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8-6 Law of Sines The Law of Sines can be used to find missing parts of triangles that are not right triangles. Theorem 8.8: Let ∆ABC be any triangle with a, b, and c representing the measures of the sides opposite the angles with measures A, B, and C, respectively. Then sin A sin B sin C a b c B c a A b C Video Link Law of Sines Let’s review the Proof for the Law of Sines on page 471. Example 1 A: If m∠B = 32, m∠C = 51, c = 12, find a. sin A sin C 180 (32 51) 97 mA sin 97 sin 51 a c a 12 a 15.3 12(sin 97) a sin 51 Example 1 B: If a = 22, b = 18, m∠A = 25, find m∠B. sin A sin B sin 25 sin B sin 25 (18) sin B 22 22 18 a b sin 25 sin 1 (18) B 22 20 B Example 2 Solving a Triangle: Finding the measures of all of the angles and sides of a triangle. Find the missing angles and sides of ∆PQR. Round angle measures to the nearest degree and side measures to the nearest tenth. A: m∠R = 66,m∠Q = 59,p = 72 Answer: m∠P = 55, q ≈ 75.3, r ≈ 80.3 B: p = 32, r = 11, m∠P = 105 Answer: m∠R = 19, m∠Q = 56, q = 27.5 Video Link Law of Sines – Missing Side Example 3 Two radar stations that are 35 miles apart located a plane at the same time. The first station indicated that the position of the plane made an angle of 37° with the line between the stations. The second station indicated that it made an angle of 54° with the same line. How far is each station from the plane? Draw a picture – label the parts – use law of sines to find the missing pieces. Answer: about 21.1 mi, about 28.3 mi Key Concepts The Law of Sines can be used to solve a triangle in the following cases. Case 1: You know the measures of two angles and any side of a triangle. (AAS or ASA) Case 2: You know the measures of two sides and an angle opposite one of these sides of the triangle. (SSA) Handout – Extra Examples for additional understanding. Homework #54 p. 475 13-29 odd, 34, 38-39