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 KeyÂ Â **

**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 180Â°. Hence, they are supplementary angles.

Example 1

Use the diagram.

A. Name a pair of vertical angles.

âˆ AFB and âˆ DFE

B. Name a pair of adjacent angles.

âˆ AFB and âˆ BFD

C. Name a pair of supplementary angles.

âˆ AFB and âˆ BFD

D. Find the measure of âˆ AFB.

Use the fact that âˆ AFB and âˆ BFD in the diagram are supplementary angles to find mâˆ AFB.

The measure of âˆ AFB is 40Â°

**Reflect**

Question 3.

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

Answer:

The measure of âˆ DFC is 90Â°

mâˆ AFB + mâˆ BFC + mâˆ DFC = 180Â°

mâˆ AFB + mâˆ BFC + 90Â° = 180Â°

Subtract 90Â° from both sides

mâˆ AFB + mâˆ BFC = 90Â°

âˆ AFB and âˆ BFC are complementary angles.

âˆ AFB and âˆ BFC are complementary angles.

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

**Draw Conclusions** Are âˆ AFC and âˆ 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:

âˆ EGF and âˆ FGB.

Question 6.

Name a pair of vertical angles.

Answer:

âˆ FGA and âˆ DGC

Question 7.

Find the measure of âˆ CGD. ______

Answer:

We see from the diagram that

mâˆ CGD + mâˆ DGE + mâˆ EGF = 180Â°.

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

mâˆ DGE + mâˆ AGB = 90Â°.

mâˆ CGD + mâˆ DGE + mâˆ EGF = 180Â° Substitute 90Â° for mâˆ DGE, and 35Â° for mâˆ EGF.

mâˆ CGD + 90Â° + 35Â° = 180Â°

mâˆ CGD + 125Â° = 180Â° Subtract 125Â° from both sides.

mâˆ CGD = 180Â° – 125Â°

mâˆ CGD = 55Â°

mâˆ CGD = 55Â°

Example 2

A. Find the measure of âˆ EHF.

Since mâˆ EHF = 2x, then mâˆ EHE = 132Â°

B. Find the measure of âˆ ZXY.

**Your Turn**

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

Find the measure of âˆ JML.

Answer:

From the diagram we see

mâˆ JML + mâˆ LMN = 180Â° Substitute 3x for mâˆ JML, and 54Â° for mâˆ LMN.

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

3x = 180Â° – 54Â°

3x = 126Â° Divide both sides by 3

mâˆ JML = 3x Substitute 41Â° for x.

mâˆ JML = 3. 41Â°

mâˆ JML = 126Â°

Question 9.

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

Answer:

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

âˆ JML + âˆ LMN = 180Â°

3x + 54Â° = 180Â°

3x + 54 – 54 = 180 – 54

3x = 126Â°

x = 42Â°

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

âˆ JML and âˆ 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

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

80 + x + 3x = 180Â°

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

4x = 180Â° – 80Â°

4x = 100Â° Divide both sides by 4.

x = 25Â°

Question 11.

Find the measures of

âˆ A and âˆ B. _____________

Answer:

x = 25Â°

mâˆ A = x Substitute 25Â° for x

mâˆ A = 25Â°

mâˆ B = 3x Substitute 25Â° for x

mâˆ B = 3 . 25Â°

mâˆ B = 75Â°

mâˆ A = 25Â°, mâˆ B = 75Â°

**Texas Go Math Grade 7 Lesson 9.1 Guided Practice Answer KeyÂ Â **

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

Question 1.

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

Answer:

Complementary angles

Question 2.

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

____ angles.

Answer:

Adjacent angles

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

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

âˆ AGB and âˆ DGE are ________________ angles, so mâˆ DGE= . (Example 1)

Answer:

– Vertical angles.

-mâˆ DGE = 30Â°.

Question 4.

Find the measure of âˆ EGF. (Example 2)

mâˆ CGD + mâˆ DGE + mâˆ EGF = 180Â°

____ + ____ + ____ = 180Â°

____ + 2x = 180Â°

2x = _____

mâˆ EGF = 2x = ___

Answer:

50Â° + 30Â° + 2x = 180Â°

80Â° + 2x = 180Â°

2x = 100Â°

mâˆ EGF = 2x = 100Â°

mâˆ EGF = 100Â°

Question 5.

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

mâˆ A + mâˆ B + mâˆ C = 180

____ + _______ + ____ = 180

2x + __ = 180

2x = _____

x = __, so, mâˆ A = ___.

x + 10 = _____, so mâˆ B = ___.

Answer:

x + x + 10Â° + 40Â° = 180Â°

2x + 50Â° = 180Â°

2x = 130Â°

x = 65Â°, so mâˆ A = 65Â°

x + 10Â° = 75Â°, so mâˆ A = 75Â°

mâˆ A = 65Â° mâˆ B = 75Â°

**Essential Question Check-In**

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

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

Answer:

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

mâˆ T = 180Â°

mâˆ T + mâˆ S = 180Â°

3(mâˆ S) + mâˆ S = 180Â°

4 . mâˆ S = 180Â° Divide both sides by 4.

mâˆ T = 3 . 45Â° = 125Â°

**Texas Go Math Grade 7 Lesson 9.1 Independent Practice Answer KeyÂ Â **

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

Question 7.

