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The top of a signal tower is 250 feet above sea level 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? A. about 81.4 ft B. about 236.4 ft C. about 726 ft D. about 804 ft 5-Minute Check 4

You used trigonometric ratios to solve right triangles. Use the Law of Sines to solve triangles. Use the Law of Cosines to solve triangles. Then/Now

Law of Sines Law of Cosines Vocabulary

Concept

Find p. Round to the nearest tenth. 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. Example 1

Cross Products Property Law of Sines (AAS or ASA) Law of Sines Cross Products Property Divide each side by sin Use a calculator. Answer: p ≈ 4.8 Example 1

Find c to the nearest tenth. B. 29.9 C. 7.8 D. 8.5 Example 1

Find x. Round to the nearest tenth. Law of Sines (ASA) Find x. Round to the nearest tenth. 6 x 57° Example 2

6 sin 50 = x sin 73 Cross Products Property Law of Sines (ASA) Law of Sines mB = 50, mC = 73, c = 6 6 sin 50 = x sin 73 Cross Products Property Divide each side by sin 73. 4.8 = x Use a calculator. Answer: x ≈ 4.8 Example 2

Find x. Round to the nearest degree. B. 10 C. 12 D. 14 43° x Example 2

Concept

Find x. Round to the nearest tenth. 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. Example 3

Take the square root of each side. Law of Cosines (SAS) Law of Cosines Simplify. Take the square root of each side. Use a calculator. Answer: x ≈ 18.9 Example 3

Find r if s = 15, t = 32, and mR = 40. Round to the nearest tenth. B. 44.5 C. 22.7 D. 21.1 Example 3

Find mL. Round to the nearest degree. Law of Cosines (SSS) Find mL. Round to the nearest degree. Law of Cosines Simplify. Example 4

Subtract 754 from each side. Law of Cosines (SSS) Subtract 754 from each side. Divide each side by –270. Solve for L. Use a calculator. Answer: mL ≈ 49 Example 4

Find mP. Round to the nearest degree. B. 51° C. 56° D. 69° Example 4

Indirect Measurement AIRCRAFT From the diagram of the plane shown, determine the approximate width of each wing. Round to the nearest tenth meter. Example 5

Use the Law of Sines to find KJ. Indirect Measurement Use the Law of Sines to find KJ. Law of Sines Cross products Example 5

Answer: The width of each wing is about 16.9 meters. Indirect Measurement Divide each side by sin . Simplify. Answer: The width of each wing is about 16.9 meters. Example 5

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. A. 93.5 in. B. 103.5 in. C. 96.7 in. D. 88.8 in. Example 5

Solve triangle PQR. Round to the nearest degree. 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 mP. p2 = r2 + q2 – 2pq cos P Law of Cosines 82 = 92 + 72 – 2(9)(7) cos P p = 8, r = 9, and q = 7 Example 6

–66 = –126 cos P Subtract 130 from each side. Solve a Triangle 64 = 130 – 126 cos P Simplify. –66 = –126 cos P Subtract 130 from each side. Divide each side by –126. Use the inverse cosine ratio. Use a calculator. Example 6

Use the Law of Sines to find mQ. Solve a Triangle Use the Law of Sines to find mQ. Law of Sines mP ≈ 58, p = 8, q = 7 Multiply each side by 7. Use the inverse sine ratio. Use a calculator. Example 6

By the Triangle Angle Sum Theorem, mR ≈ 180 – (58 + 48) or 74. Solve a Triangle By the Triangle Angle Sum Theorem, mR ≈ 180 – (58 + 48) or 74. Answer: Therefore, mP ≈ 58; mQ ≈ 48 and mR ≈ 74. Example 6

Solve ΔRST. Round to the nearest degree. A. mR = 82, mS = 58, mT = 40 B. mR = 58, mS = 82, mT = 40 C. mR = 82, mS = 40, mT = 58 D. mR = 40, mS = 58, mT = 82 Example 6

Concept

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