# Eureka Math Geometry Module 5 Lesson 13 Answer Key

## 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.

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

b. What is the measure of âˆ OAB? Explain.
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.
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.
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.
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
90Â°, angles inscribed in a diameter

b. mâˆ BCD
28Â°, inscribed angle is half measure of intercepted arc

c. mâˆ DBC
62Â°, sum of angles of a triangle is 180Â°

d. mâˆ BFG
28Â°, inscribed angle is half measure of intercepted arc

e. $$m \widehat{B C}$$
180Â°, semicircle

f. $$m \widehat{D C}$$
124Â°, intercepted arc is double inscribed angle

g. Does âˆ BGD measure 56Â°? Explain.
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
50Â°

b. mâˆ BAF
80Â°

c. mâˆ BDA
40Â°

d. mâˆ FBH
40Â°

e. mâˆ BGF
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.

(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.