13. In $\triangle A B C$, $\angle A B C=40^{\circ}, \angle A C B=20^{\circ}, N$ is a point inside the triangle, $\angle N B C=30^{\circ}, \angle N A B=20^{\circ}$, find the degree measure of $\angle N C B$.
(Mathematical Bulletin Problem 1023)
13. Let $\angle N C B=x$, then $\angle A C N=20^{\circ}-x$, from $\frac{\sin 20^{\circ}}{\sin 100^{\circ}} \cdot \frac{\sin 30^{\circ}}{\sin 10^{\circ}} \cdot \frac{\sin \left(20^{\circ}-x\right)}{\sin x}=1$, we find $x=10^{\circ}$.