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

Answer:

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

Question 8.

Name a pair of acute vertical angles.

Answer:

âˆ RUS and âˆ QUP

Question 9.

Name a pair of supplementary angles.

Answer:

âˆ TUP and âˆ PUN

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

**Justify Reasoning** Find mâˆ QUR. Justify your answer.

Answer:

âˆ QUR and âˆ RUS are supplementary angles, so

mâˆ QUR + mâˆ RUS = 180Â° Substitute 41Â° for both sides

mâˆ QUR + 41Â° = 180Â° Subtract 41Â° from both sides.

mâˆ QUR = 180Â° – 41Â°

mâˆ QUR = 139Â°

Question 11.

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

Answer:

From the diagram we see

mâˆ TUR = mâˆ TUS + mâˆ RUS

mâˆ RUQ = mâˆ NUQ + mâˆ RUN

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

mâˆ TUS = mâˆ NUQ = 90Â°

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

Since mâˆ RUS = 41Â°, we have

mâˆ RUS + mâˆ RUN = 90Â° Substitute 41Â° for âˆ RUS.

41Â° + mâˆ RUN = 90Â° Subtract 41Â° from both sides

mâˆ RUN = 90Â° – 41Â°

mâˆ RUN = 49Â°

Put these reasults into the following:

mâˆ TUR = mâˆ TUS + mâˆ RUS

mâˆ TUR = 90Â° + 41Â°

mâˆ TUR = 131Â°

mâˆ RUQ = mâˆ NUQ + mâˆ RUN

mâˆ RUQ = 90Â° + 49Â°

mâˆ RUQ = 139Â°

mâˆ TUR < mâˆ RUQ

mâˆ RUQ is greater than mâˆ TUR

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

Question 12.

x ____________

Answer:

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

4x = 84Â° Divide both sides by 4.

x = 21Â°

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

mâˆ KMH _______

Answer:

mâˆ KMH and mâˆ KMI are supplementary angles

mâˆ KMH – mâˆ KMI = 180Â° Substitute 84Â° for âˆ KMI.

mâˆ KMH + 84Â° = 180Â° Subtract 84Â° from both sides

mâˆ KMH = 180Â° – 84Â°

mâˆ KMH = 96Â°

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

Question 14.

mâˆ CBE ______

Answer:

mâˆ CBE and mâˆ EBF are supplementary angles.

mâˆ CBE + mâˆ EBF = 180Â° Substitute 62Â° for âˆ EBF.

mâˆ CBE + 62Â° = 180Â° Subtract 62Â° from both sides.

mâˆ CBE = 180Â° – 62Â°

mâˆ CBE = 118Â°

Question 15.

mâˆ ABF _____

Answer:

mâˆ ABF and mâˆ FBE are complementary angles.

mâˆ ABF + mâˆ FBE = 90Â° Substitute 62Â° for âˆ FBE.

mâˆ ABF + 62Â° = 90Â° Subtract 62Â° from both sides.

mâˆ ABF = 90Â° – 62Â°

mâˆ ABF = 28Â°

Question 16.

mâˆ CBA _____

Answer:

mâˆ CBA and mâˆ ABF are supplementary angles.

mâˆ CBA + mâˆ ABF = 180Â° Substitute 28Â° for âˆ ABF.

mâˆ CBA + 28Â° = 180Â° Subtract 28Â° from both sides.

mâˆ CBA = 180Â° – 28Â°

mâˆ CBA = 152Â°

**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 180Â°, so we have

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

25Â° + 3 + 20Â° = 180Â°

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

3x = 180Â° – 45Â°

3x = 135Â° Divide both sides by 3.

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

mâˆ Q _______

Answer:

x = 45Â°

mâˆ Q = 3x Supstitute 45Â° for z.

= 3 . 45Â°

= 135Â°

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

**Let âˆ†ABC be a right triangle with mâˆ C = 90Â°.**

Question 19.

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

Answer:

In given triangle ABC one angle is mâˆ C = 90Â° 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

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

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 45Â° right triangle wherein, the two sides have 45Â° angles. The measure of the sides are congruent and the angles are also congruent.

The 45Â° 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 45Â°-45Â° 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 90Â°. 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 38Â°. 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 180Â°

50Â° + 90Â° + J = 180Â°

140Â° + J = 180Â°

J = 180 – 140

J = 40Â°

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

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

Question 23.

**Analyze Relationships** In triangle XYZ, mâˆ X = 30Â°, 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, mâˆ X = 30Â° so the sum of remaining angles âˆ Y and âˆ Z will be 180 – 30 = 150Â°. Since âˆ Y is

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

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

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