15. In $\triangle A B C$, $\angle A B C=50^{\circ}, \angle A C B=30^{\circ}, M$ is a point inside the triangle, $\angle M C B=20^{\circ}, \angle M A C=40^{\circ}$, find the degree measure of $\angle M B C$.
(Mathematical Bulletin Problem 1208)
15. Let $\angle M B C=x$, then $\angle A B M=50^{\circ}-x$, by $\frac{\sin 60^{\circ}}{\sin 40^{\circ}} \cdot \frac{\sin x}{\sin \left(50^{\circ}-x\right)} \cdot \frac{\sin 10^{\circ}}{\sin 20^{\circ}}=1$ and $\sin 60^{\circ} \cdot \sin x=\frac{1}{2 \sin 20^{\circ}} \cdot 4 \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \cdot \sin \left(50^{\circ}-x\right)=\frac{\sin 60^{\circ} \cdot \sin \left(50^{\circ}-x\right)}{2 \sin 20^{\circ}}$, we find $x=30^{\circ}$.