562 CHAPTER 8 Applications of Trigonometric Functions Using the Law of Sines to Solve an SSA Triangle (No Solution) Solve the triangle: = = = ° a c C 2, 1, 50 Solution EXAMPLE 3 Because = = a c 2, 1, and = ° C 50 are known, use the Law of Sines to find the angle A. = = ° = ° ≈ A a C c A A sin sin sin 2 sin50 1 sin 2 sin50 1.53 Since there is no angle A for which > A sin 1, there is no triangle with the given measurements. Figure 28 illustrates the measurements given. Note that no matter how side c is positioned, it will never touch side b to form a triangle. Figure 28 a 5 2 c 5 1 b 508 Two Triangles If > = a h b A sin and < a b, then two distinct triangles can be formed from the given information. See Figure 26. One Triangle If ≥ a b, only one triangle can be formed. See Figure 27. Figure 26 > a b A sin and < a b a a A b a A b h 5 b sin A Figure 27 ≥ a b Fortunately, it is not necessary to rely on a figure or on complicated relationships to draw the correct conclusion in the ambiguous case. The Law of Sines leads us to the correct determination. Let’s see how. Using the Law of Sines to Solve an SSA Triangle (One Solution) Solve the triangle: = = = ° a b A 3, 2, 40 Solution EXAMPLE 4 See Figure 29(a). Because = = a b 3, 2, and = ° A 40 are known, use the Law of Sines to find the angle B. = A a B b sin sin Then ° = = ° ≈ B B sin40 3 sin 2 sin 2 sin40 3 0.43 There are two angles ° < < ° B B , 0 180 , for which ≈ B sin 0.43. ≈ ° ≈ °− ° = ° B B 25.4 and 180 25.4 154.6 1 2 The angle ≈ ° B 154.6 2 is discarded because the sum of the angles in a triangle must equal ° 180 and + ≈ ° + ° = ° > ° A B 40 154.6 194.6 180 . 2 Using ≈ ° B 25.4 1 gives = °−− ≈ °−°− °= ° C A B 180 180 40 25.4 114.6 1 NOTE The angle B1 was determined by finding the value of sin 2sin40 3 . 1( ) ° − Using the rounded value and evaluating sin 0.43 1( ) − will yield a slightly different result. j Figure 29(a) 408 2 3 c C B
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