## 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(\(\frac{\pi}{3}\))?

Answer:

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(\(\frac{\pi}{3}\))?

Answer:

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(\(\frac{\pi}{3}\))?

Answer:

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(\(\frac{\pi}{3}\))?

Answer:

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.

Answer:

Sample solution:

Exercise 6.

Analyze the construction shown below. Argue that the lines shown are tangent to the circle with center B.

Answer:

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.

Answer:

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 \(\overline{E F}\)?

Answer:

The length of \(\overline{E F}\) represents the cotangent of \(\frac{\pi}{3}\).

Exercise 9.

Which trigonometric function corresponds to the length of \(\overline{O F}\)?

Answer:

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?

Answer:

cot^{2} (\(\frac{\pi}{3}\)) + 1^{2} = csc^{2} (\(\frac{\pi}{3}\))

Exercise 11.

Which trigonometric identities give the relationships between the corresponding sides of △OEF and △OGA?

Answer:

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.

Answer:

Let x = EF. Then we have x^{2} + 1^{2} = (2x)^{2} = 4x^{2}, which means 3x^{2} = 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.

Answer:

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.

Answer:

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.

Answer:

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 \(\overline{P A}\) ≅ \(\overline{P B}\).

Answer:

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.

Answer:

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.

Answer:

Question 2.

Explain why your construction produces lines that are indeed tangent to the given circle.

Answer:

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.