In $\triangle A B C$, $3 \sin A+4 \cos B=6,4 \sin B+3 \cos A=1$, then the degree of $\angle C$ is ( )
(A) $30^{\circ}$
(B) $60^{\circ}$
(C) $120^{\circ}$
(D) $30^{\circ}$ or $150^{\circ}$
5. (A).
$(3 \sin A+4 \cos B)^{2}+(1 \sin B+3 \cos A)^{2}=37, \sin (A+B)=\frac{1}{2}$, so $\sin C=\frac{1}{2}, \angle C$ $=30^{\circ}$ or $150^{\circ}$.
If $\angle C=150^{\circ}$, then $\angle A<30^{\circ}$, thus $3 \sin A+4 \cos B<3 \cdot \frac{1}{2}+4<6$, which is a contradiction. Therefore, $\angle C=30^{\circ}$.