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Pre-Calc Exam Notes 47

# Pre-Calc Exam Notes 47 - A we get a 2 = b 2 c 2 − 2 bc...

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The Law of Cosines Section 2.2 47 It remains to solve a triangle in Case 4, i.e. given three sides. We will now see how to use the Law of Cosines for that case. Example2.6 b = 3 a = 2 c = 4 A B C Case 4: Three sides. Solve the triangle ABC given a = 2, b = 3, and c = 4. Solution: We will use the Law of Cosines to find B and C , then use A = 180 B C . First, we use the formula for b 2 to find B : b 2 = c 2 + a 2 2 ca cos B cos B = c 2 + a 2 b 2 2 ca cos B = 4 2 + 2 2 3 2 2(4)(2) = 0 . 6875 B = 46 . 6 Now we use the formula for c 2 to find C : c 2 = a 2 + b 2 2 ab cos C cos C = a 2 + b 2 c 2 2 ab cos C = 2 2 + 3 2 4 2 2(2)(3) = − 0 . 25 C = 104 . 5 Thus, A = 180 B C = 180 46 . 6 104 . 5 A = 28 . 9 . It may seem that there is always a solution in Case 4 (given all three sides), but that is not true, as the following example shows. Example2.7 b = 3 a = 2 c = 6 A B C Case 4: Three sides. Solve the triangle ABC given a = 2, b = 3, and c = 6. Solution: If we blindly try to use the Law of Cosines to find A , we get
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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 op-posite 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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