# Eureka Math Precalculus Module 4 Lesson 5 Answer Key

## Engage NY Eureka Math Precalculus Module 4 Lesson 5 Answer Key

### Eureka Math Precalculus Module 4 Lesson 5 Exercise Answer Key

Exercises 1–12
The circle shown to the right is a unit circle, and the length of $$\widehat{D A}$$ is $$\frac{\pi}{3}$$ radians. Exercise 1.
Which segment in the diagram has length sin⁡($$\frac{\pi}{3}$$)?
The sine of $$\frac{\pi}{3}$$ is the vertical component of point A, so $$\overline{A B}$$ has length sin($$\frac{\pi}{3}$$).

Exercise 2.
Which segment in the diagram has length cos⁡($$\frac{\pi}{3}$$)?
The cosine of $$\frac{\pi}{3}$$ is the horizontal component of point A, so $$\overline{O B}$$ has length cos($$\frac{\pi}{3}$$).

Exercise 3.
Which segment in the diagram has length tan⁡($$\frac{\pi}{3}$$)?
Since △OAB is similar to △OCD, $$\frac{C D}{1}$$ = $$\frac{A B}{O B}$$ = $$\frac{\sin \left(\frac{\pi}{3}\right)}{\cos \left(\frac{\pi}{3}\right)}$$. Thus, $$\overline{C D}$$ has
length tan($$\frac{\pi}{3}$$).

Exercise 4.
Which segment in the diagram has length sec⁡($$\frac{\pi}{3}$$)?
Since △OAB is similar to △OCD, $$\frac{OD}{OB}$$ = $$\frac{OC}{OA}$$. Since OA = OD = 1, we have OC = $$\frac{1}{OB}$$ = $$\frac{1}{\cos \left(\frac{\pi}{3}\right)}$$. Thus, $$\overline{O C}$$ has length sec($$\frac{\pi}{3}$$).

Exercise 5.
Use a compass to construct the tangent lines to the given circle that pass through the given point. Sample solution: Exercise 6.
Analyze the construction shown below. Argue that the lines shown are tangent to the circle with center B. Since point D is on circle C, we know that ∠BDA is a right angle. Since point D is on circle B and ∠BDA is a right angle, it follows that (AD) ⃡ is tangent to circle B. In the same way, we can show that $$\overleftrightarrow{A E}$$ is tangent to circle B.

Exercise 7.
Use a compass to construct a line that is tangent to the circle below at point F. Then choose a point G on the tangent line, and construct another tangent to the circle through G. Sample solution: Exercise 8.
The circles shown below are unit circles, and the length of $$\widehat{D A}$$ is $$\frac{\pi}{3}$$ radians. Which trigonometric function corresponds to the length of $$\overline{E F}$$?
The length of $$\overline{E F}$$ represents the cotangent of $$\frac{\pi}{3}$$.

Exercise 9.
Which trigonometric function corresponds to the length of $$\overline{O F}$$?
The length of $$\overline{O F}$$ represents the cosecant of $$\frac{\pi}{3}$$.

Exercise 10.
Which trigonometric identity gives the relationship between the lengths of the sides of △OEF?
cot2 ($$\frac{\pi}{3}$$) + 12 = csc2 ($$\frac{\pi}{3}$$)

Exercise 11.
Which trigonometric identities give the relationships between the corresponding sides of △OEF and △OGA?
We have $$\frac{1}{\sin \left(\frac{\pi}{3}\right)}$$ = $$\frac{\cot \left(\frac{\pi}{3}\right)}{\cos \left(\frac{\pi}{3}\right)}$$ and $$\frac{1}{\sin \left(\frac{\pi}{3}\right)}$$ = $$\frac{\csc \left(\frac{\pi}{3}\right)}{1}$$.

Exercise 12.
What is the value of csc($$\frac{\pi}{3}$$)? What is the value of cot($$\frac{\pi}{3}$$)? Use the Pythagorean theorem to support your answers.
Let x = EF. Then we have x2 + 12 = (2x)2 = 4x2, which means 3x2 = 1; therefore, x = $$\sqrt{\frac{1}{3}}$$. So cot($$\frac{\pi}{3}$$) = $$\sqrt{\frac{1}{3}}$$
and csc($$\frac{\pi}{3}$$) = 2$$\sqrt{\frac{1}{3}}$$.

