Unformatted text preview: A , we get a 2 = b 2 + c 2 − 2 bc cos A ⇒ cos A = b 2 + c 2 − a 2 2 bc = 3 2 + 6 2 − 2 2 2(3)(6) = 1.139 , which is impossible since  cos A ≤ 1. Thus, there is no solution . b = 3 c = 6 a = 2 We could have saved ourselves some effort by recognizing that the length of one of the sides ( c = 6) is greater than the sums of the lengths of the remaining sides ( a = 2 and b = 3), which (as the picture on the right shows) is impossible in a triangle. The Law of Cosines can also be used to solve triangles in Case 2 (two sides and one opposite angle), though it is less commonly used for that purpose than the Law of Sines. The following example gives an idea of how to do this....
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 Fall '11
 Dr.Cheun
 Calculus, Law Of Cosines, Law of sines, Cos, triangle

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