$20 \cdot 6$ As shown in the figure, points $A, B$ are on a semicircle with diameter $M N$ and center $C$, point $P$ is on $C N$ and $\angle C A P=\angle C B P=10^{\circ}$. If $\overparen{M A} \stackrel{\mathrm{m}}{=} 40^{\circ}$, then $\widehat{B N}$ equals
(A) $10^{\circ}$.
(B) $15^{\circ}$.
(C) $20^{\circ}$.
(D) $25^{\circ}$.
(E) $30^{\circ}$.
(34th American High School Mathematics Examination, 1983)
[Solution 1] From the problem, we have $\angle C P A$ and $\angle C P B$ are supplementary.
In $\triangle A C P$,
$$
\angle C P A=\angle M C A-\angle C A P=40^{\circ}-10^{\circ}=30^{\circ} \text {, }
$$

Therefore,
$$
\angle C P B=150^{\circ} .
$$

Thus, $\overparen{B N} \stackrel{\mathrm{m}}{=} \angle P C B=180^{\circ}-10^{\circ}-150^{\circ}=20^{\circ}$.
Hence, the answer is (C).
[Solution 2] After finding $\angle C P A=30^{\circ}$, apply the Law of Sines in $\triangle A C P$ and $\triangle B C P$, we have
$$
\frac{\sin 10^{\circ}}{C P}=\frac{\sin 30^{\circ}}{A C} \text {, and } \frac{\sin 10^{\circ}}{C P}=\frac{\sin \angle C P B}{B C} \text {. }
$$

Since $A C=B C$, from the above two equations, we get $\sin \angle C P B=\frac{1}{2}$.
Since $\angle C P B \neq \angle C P A$, it must be that $\angle C P B=150^{\circ}$, so $\overparen{B N}=20^{\circ}$. Hence, the answer is $(C)$.