Prove that in a triangle ABC,:

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

implies:

A = B = 45° and C = 90°.

If C=90, then we can show A=B=45 pretty easily:

1=
sin(A)sin(B)sin(C)+cos(A)cos(B) =
sin(A)sin(90-A)+cos(A)cos(90-A) =
sin(A)cos(A)+cos(A)sin(A)=
2sin(A)cos(A)=
sin(2A)

so A=45.

Posted by Brian Wainscott on 2004-02-26 16:13:30