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

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

Example

a. Find x. Justify your answer.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

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.

Answer:

x = 115, y = 65

Exercise 4.

Answer:

x = 59, y = 76

Exercise 5.

Answer:

x = 34, y = 146

Exercise 6.

In the circle shown, \(\overline{B C}\) is a diameter. Find x and y.

Answer:

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.

Answer:

\(\overline{B C}\) 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.

Answer:

x = 85

Question 2.

Answer:

x = 65

Question 3.

Answer:

x = 7

Question 4.

Answer:

x = 9

Question 5.

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

Answer:

60 = \(\frac{1}{2}\)(y + 20)

\(\frac{x + (85 + x)}{2}\) = 90

x = 70, y = 100

Question 6.

Find the ratio of \(m \widehat{E C}\) : \(m \widehat{D B}\).

Answer:

3:4

Question 7.

\(\overline{B C}\) is a diameter of circle A. Find x.

Answer:

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.)

Answer:

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?

Answer:

Sandra is correct. We would need more information to determine the answer. Lowell is incorrect because ∠DFC is not an inscribed angle.

Question 2.

If m∠EFC = 9°, find and explain how you determined your answer.

a. m∠BFE

Answer:

81°, m∠EFC + m∠BFE = 180° (supplementary angles), so 180° – 99° = m∠BFE.

b. \(m \widehat{B E}\)

Answer:

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