Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 9.1 Answer Key Angle Relationships.

## Texas Go Math Grade 7 Lesson 9.1 Answer Key Angle Relationships

**Essential Question**

How can you use angle relationships to solve problems?

**Texas Go Math Grade 7 Lesson 9.1 Explore Activity Answer Key **

**Measuring Angles**

It is useful to work with pairs of angles and to understand how pairs of angles relate to each other. Congruent angles are angles that have the same measure.

Step 1

Using a ruler, draw a pair of intersecting lines. Label each angle from 1 to 4.

Step 2

Use a protractor to help you complete the chart.

**Reflect**

Question 1.

Conjecture Share your results with other students. Make a conjecture about pairs of angles that are opposite each other.

Answer:

The measures of the vertically opposite angles are equal, so we have two pairs of angles with the same measure.

**Go Math Answer Key Grade 7 Practice and Homework Lesson 9.1 Question 2.**

Conjecture When two lines intersect to form two angles, what conjecture can you make about the pairs of angles that are next to each other?

Answer:

Every pair of angles, obtained by the intersection of two lines, that are next to each other form a straight line, or an angle of 180°. Hence, they are supplementary angles.

Example 1

Use the diagram.

A. Name a pair of vertical angles.

∠AFB and ∠DFE

B. Name a pair of adjacent angles.

∠AFB and ∠BFD

C. Name a pair of supplementary angles.

∠AFB and ∠BFD

D. Find the measure of ∠AFB.

Use the fact that ∠AFB and ∠BFD in the diagram are supplementary angles to find m∠AFB.

The measure of ∠AFB is 40°

**Reflect**

Question 3.

**Analyze Relationships** What is the relationship between ∠AFB and ∠BFC? Explain.

Answer:

The measure of ∠DFC is 90°

m∠AFB + m∠BFC + m∠DFC = 180°

m∠AFB + m∠BFC + 90° = 180°

Subtract 90° from both sides

m∠AFB + m∠BFC = 90°

∠AFB and ∠BFC are complementary angles.

∠AFB and ∠BFC are complementary angles.

**Lesson 9.1 Practice A Geometry Answers Go Math Grade 7 Question 4.**

**Draw Conclusions** Are ∠AFC and ∠BFC adjacent angles? Why or why not?

Answer:

Yes, they are, because they have a common side, a common vertex(corner point), and they don’t overlap.

**Your Turn**

**Use the diagram.**

Question 5.

Name a pair of supplementary angles.

Answer:

∠EGF and ∠FGB.

Question 6.

Name a pair of vertical angles.

Answer:

∠FGA and ∠DGC

Question 7.

Find the measure of ∠CGD. ______

Answer:

We see from the diagram that

m∠CGD + m∠DGE + m∠EGF = 180°.

∠DGE and ∠AGB are vertically opposite angles, so their measures are equal.

m∠DGE + m∠AGB = 90°.

m∠CGD + m∠DGE + m∠EGF = 180° Substitute 90° for m∠DGE, and 35° for m∠EGF.

m∠CGD + 90° + 35° = 180°

m∠CGD + 125° = 180° Subtract 125° from both sides.

m∠CGD = 180° – 125°

m∠CGD = 55°

m∠CGD = 55°

Example 2

A. Find the measure of ∠EHF.

Since m∠EHF = 2x, then m∠EHE = 132°

B. Find the measure of ∠ZXY.

**Your Turn**

**Go Math Grade 7 Answer Key Pdf Lesson 9.1 Question 8.**

Find the measure of ∠JML.

Answer:

From the diagram we see

m∠JML + m∠LMN = 180° Substitute 3x for m∠JML, and 54° for m∠LMN.

3x + 54° = 180° Subtract 54° from both sides.

3x = 180° – 54°

3x = 126° Divide both sides by 3

m∠JML = 3x Substitute 41° for x.

m∠JML = 3. 41°

m∠JML = 126°

Question 9.

**Critique Reasoning** Cory says that to find m∠JML above you can stop when you get to the solution step 3x = 126°. Explain why this works.

Answer:

Both angle ∠JML and ∠LMN are on same straight line and the angle of straight line is 180° So sum of the ∠JML and ∠LMN will be 180° this means both angle are supplementary to each other.

∠JML + ∠LMN = 180°

3x + 54° = 180°

3x + 54 – 54 = 180 – 54

3x = 126°

x = 42°

Hence, 3x = 126° works both angle 3x and 54° are supplementary to each other. So the sum of both the angle will be equal to the 180°

∠JML and ∠LMN are supplementary to each other.

Example 3

The front of the top story of a house is shaped like an isosceles triangle. The measure of the angle at the top of the triangle is 70. Find the measure of each of the base angles.

**Your Turn**

**Use the diagram.**

**Lesson 9.1 Answer Key 7th Grade Angle Relationships Question 10.**

Find the value of x. ____________

Answer:

The sum of the measures of angles in a triangle is

∠C + ∠A + ∠B = 180° Substitute 800° for ∠C, x for ∠A, and 3x for ∠B.

