47 If the acute angles $A, B$ of triangle $ABC$ satisfy the equation $\sin ^{2} A+\sin ^{2} B=\sin (A+B)$, then this triangle is
A. Acute triangle
B. Right triangle
C. Obtuse triangle
D. Any triangle
47 B. From the given, we have $\sin A(\sin A-\cos B)=\sin B(\cos A-\sin B)$. If $\sin A>\cos B$, $\cos A>\sin B$, then knowing that $A, B$ are acute angles, we get $\sin ^{2} A+\cos ^{2} A<\cos ^{2} B+\sin ^{2} B=1$, which is impossible. Similarly, it is also impossible to have $\sin A<\cos B$ and $\cos A<\sin B$. Therefore, it can only be that $\sin A=\cos B, \cos A=\sin B$, thus $\sin (A+B)=1, A+B=\frac{\pi}{2}$.