$A=120,B=C=30$

Since $2\sin B\sin (A+B)=2\sin B\sin C=\cos (C-B)-\cos ( C+B)=\cos (C-B)+\cos A$ Hence: $\cos (C-B)=1$ , so $B=C$ we have $4\sin^2C \cos C= \cos C$ , becasue $C<\frac{\pi}{2}$ , then $\sin C=\frac{1}{2} $ , so$A=120,B=C=30 $

lssl wrote: Since $2\sinh\sin(A+B)=2\sinh\sinh=\cos(C-B)-\cos(C+B)=\cos(A-B)+\cos A$ Seems there are some mistakes: I can't see any equality in fact.

$SinA=–1/2$ neglected bcoz of acute angle