Prove that in a triangle ABC,:

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

implies:

A = B = 45° and C = 90°.

Take a comparism

Sin(A)Sin(A) + Cos(A)Cos(A) = 1

This is standard trig.

So

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

for this to hold true

sin(C)=1

therefore C = 90°

and sin(B) = sin(A) giving A = B = 45°

Posted by Nebo on 2004-08-19 10:44:27