# 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.  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 = m∠BAC and b is the angle measure of the arc intercepted by ∠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 ∠AOC? Explain.
b°. The central angle is equal to the degree measure of the arc it intercepts.

b. What is the measure of ∠OAB? Explain.
90°. 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)° since m∠OAC + m∠CAB = 90°.

d. What is the measure of ∠AOC in terms of a? Show how you got the answer.
The sum of the angles of a triangle is 180°, 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, $$m \widehat{B D}$$ = 56°, and (BC) ̅ is a diameter. Find the listed measure, and explain your answer. a. m∠BDC
90°, angles inscribed in a diameter

b. m∠BCD
28°, inscribed angle is half measure of intercepted arc

c. m∠DBC
62°, sum of angles of a triangle is 180°

d. m∠BFG
28°, inscribed angle is half measure of intercepted arc

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

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

g. Does ∠BGD measure 56°? Explain.
No, the central angle of $$would be 56°. ∠BGD is not a central angle because its vertex is not the center of the circle. h. How do you think we could determine the measure of ∠BGD? Answer: Answers will vary. This leads to today’s lesson. Exercises Exercise 1. Find x, y, a, b, and/or c. Answer: a = 34°, b = 56°, c = 52° Exercise 2. Answer: a = 16°, b = 148° Exercise 3. Answer: a = 86°, b = 43° Exercise 4. Answer: 2(3x + 4y) = 7x + 6y 65 + 65 + (7x + 6y) = 180 x = 5, y = 2.5 Exercise 5. Answer: a = 60° ### Eureka Math Geometry Module 5 Lesson 13 Problem Set Answer Key In Problems 1–9, solve for a, b, and/or c. Question 1. Answer: a = 67° Question 2. Answer: a = 67° Question 3. Answer: a = 67° Question 4. Answer: a = 116° Question 5. Answer: a = 116° Question 6. Answer: a = 116°, b = 64°, c = 26° Question 7. Answer: a = 45°, b = 45° Question 8. Answer: a = 47°, b = 47° Question 9. Answer: a = 57° Question 10. is tangent to circle A. [latex]\overline{D F}$$ is a diameter. Find the angle measurements. a. m∠BCD
50°

b. m∠BAF
80°

c. m∠BDA
40°

d. m∠FBH
40°

e. m∠BGF
98°

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) m∠EBG = 90°    Tangent perpendicular to radius
f = 90 – e     Sum of angles
m∠ECB = 90°     Angle inscribed in semicircle
In △ECB,
b + 90 + e = 180     Sum of angles of a triangle
b = 90 – e
a = b     Angles inscribed in same arc are congruent
a = f     Substitution

(ii) a + c = 180     Inscribed in opposite arcs
a = f      Inscribed in same arc
f + d = 180      Linear pairs form supplementary angles
c + f = f + d    Substitution
c = d

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

Question 1.
Find a, b, and c. 