### Eureka Math Precalculus Module 4 Lesson 5 Problem Set Answer Key

Question 1.
Prove Thales’ theorem: If A, B, and P are points on a circle where (AB) ̅ is a diameter of the circle, then ∠APB is a right angle. Since OA = OP = OB, △OPA and △OPB are isosceles triangles. Therefore, m∠OAP = m∠OPA, and m∠OPB = m∠OBP.
Let m∠OPA = α and m∠OPB = β. The sum of three internal angles of △APB equals 180°.
Therefore, α + (α + β) + β = 180°, so 2α + 2β = 180°, and α + β = 90°. Since m∠APB = α + β, we have m∠APB = 90°, so ∠APB is a right angle.

Question 2.
Prove the converse of Thales’ theorem: If $$\overline{A B}$$ is a diameter of a circle and P is a point so that ∠APB is a right angle, then P lies on the circle for which $$\overline{A B}$$ is a diameter. Construct the right triangle, △APB.
Construct the line h that is parallel to $$\overline{P B}$$ through point A.
Construct the line g that is parallel to $$\overline{A P}$$ through point B.
Let C be the intersection of lines h and g.
The quadrilateral ACBP forms a parallelogram by construction.
By the properties of parallelograms, the adjacent angles are supplementary. Since ∠APB is a right angle, it follows that angles ∠CAP, ∠BCA, and ∠PBC are also right angles. Therefore, the quadrilateral ACBP is a rectangle.

Let O be the intersection of the diagonals $$\overline{A B}$$ and $$\overline{C P}$$. Then, by the properties of parallelograms, point O is the midpoint of $$\overline{A B}$$ and $$\overline{C P}$$, so OA = OB = OC = OP. Therefore, O is the center of the circumscribing circle, and the hypotenuse of △APB, $$\overline{A B}$$, is a diameter of the circle.

Question 3.
Construct the tangent lines from point P to the circle given below. Mark any three points A, B, and C on the circle, and construct perpendicular bisectors of $$\overline{A B}$$ and $$\overline{B C}$$.
Let O be the intersection of the two perpendicular bisectors. Construct the midpoint H of $$\overline{O P}$$. Construct a circle with center H and radius OH.
The circle centered at H will intersect the original circle O at points A and B.
Construct two tangent lines $$\overleftrightarrow{P A}$$ and $$\overleftrightarrow{P B}$$. Question 4.
Prove that if segments from a point P are tangent to a circle at points A and B, then $$\overline{P A}$$ ≅ $$\overline{P B}$$. Let P be a point outside of a circle with center O, and let A and B be points on the circle so that $$\overline{P A}$$ and $$\overline{P B}$$ are tangent to the circle. Then, OA = OB, OP = OP, and m∠OAP = m∠OBP = 90°, so △PAO≅ △PBO by the Hypotenuse Leg congruence criterion. Therefore, $$\overline{P A}$$ ≅ $$\overline{P B}$$ because corresponding parts of congruent triangles are congruent.

Question 5.
Given points A, B, and C so that AB = AC, construct a circle so that $$\overline{A B}$$ is tangent to the circle at B and $$\overline{A C}$$ is tangent to the circle at C. Construct a perpendicular bisector of $$\overline{A B}$$.
Construct a perpendicular bisector of $$\overline{A C}$$.
The perpendicular bisectors will intersect at point H.
Construct a line through points A and H.
Construct a circle with center H and radius $$\overline{H A}$$.
The circle centered at H will intersect $$\overleftrightarrow{H A}$$ at I.
Construct a circle centered at I with radius $$\overline{I B}$$. ### Eureka Math Precalculus Module 4 Lesson 5 Exit Ticket Answer Key

Question 1.
Use a compass and a straightedge to construct the tangent lines to the given circle that pass through the given point.  Since points D and E are on circle C, ∠BDA and ∠BEA are right angles. Thus, $$\overleftrightarrow{A D}$$ and $$\overleftrightarrow{A E}$$ are tangent to circle B.