Prove that in a triangle ABC,:

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

implies:

A = B = 45° and C = 90°.

(In reply to solution by jacob)

sin(a)sin(b)sin(c) <= sin(a)sin(b) only if
sin(a)sin(b)>=0. I like the style of the
proof, but this should be fixed.

Posted by Federico Kereki on 2004-02-27 06:43:35