80 + x + 3x = 180°

4x + 80° = 180° Subtract 80° from both sides.

4x = 180° – 80°

4x = 100° Divide both sides by 4.

x = 25°

Question 11.

Find the measures of

∠A and ∠B. _____________

Answer:

x = 25°

m∠A = x Substitute 25° for x

m∠A = 25°

m∠B = 3x Substitute 25° for x

m∠B = 3 . 25°

m∠B = 75°

m∠A = 25°, m∠B = 75°

**Texas Go Math Grade 7 Lesson 9.1 Guided Practice Answer Key **

**For Exercises 1-2, use the figure. (Example 1)**

Question 1.

**Vocabulary** The sum of the measures of ∠UWV and ∠UWZ is 90°, so ∠UWV and ∠UWZ are ____ angles.

Answer:

Complementary angles

Question 2.

**Vocabulary** ∠UWV and ∠VWX share a vertex and one side. They do not overlap, so ∠UWV and ∠VWX are

____ angles.

Answer:

Adjacent angles

**For Exercises 3-4, use the figure.**

**Angle Relationship Texas Go Math Grade 7 Answer Key Question 3.**

∠AGB and ∠DGE are ________________ angles, so m∠DGE= . (Example 1)

Answer:

– Vertical angles.

-m∠DGE = 30°.

Question 4.

Find the measure of ∠EGF. (Example 2)

m∠CGD + m∠DGE + m∠EGF = 180°

____ + ____ + ____ = 180°

____ + 2x = 180°

2x = _____

m∠EGF = 2x = ___

Answer:

50° + 30° + 2x = 180°

80° + 2x = 180°

2x = 100°

m∠EGF = 2x = 100°

m∠EGF = 100°

Question 5.

Find the measures of ∠A and ∠B. (Example 3)

m∠A + m∠B + m∠C = 180

____ + _______ + ____ = 180

2x + __ = 180

2x = _____

x = __, so, m∠A = ___.

x + 10 = _____, so m∠B = ___.

Answer:

x + x + 10° + 40° = 180°

2x + 50° = 180°

2x = 130°

x = 65°, so m∠A = 65°

x + 10° = 75°, so m∠A = 75°

m∠A = 65° m∠B = 75°

**Essential Question Check-In**

**Go Math Lesson 9.1 Angle Relationships Answers Question 6.**

Suppose that you know that ∠T and ∠S are supplementary and that m∠T = 3 . (m∠S). How can you find m∠T?

Answer:

∠T and ∠S are supplementary, so they form an angle of 180°

m∠T = 180°

m∠T + m∠S = 180°

3(m∠S) + m∠S = 180°

4 . m∠S = 180° Divide both sides by 4.

m∠T = 3 . 45° = 125°

**Texas Go Math Grade 7 Lesson 9.1 Independent Practice Answer Key **

**For Exercises 7-11, use the figure.**

Question 7.

Name a pair of adjacent angles. Explain why they are adjacent.

Answer:

∠RUS and ∠SUT because they have a common side and a common vertex.

Question 8.

Name a pair of acute vertical angles.

Answer:

∠RUS and ∠QUP

Question 9.

Name a pair of supplementary angles.

Answer:

∠TUP and ∠PUN

**Lesson 9.1 Understand Angle Relationships Answer Key Question 10.**

**Justify Reasoning** Find m∠QUR. Justify your answer.

Answer:

∠QUR and ∠RUS are supplementary angles, so

m∠QUR + m∠RUS = 180° Substitute 41° for both sides

m∠QUR + 41° = 180° Subtract 41° from both sides.

m∠QUR = 180° – 41°

m∠QUR = 139°

Question 11.

**Draw Conclusions** Which is greater, m∠TUR or m∠RUQ? Explain.

Answer:

From the diagram we see

m∠TUR = m∠TUS + m∠RUS

m∠RUQ = m∠NUQ + m∠RUN

∠TUS and ∠NUQ are vertical opposite angles, so their measures are equal.

m∠TUS = m∠NUQ = 90°

∠RUN and ∠RUS are complementary angles, so they form an angle of 90°

Since m∠RUS = 41°, we have

m∠RUS + m∠RUN = 90° Substitute 41° for ∠RUS.

41° + m∠RUN = 90° Subtract 41° from both sides

m∠RUN = 90° – 41°

m∠RUN = 49°

Put these reasults into the following:

m∠TUR = m∠TUS + m∠RUS

m∠TUR = 90° + 41°

m∠TUR = 131°

m∠RUQ = m∠NUQ + m∠RUN

m∠RUQ = 90° + 49°

m∠RUQ = 139°

m∠TUR < m∠RUQ

m∠RUQ is greater than m∠TUR

**Solve for each indicated angle measure or variable in the figure.**

Question 12.

x ____________

Answer:

∠KMI and ∠HMG are vertical opposite angles, so they have the same measure.

