# Eureka Math Geometry Module 5 Lesson 14 Answer Key

## Engage NY Eureka Math Geometry Module 5 Lesson 14 Answer Key

### Eureka Math Geometry Module 5 Lesson 14 Example Answer Key

Example

80°. If you draw △BDG, m∠DBG = 20° and m∠BDG = 60° because they are half of the measures of their inscribed arcs. That means m∠BGD = 100° because the sums of the angles of a triangle total 180°. ∠DGB and ∠BGE are supplementary, so m∠BGE = 80°.

b. Find x.

31.5°

### Eureka Math Geometry Module 5 Lesson 14 Exercise Answer Key

Opening Exercise
$$\overleftrightarrow{D B}$$ is tangent to the circle as shown.

a. Find the values of a and b.
a = 13, b = 80

b. Is $$\overline{C B}$$ a diameter of the circle? Explain.
No, if $$\overline{C B}$$ was a diameter, then m∠CEB would be 90°.

Exercises 1–2

Exercise 1.
In circle P, $$\overline{P O}$$ is a radius, and $$m \widehat{M O}$$ = 142°. Find m∠MOP, and explain how you know.

m∠MOP = 19°
Since $$\overline{P O}$$ is a radius and extends to a diameter, the measure of the arc intercepted by the diameter is 180°. $$m \widehat{M O}$$ = 142°, so the arc intercepted by ∠MOP is 180° – 142° or 38°. ∠MOP is inscribed in this arc, so its measure is half the degree measure of the arc or $$\frac{1}{2}$$ (38°) = 19°.

Exercise 2.
In the circle shown, $$m \widehat{C E}$$ = 55°. Find m∠DEF and $$m \widehat{E G}$$. Explain your answer.

m∠DEF = 27.5°
$$m \widehat{E G}$$ = 55°
$$m \widehat{C E}$$ = $$m \widehat{D F}$$ and $$m \widehat{D F}$$ = $$m \widehat{E G}$$ because arcs between parallel lines are equal in measure.
By substitution, $$m \widehat{E G}$$ = 55°.
$$m \widehat{D F}$$ = 55°, so m∠DEF = $$\frac{1}{2}$$ (55°) = 27.5° because it is inscribed in a 55° arc.

Exercises 3–7
In Exercises 3–5, find x and y.

Exercise 3.

x = 115, y = 65

Exercise 4.

x = 59, y = 76

Exercise 5.

x = 34, y = 146

Exercise 6.
In the circle shown, $$\overline{B C}$$ is a diameter. Find x and y.

x = 24, y = 53

Exercise 7.
In the circle shown, $$\overline{B C}$$is a diameter. DC:BE = 2:1. Prove y = 180 – $$\frac{3}{2}$$ x using a two – column proof.

$$\overline{B C}$$ is a diameter of circle A Given
m∠DBC = x° Given
$$m \widehat{D C}$$ = 2x°    Arc is double angle measure of inscribed angle
$$m \widehat{B E}$$ = x°   DC : BE = 2 : 1
$$m \widehat{B D C}$$ = $$m \widehat{B E C}$$ = 180°   Semicircle measures 180°
$$m \widehat{D C}$$ = 180° – 2x°    Arc addition
$$m \widehat{E C}$$ = 180° – x°     Arc addition
m∠BFD = $$\frac{1}{2}$$(180° – 2x° + 180° – x°)    Measure of angle whose vertex lies in a circle is half the angle measures of arcs intercepted by it and its vertical angles
y° = 180° – $$\frac{3}{2}$$ x°    Substitution and simplification

### Eureka Math Geometry Module 5 Lesson 14 Problem Set Answer Key

In Problems 1–4, find x.
Question 1.

x = 85

Question 2.

x = 65

Question 3.

x = 7

Question 4.

x = 9

Question 5.
Find x ($$m \widehat{C E}$$) and y ($$m \widehat{D G}$$).

60 = $$\frac{1}{2}$$(y + 20)
$$\frac{x + (85 + x)}{2}$$ = 90
x = 70, y = 100

Question 6.
Find the ratio of $$m \widehat{E C}$$ : $$m \widehat{D B}$$.

3:4

Question 7.
$$\overline{B C}$$ is a diameter of circle A. Find x.

x = 108

Question 8.
Show that the general formula we discovered in Example 1 also works for central angles. (Hint: Extend the radii to form two diameters, and use relationships between central angles and arc measure.)

Extend the radii to form two diameters.
Let the measure of the central angle be equal to x°.
The measure $$\widehat{B C}$$ = x° because the angle measure of the arc intercepted by a central angle is equal to the measure of the central angle.
The measure of the vertical angle is also x° because vertical angles are congruent.
The angle of the arc intercepted by the vertical angle is also x°.
The measure of the central angle is half the sum of the angle measures of the arcs intercepted by the central angle and its vertical angle.
x = $$\frac{1}{2}$$ (x + x)
This formula also works for central angles.

### Eureka Math Geometry Module 5 Lesson 14 Exit Ticket Answer Key

Question 1.
Lowell says that m∠DFC = $$\frac{1}{2}$$ (123) = 61° because it is half of the intercepted arc. Sandra says that you cannot determine the measure of ∠DFC because you do not have enough information. Who is correct and why?

b. $$m \widehat{B E}$$
39°, 81° = $$\frac{1}{2}$$($$m \widehat{B E}$$ + 123°) using the formula for an angle with vertex inside a circle.