Stitz-Zeager_College_Algebra_e-book

B we put theorem 1110 to good use in the following

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online