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 re: solution by Federico Kereki)

Since the angles of a triangle add up to 180 degrees, no one angle can be greater than 180 degrees, sin(a)sin(b)>=0.  In fact equality is achieved only in the degenerate triangle where a and b are each zero degrees and c is 180, which is another "solution" if degenerate triangles are allowed.

Posted by Charlie on 2004-02-27 07:22:17