4x = 84° Divide both sides by 4.

x = 21°

**Go Math Answer Key Angle Pair Relationships Worksheet Pdf Question 13.**

m∠KMH _______

Answer:

m∠KMH and m∠KMI are supplementary angles

m∠KMH – m∠KMI = 180° Substitute 84° for ∠KMI.

m∠KMH + 84° = 180° Subtract 84° from both sides

m∠KMH = 180° – 84°

m∠KMH = 96°

Solve for each indicated angle measure or variable in the figure.

Question 14.

m∠CBE ______

Answer:

m∠CBE and m∠EBF are supplementary angles.

m∠CBE + m∠EBF = 180° Substitute 62° for ∠EBF.

m∠CBE + 62° = 180° Subtract 62° from both sides.

m∠CBE = 180° – 62°

m∠CBE = 118°

Question 15.

m∠ABF _____

Answer:

m∠ABF and m∠FBE are complementary angles.

m∠ABF + m∠FBE = 90° Substitute 62° for ∠FBE.

m∠ABF + 62° = 90° Subtract 62° from both sides.

m∠ABF = 90° – 62°

m∠ABF = 28°

Question 16.

m∠CBA _____

Answer:

m∠CBA and m∠ABF are supplementary angles.

m∠CBA + m∠ABF = 180° Substitute 28° for ∠ABF.

m∠CBA + 28° = 180° Subtract 28° from both sides.

m∠CBA = 180° – 28°

m∠CBA = 152°

**Solve for each indicated angle measure or variable in the figure.**

Question 17.

x ___________

Answer:

The sum of the measures of angles in a triangle is 180°, so we have

∠P + ∠Q + ∠R = 180° Substitute 25° for ∠P, 3x for ∠Q and 20° for ∠R.

25° + 3 + 20° = 180°

3x + 45° = 180° Subtract 45° from both sides.

3x = 180° – 45°

3x = 135° Divide both sides by 3.

**Texas Go Math Grade 7 Solutions Lesson 9.1 Answer Key Question 18.**

m∠Q _______

Answer:

x = 45°

m∠Q = 3x Supstitute 45° for z.

= 3 . 45°

= 135°

**Texas Go Math Grade 7 Lesson 9.1 H.O.T. Focus on Higher Order Thinking Answer Key **

**Let ∆ABC be a right triangle with m∠C = 90°.**

Question 19.

**Critical Thinking** An equilateral triangle has three congruent sides and three congruent angles. Can ∆ABC be an equilateral triangle? Explain your reasoning.

Answer:

In given triangle ABC one angle is m∠C = 90° so the triangle ABC cannot be a equilateral triangle. Because in

equilateral triangle all three sides and angle are congruent to each other. So each angle of equilateral triangle is

60° because sum of all the interior angle of triangle is 180°. Hence each angle of equilateral triangle is \(\frac{180}{3}\) = 60°.

Thus the given triangle ABC cannot be a equilateral triangle.

Given triangle ABC cannot be a equilateral triangle.

Question 20.

**Counterexample** An isosceles triangle has two congruent sides, and the angles opposite those sides are congruent. River says that right triangle ABC cannot be an isosceles triangle. Give a counterexample to show that his statement is incorrect.

Answer:

There is a special right triangle that can be considered as an isosceles triangle. This is the 45° right triangle wherein, the two sides have 45° angles. The measure of the sides are congruent and the angles are also congruent.

The 45° right triangle is an isosceles triangle.

**Angle Relationships Practice Answer Key Lesson 9.1 Go Math 7th Grade Question 21.**

**Make a Conjecture** In a scalene triangle, no two sides have the same length, and no two angles have the same measure. Do you think a right triangle can be a scalene triangle? Explain your reasoning.

Answer:

Any right triangle can be a scalene triangle except for the 45°-45° right triangle. This is because the two angles of

a right triangle can be of any measure for as long as the total of the two angles is 90°. The measurement of the

sides of the right triangle will also vary.

Yes

Question 22.

**Represent Real-World Problems** The railroad tracks meet the road as shown. The town will allow a parking lot at angle J if the measure of angle J is greater than 38°. Can a parking lot be built at an angle? Why or why not?

Answer:

All the upper angles are on a straight line so they must be supplementary to each other, thus their sum will be equal to 180°

50° + 90° + J = 180°

140° + J = 180°

J = 180 – 140

J = 40°

It is given in the problem that the minimum required angle J for the parking lot should be greater than 38° and we have found that angle J is 40°. So a parking lot can be built there.

Hence, the parking lot can be built there because angle J is 40°

Question 23.

**Analyze Relationships** In triangle XYZ, m∠X = 30°, and all the angles have measures that are whole numbers. Angle Y is an obtuse angle. What is the greatest possible measure that angle Z can have? Explain your answer.

Answer:

In triangle XYZ, m∠X = 30° so the sum of remaining angles ∠Y and ∠Z will be 180 – 30 = 150°. Since ∠Y is

an obtuse angle and all angle measures the whole number, so minimum value of ∠Y will be 91°. And in equation

∠Y + ∠Z = 180° if ∠Y is minimum then ∠Z must be maximum. Thus the maximum value of ∠Y = 150 – 91 = 59°.

Hence, the greatest possible measure of angle Z is 59°.