861 CHAPTER 8 Test Prep Solve triangle ABC, given A = 44.5°, a = 11.0 in., and c = 7.0 in. Find angle C. sin C 7.0 = sin 44.5° 11.0 Law of sines sin C≈0.4460 Solve for sin C. C≈26.5° Use the inverse sine function. Another angle with this sine value is 180° - 26.5° ≈153.5°. However, 153.5° + 44.5° 7180°, so there is only one triangle. B = 180° - 44.5° - 26.5° Angle sum formula B = 109° Subtract. Use the law of sines again to solve for b. b ≈14.8 in. Law of Cosines In any triangle ABC, with sides a, b, and c, the following hold true. a2 =b2 +c2 −2bc cos A b2 =a2 +c2 −2 ac cos B c2 =a2 +b2 −2 ab cos C Heron’s Area Formula If a triangle has sides of lengths a, b, and c, with semi- perimeter s = 1 2 1 a +b +c2, then the area of the triangle is given by the following. =!s1s −a2 1s −b2 1s −c2 Ambiguous Case If we are given the lengths of two sides and the angle opposite one of them (for example, A, a, and b in triangle ABC), then it is possible that zero, one, or two such triangles exist. If A is acute, h is the altitude from C, and • a 6h 6b, then there is no triangle. • a = h and h 6b, then there is one triangle (a right triangle). • a Ú b, then there is one triangle. • h 6a 6b, then there are two triangles. If A is obtuse and • a … b, then there is no triangle. • a 7b, then there is one triangle. See the guidelines in this section that illustrate the possible outcomes. In triangle ABC, find C if a = 11 units, b = 13 units, and c = 20 units. Then find its area. c2 = a2 + b2 - 2ab cos C Law of cosines 202 = 112 + 132 - 211121132 cos C Substitute. 400 = 121 + 169 - 286 cos C Square and multiply. cos C = 400 - 121 - 169 -286 Solve for cos C. cos C≈ -0.38461538 Use a calculator. C≈113° Use the inverse cosine function. The semiperimeter s of the above triangle is s = 1 2 111 + 13 + 202 = 22, so the area is = 222122 - 112122 - 132122 - 202 = 66 sq units. 8.2 The Law of Cosines Concepts Examples
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