## Engage NY Eureka Math Geometry Module 5 Lesson 13 Answer Key

### Eureka Math Geometry Module 5 Lesson 13 Exploratory Challenge Answer Key

Exploratory Challenge

Examine the diagrams shown. Develop a conjecture about the relationship between a and b.

a = \(\frac{1}{2}\) b

Test your conjecture by using a protractor to measure a and b.

Answer:

Do your measurements confirm the relationship you found in your homework?

If needed, revise your conjecture about the relationship between a and b:

Now, test your conjecture further using the circle below.

Now, we will prove your conjecture, which is stated below as a theorem.

THE TANGENT – SECANT THEOREM: Let A be a point on a circle, let \(\overleftrightarrow{A B}\) be a tangent ray to the circle, and let C be a point on the circle such that \(\overleftrightarrow{A C}\) is a secant to the circle. If a = mâˆ BAC and b is the angle measure of the arc intercepted by âˆ BAC, then a = \(\frac{1}{2}\) b.

Given circle O with tangent \(\overleftrightarrow{A B}\), prove what we have just discovered using what you know about the properties of a circle and tangent and secant lines.

a. Draw triangle AOC. What is the measure of âˆ AOC? Explain.

Answer:

bÂ°. The central angle is equal to the degree measure of the arc it intercepts.

b. What is the measure of âˆ OAB? Explain.

Answer:

90Â°. The radius is perpendicular to the tangent line at the point of tangency.

c. Express the measure of the remaining two angles of triangle AOC in terms of a and explain.

Answer:

The angles are congruent because the triangle is isosceles. Each angle has a measure of (90 – a)Â° since mâˆ OAC + mâˆ CAB = 90Â°.

d. What is the measure of âˆ AOC in terms of a? Show how you got the answer.

Answer:

The sum of the angles of a triangle is 180Â°, so 90 – a + 90 – a + b = 180. Therefore, b = 2a or a = 1/2 b.

e. Explain to your neighbor what we have just proven.

Answer:

An inscribed angle formed by a secant and tangent line is half of the angle measure of the arc it intercepts.

### Eureka Math Geometry Module 5 Lesson 13 Exercise Answer Key

Opening Exercise

In circle A, \(m \widehat{B D}\) = 56Â°, and (BC) Ì… is a diameter. Find the listed measure, and explain your answer.

a. mâˆ BDC

Answer:

90Â°, angles inscribed in a diameter

b. mâˆ BCD

Answer:

28Â°, inscribed angle is half measure of intercepted arc

c. mâˆ DBC

Answer:

62Â°, sum of angles of a triangle is 180Â°

d. mâˆ BFG

Answer:

28Â°, inscribed angle is half measure of intercepted arc

e. \(m \widehat{B C}\)

Answer:

180Â°, semicircle

f. \(m \widehat{D C}\)

Answer:

124Â°, intercepted arc is double inscribed angle

g. Does âˆ BGD measure 56Â°? Explain.

Answer:

No, the central angle of \( would be 56Â°. âˆ BGD is not a central angle because its vertex is not the center of the circle.

h. How do you think we could determine the measure of âˆ BGD?

Answer:

Answers will vary. This leads to todayâ€™s lesson.

Exercises

Exercise 1.

Find x, y, a, b, and/or c.

Answer:

a = 34Â°, b = 56Â°, c = 52Â°

Exercise 2.

Answer:

a = 16Â°, b = 148Â°

Exercise 3.

Answer:

a = 86Â°, b = 43Â°

Exercise 4.

Answer:

2(3x + 4y) = 7x + 6y

65 + 65 + (7x + 6y) = 180

x = 5, y = 2.5

Exercise 5.

Answer:

a = 60Â°

### Eureka Math Geometry Module 5 Lesson 13 Problem Set Answer Key

In Problems 1â€“9, solve for a, b, and/or c.

Question 1.

Answer:

a = 67Â°

Question 2.

Answer:

a = 67Â°

Question 3.

Answer:

a = 67Â°

Question 4.

Answer:

a = 116Â°

Question 5.

Answer:

a = 116Â°

Question 6.

Answer:

a = 116Â°, b = 64Â°, c = 26Â°

Question 7.

Answer:

a = 45Â°, b = 45Â°

Question 8.

Answer:

a = 47Â°, b = 47Â°

Question 9.

Answer:

a = 57Â°

Question 10.

is tangent to circle A. [latex]\overline{D F}\) is a diameter. Find the angle measurements.

a. mâˆ BCD

Answer:

50Â°

b. mâˆ BAF

Answer:

80Â°

c. mâˆ BDA

Answer:

40Â°

d. mâˆ FBH

Answer:

40Â°

e. mâˆ BGF

Answer:

98Â°

Question 11.

\(\overline{B G}\) is tangent to circle A. \(\overline{B E}\) is a diameter. Prove: (i) f = a and (ii) d = c.

Answer:

(i) mâˆ EBG = 90Â°Â Â Tangent perpendicular to radius

f = 90 – eÂ Â Â Sum of angles

mâˆ ECB = 90Â°Â Â Â Angle inscribed in semicircle

In â–³ECB,

b + 90 + e = 180Â Â Â Sum of angles of a triangle

b = 90 – e

a = bÂ Â Â Angles inscribed in same arc are congruent

a = fÂ Â Â Substitution

(ii) a + c = 180Â Â Â Inscribed in opposite arcs

a = fÂ Â Â Inscribed in same arc

f + d = 180Â Â Â Linear pairs form supplementary angles

c + f = f + dÂ Â Substitution

c = d

### Eureka Math Geometry Module 5 Lesson 13 Exit Ticket Answer Key

Question 1.

Find a, b, and c.

Answer:

a = 56, b = 63, c = 61