777 8.1 The Law of Sines 558 409 b = 8.94 a = 25.1 B Figure 5 EXAMPLE 4 Solving the Ambiguous Case (No SuchTriangle) Solve triangle ABC if B = 55° 40′, b = 8.94 m, and a = 25.1 m. SOLUTION We are given B, b, and a. We use the law of sines to find angle A. sin A a = sin B b Law of sines (alternative form) sin A 25.1 = sin 55° 40′ 8.94 Substitute the given values. sin A = 25.1 sin 55° 40′ 8.94 Multiply by 25.1. sin A ≈2.3184379 Use a calculator. Because sin A cannot be greater than 1, there can be no such angle A — and thus no triangle with the given information. An attempt to sketch such a triangle leads to the situation shown in Figure 5. S Now Try Exercise 37. Choose a form that has the unknown variable in the numerator. NOTE In the ambiguous case, we are given two sides and an angle opposite one of the sides (SSA). For example, suppose b, c, and angle C are given. This situation represents the ambiguous case because angle C is opposite side c. y x 63.9° 63.9° 116.1° –1 –1 1 (20.4399, 0.8979) 1 (0.4399, 0.8979) Figure 6 EXAMPLE 5 Solving the Ambiguous Case (TwoTriangles) Solve triangle ABC if A = 55.3°, a = 22.8 ft, and b = 24.9 ft. SOLUTION To begin, use the law of sines to find angle B. sin A a = sin B b sin 55.3° 22.8 = sin B 24.9 Substitute the given values. sin B = 24.9 sin 55.3° 22.8 Multiply by 24.9 and rewrite. sin B ≈0.8978678 Use a calculator. There are two angles B between 0° and 180° that satisfy this condition. See Figure 6. Because sin B ≈0.8978678, one value of angle B, to the nearest tenth, is B1 = 63.9°. Use the inverse sine function. Supplementary angles have the same sine value, so another possible value of B is B2 = 180° - 63.9° = 116.1°. To see whether B2 = 116.1° is a valid possibility, add 116.1° to the measure of A, 55.3°. Because 116.1° + 55.3° = 171.4°, and this sum is less than 180°, it is a valid angle measure for this triangle. Solve for sin B.
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