$20 \cdot 35$ As shown, $MN$ is the diameter of semicircle $O$. If $\angle K=20^{\circ}, \angle PMQ=40^{\circ}$, then $\angle MQP$ equals
(A) $30^{\circ}$.
(B) $35^{\circ}$.
(C) $40^{\circ}$.
(D) $50^{\circ}$.
[Solution] Draw $QS // NM$, intersecting the semicircle at $S$, then $\widehat{QN}=\overparen{SM}$.
But $PM - QN \stackrel{\mathrm{m}}{=} 40^{\circ} (\text{twice the measure of } \angle K)$.
Therefore, $\quad PS \stackrel{\mathrm{m}}{=} 40^{\circ}$.
Also, $PQ \stackrel{\mathrm{m}}{=} 80^{\circ}. (\text{twice the measure of } \angle PMQ)$
$$
\therefore \overparen{SM} \stackrel{m}{=} \frac{1}{2}\left(180^{\circ}-40^{\circ}-80^{\circ}\right)=
$$
$30^{\circ}$,
and
$$
\angle PQS=20^{\circ}, \angle SQM=15^{\circ},
$$
$\therefore$
$$
\angle PQM=\angle PQS+\angle SQM=35^{\circ}.
$$

Therefore, the answer is $(B)$.