Example 5. As shown in the figure, on a semicircle with center $C$ and diameter $M N$, there are two distinct points $A$ and $B$. $P$ is on $C N$, and $\angle C A P=\angle C B P=10^{\circ}, \angle M C A=40^{\circ}$. Find $\angle B C N$. (34th American Competition Problem)
Solution: Clearly, $\angle C P A = 30^{\circ}$. Therefore, applying the Law of Sines in $\triangle A C P$ and $\triangle B C P$, we have:
$$
\begin{array}{l}
\frac{\sin 10^{\circ}}{C P}=\frac{\sin 30^{\circ}}{A C} \text { and } \frac{\sin 10^{\circ}}{C P}=\frac{\sin \angle C P B}{B C}. \\
\therefore \frac{\sin 30^{\circ}}{A C}=\frac{\sin \angle C P B}{B C}, \sin \angle C P B=\frac{1}{2}. \\
\because A C=B C, \angle C P B \neq \angle C P A, \\
\therefore \angle C P B=150^{\circ}.
\end{array}
$$

Thus, $\angle B C N=180^{\circ}-10^{\circ}-150^{\circ}=20^{\circ}$.