779 8.1 The Law of Sines EXAMPLE 6 Solving the Ambiguous Case (OneTriangle) Solve triangle ABC, given A = 43.5°, a = 10.7 in., and c = 7.2 in. SOLUTION Find angle C. sin C c = sin A a Law of sines (alternative form) sin C 7.2 = sin 43.5° 10.7 Substitute the given values. sin C = 7.2 sin 43.5° 10.7 Multiply by 7.2. sin C≈0.46319186 Use a calculator. C≈27.6° Use the inverse sine function. There is another angle C that has sine value 0.46319186. It is C = 180° - 27.6° = 152.4°. However, notice in the given information that c 6a, meaning that in the triangle, angle C must have measure less than angle A. Notice also that when we add this obtuse value to the given angle A = 43.5°, we obtain 152.4° + 43.5° = 195.9°, which is greater than 180°. Thus either of these approaches shows that there can be only one triangle. See Figure 8. The measure of angle B can be found next. B = 180° - 27.6° - 43.5° Substitute. B = 108.9° Subtract. We can find side b with the law of sines. b sin B = a sin A Law of sines b sin 108.9° = 10.7 sin 43.5° Substitute known values. b = 10.7 sin 108.9° sin 43.5° Multiply by sin 108.9°. b ≈14.7 in. Use a calculator. S Now Try Exercise 41. When solving triangles, it is important to analyze the given information to determine whether it forms a valid triangle. A B b C 7.2 in. 10.7 in. 43.58 27.68 Figure 8 EXAMPLE 7 Analyzing Data Involving an Obtuse Angle Without using the law of sines, explain why A = 104°, a = 26.8 m, and b = 31.3 m cannot be valid for a triangle ABC. SOLUTION Because A is an obtuse angle, it is the largest angle, and so the longest side of the triangle must be a. However, we are given b 7a. Thus, B7A, which is impossible if A is obtuse. Therefore, no such triangle ABC exists. S Now Try Exercise 53.
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