$20 \cdot 4$ As shown in the figure, $A, B, Q, D, C$ are on the circle, and the measures of arcs $B Q$ and $Q D$ are $42^{\circ}$ and $38^{\circ}$, respectively. The sum of $\angle P$ and $\angle Q$ is 
(A) $80^{\circ}$.
(B) $62^{\circ}$.
(C) $40^{\circ}$.
(I) $46^{\circ}$.
(E) None of these.
(22nd American High School Mathematics Examination, 1971)
[Solution 1] $\angle P+\angle Q$
$$
\begin{array}{l}
=360^{\circ}-(\angle P A Q+\angle P C Q) \\
=360^{\circ}-\left(180^{\circ}-21^{\circ}\right)-\left(180^{\circ}-19^{\circ}\right) \\
=40^{\circ} .
\end{array}
$$

Therefore, the answer is $(C)$.
[Solution 2] According to the given, $\angle P \stackrel{m}{=} \frac{1}{2}(\overparen{B D}-\overparen{A C}), \angle Q \stackrel{m}{=} \frac{1}{2} \overparen{A C}$. Therefore, $\angle P+\angle Q \stackrel{\mathrm{m}}{=} \frac{1}{2} \overparen{B D} \stackrel{\mathrm{m}}{=} \frac{1}{2}\left(42^{\circ}+38^{\circ}\right)=40^{\circ}$. Therefore, the answer is $(C)$.