ss_7_3

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
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This note was uploaded on 12/11/2011 for the course MAC 1140 taught by Professor Kutter during the Fall '11 term at FSU.

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

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