$12 \cdot 28$ If $A, B$ are two interior angles of an acute $\triangle ABC$, then the complex number
$$
z=(\cos B-\sin A)+i(\sin B-\cos A)
$$

in the complex plane corresponds to a point located in
(A) the first quadrant. (B) the second quadrant. (C) the third quadrant. (D) the fourth quadrant. (China High School Mathematics League, 1989)
[Solution] $\because A+B>\frac{\pi}{2}$, then $0<\frac{\pi}{2}-A<B<\frac{\pi}{2}$. Since $\sin \left(\frac{\pi}{2}-A\right)=\cos A$, it follows that $\cos B-\sin A<0$. Knowing that $z$ is in the second quadrant, the answer is $(B)$.