Stitz-Zeager_College_Algebra_e-book

# B we put theorem 1110 to good use in the following

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Unformatted text preview: manipulating the expression for A2 . 11.3 The Law of Cosines A2 = = = 777 1 ab sin(γ ) 2 2 1 22 2 a b sin (γ ) 4 a2 b2 1 − cos2 (γ ) 4 Using the Law of Cosines, we have cos(γ ) = A2 = since sin2 (γ ) = 1 − cos2 (γ ). a2 +b2 −c2 . 2ab Substituting yields a2 b2 1 − cos2 (γ ) 4 2 a2 + b2 − c2 2ab = a2 b2 1− 4 = a2 + b2 − c2 a2 b2 1− 4 4a2 b2 = a2 b2 4 = 4a2 b2 − a2 + b2 − c2 16 2 4a2 b2 − a2 + b2 − c2 4a2 b2 2 2 2 = = = = = = = = (2ab)2 − a2 + b2 − c2 16 2 + b2 − c2 2ab − a 2ab + a2 + b2 − c2 16 2 − a2 + 2ab − b2 c a2 + 2ab + b2 − c2 16 2 − a2 − 2ab + b2 c a2 + 2ab + b2 − c2 16 2 − (a − b)2 c (a + b)2 − c2 16 (c − (a − b))(c + (a − b))((a + b) − c)((a + b) + c) 16 (b + c − a)(a + c − b)(a + b − c)(a + b + c) 16 (b + c − a) (a + c − b) (a + b − c) (a + b + c) · · · 2 2 2 2 diﬀerence of squares. perfect square trinomials. At this stage, we recognize the last factor as the semiperimeter, s = diﬀerence of sq...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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