1. In $\triangle A B C$, it is known that $2 \sin A+3 \cos B=4,3 \sin B+2 \cos A=\sqrt{3}$, then the degree of $\angle C$ is
1. $\because(2 \sin A+3 \cos B)^{2}+(\sin B+2 \cos A)^{2}=19, \sin (A+B)=\frac{1}{2}$,
$\therefore \sin C=\frac{1}{2}, \angle C=30^{\circ}$ or $150^{\circ}$.
If $\angle C=150^{\circ}$, then $\angle A<30^{\circ}$, at this time $2 \sin A+3 \cos B<2 \cdot \frac{1}{2}+3=4$, contradiction. Therefore, the answer is $30^{\circ}$.