$7 \cdot 30$ In $\triangle A B C$, if $3 \sin A+4 \cos B=6$ and $4 \sin B+3 \cos A=1$, then the degree measure of $\angle C$ is
(A) 30 .
(B) 60 .
(C) 90 .
(D) 120 .
(E) 150 .
(50th American High School Mathematics Examination, 1999)
[Sol] Squaring both sides of the given equations and then adding them respectively, we get:
$$
\begin{array}{l}
9\left(\sin ^{2} A+\cos ^{2} A\right)+24(\sin A \cos B+\cos A \sin B)+16\left(\sin ^{2} B+\cos ^{2} B\right) \\
=37,
\end{array}
$$

Simplifying, we obtain $\sin (A+B)=\frac{1}{2}$.
Solving, we get $A+B=30^{\circ}$ or $150^{\circ}$. The former is not appropriate, so we know $A+B=150^{\circ}$, which implies $C=$ $30^{\circ}$.
Therefore, the answer is $(A)$.