 15: analytic geometry, unit 16: geometric constructions, unit 17: miscellaneous.

IMAGES

  1. Lesson 3 Homework Practice Angles Of Triangles

    angles and triangles homework 3

  2. Angles In A Triangle Worksheet

    angles and triangles homework 3

  3. 7th Grade Constructing Triangles from Angles Lesson: FOLDABLE & Homework

    angles and triangles homework 3

  4. 👉 Angles in Triangles Worksheet PDF

    angles and triangles homework 3

  5. KS3: Angles in Triangles

    angles and triangles homework 3

  6. Lesson 3 Extra Practice Angles Of Triangles Answer Key

    angles and triangles homework 3

VIDEO

  1. Angles And Triangles Part -1

  2. Angles of a triangle and complementary angles in right triangle

  3. Lesson 6-3 Interior Angles Of Triangles

  4. ANGLES IN TRIANGLE: Find The Measure Of Internal Angles

  5. Lesson 6-4 Exterior Angles Of Triangles

  6. Interior Angles of Triangles (Example 3)

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