778 CHAPTER 8 Applications of Trigonometry 55.38 B1 A c 1 b = 24.9 a = 22.8 C1 63.98 A a = 22.8 b = 24.9 B2 C2 c2 55.38 116.18 Figure 7 Now separately solve triangles AB1C1 and AB2C2 shown in Figure 7. Begin with AB1C1. Find angle C1 first. C1 = 180° - A - B1 Angle sum formula, solved for C1 C1 = 180° - 55.3° - 63.9° Substitute. C1 = 60.8° Subtract. Now use the law of sines to find side c1. a sin A = c1 sin C1 22.8 sin 55.3° = c1 sin 60.8° Substitute. c1 = 22.8 sin 60.8° sin 55.3° Multiply by sin 60.8° and rewrite. c1 ≈24.2 ft Use a calculator. To solve triangle AB2C2, first find angle C2. C2 = 180° - A - B2 Angle sum formula, solved for C2 C2 = 180° - 55.3° - 116.1° Substitute. C2 = 8.6° Subtract. Use the law of sines to find side c2. a sin A = c2 sin C2 22.8 sin 55.3° = c2 sin 8.6° Substitute. c2 = 22.8 sin 8.6° sin 55.3° Multiply by sin 8.6° and rewrite. c2 ≈4.15 ft Use a calculator. S Now Try Exercise 45. Solve for c1. Solve for c2 . The ambiguous case results in zero, one, or two triangles. The following guidelines can be used to determine how many triangles there are. Number ofTriangles Satisfying the Ambiguous Case (SSA) Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate the value of sin B.) 1. If applying the law of sines results in an equation having sin B71, then no triangle satisfies the given conditions. 2. If sin B = 1, then one triangle satisfies the given conditions and B = 90°. 3. If 0 6sin B61, then either one or two triangles satisfy the given conditions. (a) If sin B = k, then let B1 = sin-1 k and use B 1 for B in the first triangle. (b) Let B2 = 180° - B1. If A + B2 6180°, then a second triangle exists. In this case, use B2 for B in the second triangle.
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