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 December 1st, 2011, 04:32 AM #1 Member   Joined: Dec 2011 Posts: 61 Thanks: 0 perimeter The triangle ABC has sides of length a,b and c.Show that perimeter of triangle ABC satisfies p^2 = 2bc(1 + cos A) + 2ac(1 + cos B) + 2ab(1 + cos C)
 December 1st, 2011, 05:14 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: perimeter The perimeter p is p = a + b + c. By the law of cosines, we may state: $2ab\cos C=a^2+b^2-c^2$ $2ac\cos B=a^2+c^2-b^2$ $2bc\cos A=b^2+c^2-a^2$ Adding, we get: $2ab\cos C+2ac\cos B+2bc\cos A=a^2+b^2+c^2$ Add 2ab + 2ac + 2bc to both sides: $2ab\cos C+2ac\cos B+2bc\cos A+2ab+2ac+2bc=a^2+b^2+c^2+2ab+2ac+2bc$ $2ab$$\cos C+1$$+2ac$$\cos B+1$$+2bc$$\cos A+1$$=$$a+b+c$$^2$ $p^2=2bc(1+\cos A)+2ac(1+\cos B)+2ab(1+\cos C)$

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