9. In $\triangle A B C$, $\angle A B C=\angle A C B=40^{\circ}, P$ is a point inside the triangle, $\angle P A C=20^{\circ}, \angle P C B=30^{\circ}$, find the degree measure of $\angle P B C$.
9. Let $\angle P B C=x$, then $\angle A B P=40^{\circ}-x$, from $\frac{\sin 80^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sin x}{\sin \left(40^{\circ}-x\right)} \cdot \frac{\sin 10^{\circ}}{\sin 30^{\circ}}=1$, we find that $x=20^{\circ}$.