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

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## 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

Reflect

Question 1.
Conjecture Share your results with other students. Make a conjecture about pairs of angles that are opposite each other.
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?
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.
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?
Yes, they are, because they have a common side, a common vertex(corner point), and they don’t overlap.

Use the diagram.

Question 5.
Name a pair of supplementary angles.
âˆ EGF and âˆ FGB.

Question 6.
Name a pair of vertical angles.
âˆ FGA and âˆ DGC

Question 7.
Find the measure of âˆ CGD. ______
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.

Find the measure of âˆ JML.

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.
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.

Use the diagram.

Find the value of x. ____________
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. _____________
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.

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.

For Exercises 3-4, use the figure.

âˆ AGB and âˆ DGE are ________________ angles, so mâˆ DGE= . (Example 1)
– 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 = ___
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 = ___.
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?
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.
âˆ RUS and âˆ SUT because they have a common side and a common vertex.

Question 8.
Name a pair of acute vertical angles.
âˆ RUS and âˆ QUP

Question 9.
Name a pair of supplementary angles.
âˆ TUP and âˆ PUN

Lesson 9.1 Understand Angle Relationships Answer Key Question 10.
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.
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 ____________
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 _______
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 ______
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 _____
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 _____
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 ___________
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 _______
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.
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.
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.
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?

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.