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ss_7_3 - Section 7.3 The Law of Cosines In this section we...

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Section 7.3 The Law of Cosines In this section, we introduce the Law of Cosines, and we use the Law of Cosines to solve triangles that could not be solved with the Law of Sines in the previous section. If a , b , c are the sides of any triangle and α , β , γ are the angles opposite these sides, then c 2 = a 2 + b 2 - 2 ab cos γ Law of Cosines b 2 = a 2 + c 2 - 2 ac cos β Law of Cosines a 2 = b 2 + c 2 - 2 bc cos α Law of Cosines We will use the Law of Cosines to solve triangles of the form SAS and SSS Case of SAS Example. Solve the triangle if a = 3, b = 4, γ = 60 o . Solution. This triangle is of the form SAS. c 2 = a 2 + b 2 - 2 ab cos γ implies c 2 = 3 2 + 4 2 - 2(3)(4) cos 60 o = 9 + 16 - 24( 1 2 ) = 13. Thus c = 13. By the law of cosines, a 2 = b 2 + c 2 - 2 bc cos α . Hence, 9 = 16 + 13 - 2(4) 13 cos α So, cos α = 20 8 13 = 5 2 13 Hence, α = arccos 5 2 13 = 46 . 10 o . β = 180 o - ( α + γ ) = 180 o - (46 . 10 o + 60 o ) = 73 . 9 o . Case of SSS Example. Solve the triangle if a = 2, b = 3, c = 4. Solution. This triangle is of the form SSS. c 2 = a 2 + b 2 - 2 ab cos γ implies 4 2 = 2 2 + 3 2 - 2(2)(3) cos γ . So cos γ = - 3 12 = - 1 4 and γ = 104 . 48 o . a 2 = b 2 + c 2 - 2 bc cos α implies 2 2 = 3 2 + 4 2 - 2(3)(4) cos α . So cos α = 21 24 = 7 8 and α = 28 . 96
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