Prove that in a triangle ABC,:

sin(A)sin(B)sin(C) + cos(A)cos(B) = 1

implies:

A = B = 45° and C = 90°.

replace sin(c) with sin(180-(a+b))= sin(a+b)=

cos(a)sin(b)+    sin(a)cos(b)

take partial with respect to A,B and set =0

probably leads to angle A= angle B etc.

Posted by red_sox_fan_032003 on 2004-02-26 16:19:24