Five-Minute Check (over Lesson 8–5)

NGSSS

Then/Now

New Vocabulary

Theorem 8.10: Law of Sines

Example 1: Law of Sines (AAS or ASA)

Example 2: Law of Sines (SSA)

Theorem 8.11: Law of Cosines

Example 3: Law of Cosines (SAS)

Example 4: Law of Cosines (SSS)

Example 5: Real-World Example: Indirect Measurement

Example 6: Solve a Triangle

Concept Summary: Solving a Triangle

Over Lesson 8–5

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. URT

B. SRT

C. RST

D. SRU

Name the angle of depression in the figure.

Over Lesson 8–5

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. about 70.6°

B. about 60.4°

C. about 29.6°

D. about 19.4°

Find the angle of elevation of the Sun when a 6-meter flagpole casts a 17-meter shadow.

Over Lesson 8–5

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. about 1.8°

B. about 2.4°

C. about 82.4°

D. about 88.6°

After flying at an altitude of 575 meters, a helicopter starts to descend when its ground distance from the landing pad is 13.5 kilometers. What is the angle of depression for this part of the flight?

Over Lesson 8–5

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. about 81.4 ft

B. about 236.4 ft

C. about 726 ft

D. about 804 ft

The top of a signal tower is 250 feet above sea level. The angle of depression from the top of the tower to a passing ship is 19°. How far is the foot of the tower from the ship?

Over Lesson 8–5

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. 50 ft

B. 104 ft

C. 1060 ft

D. 4365 ft

Jay is standing 50 feet away from the Eiffel Tower and measures the angle of elevation to the top of the tower as 87.3°. Approximately how tall is the Eiffel Tower?

MA.912.T.2.1 Define and use the trigonometric ratios (sine, cosine, tangent, cotangent, secant, cosecant) in terms of angles of right triangles.

You used trigonometric ratios to solve right triangles. (Lesson 8–4)

• Use the Law of Sines to solve triangles.

• Use the Law of Cosines to solve triangles.

• Law of Sines

• Law of Cosines

Law of Sines (AAS or ASA)

Find p. Round to the nearest tenth.

We are given measures of two angles and a nonincluded side, so use the Law of Sines to write a proportion.

Law of Sines (AAS or ASA)

Law of Sines

Use a calculator.

Divide each side by sin

Cross Products Property

Answer: p ≈ 4.8

A. A

B. B

C. C

D. D A B C D

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A. 4.6

B. 29.9

C. 7.8

D. 8.5

Find c to the nearest tenth.

Law of Sines (SSA)

Find x. Round to the nearest degree.

Law of Sines (SSA)

Law of Sines

mB = 50, b = 10, a = 11

Cross Products Property

Divide each side by 10.

Use a calculator.

Use the inverse sine ratio.

Answer: x ≈ 57.4

A. A

B. B

C. C

D. D A B C D

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A. 39

B. 43

C. 46

D. 49

Find x. Round to the nearest degree.

Law of Cosines (SAS)

Find x. Round to the nearest tenth.

Use the Law of Cosines since the measures of two sides and the included angle are known.

Law of Cosines (SAS)

Answer: x ≈ 18.9

Simplify.

Take the square root of each side.

Law of Cosines

Use a calculator.

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. 25.1

B. 44.5

C. 22.7

D. 21.1

Find r if s = 15, t = 32, and mR = 40. Round to the nearest tenth.

Law of Cosines (SSS)

Find mL. Round to the nearest degree.

Law of Cosines

Simplify.

Law of Cosines (SSS)

Answer: mL ≈ 49

Solve for L.

Use a calculator.

Subtract 754 from each side.

Divide each side by –270.

A. A

B. B

C. C

D. D A B C D

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A. 44°

B. 51°

C. 56°

D. 69°

Find mP. Round to the nearest degree.

Indirect Measurement

AIRCRAFT From the diagram of the plane shown, determine the approximate width of each wing. Round to the nearest tenth meter.

Indirect Measurement

Cross products

Law of Sines

Use the Law of Sines to find KJ.

Indirect Measurement

Answer: The width of each wing is about 16.9 meters.

Simplify.

Divide each side by sin .

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. 93.5 in.

B. 103.5 in.

C. 96.7 in.

D. 88.8 in.

The rear side window of a station wagon has the shape shown in the figure. Find the perimeter of the window if the length of DB is 31 inches. Round to the nearest tenth.

Solve a Triangle

Solve triangle PQR. Round to the nearest degree.

Since the measures of three sides are given (SSS), use the Law of Cosines to find mP.

p2 = r2 + q2 – 2pq cos P Law of Cosines

82 = 92 + 72 – 2(9)(7) cos P p = 8, r = 9, and q = 7

Solve a Triangle

64 = 130 – 126 cos P Simplify.

–66 = –126 cos P Subtract 130 fromeach side.

Divide each side by –126.

Use the inversecosine ratio.

Use a calculator.

Solve a Triangle

Use the Law of Sines to find mQ.

Law of Sines

Multiply each side by 7.

Use the inversesine ratio.

Use a calculator.

mP ≈ 58, p = 8,q = 7.

Solve a Triangle

Answer: Therefore, mP ≈ 58; mQ ≈ 48 andmR ≈ 74.

By the Triangle Angle Sum Theorem, mR ≈ 180 – (58 + 48) or 74.

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. mR = 82, mS = 58, mT = 40

B. mR = 58, mS = 82, mT = 40

C. mR = 82, mS = 40, mT = 58

D. mR = 40, mS = 58, mT = 82

Solve ΔRST. Round to the nearest degree.