Points $A,B,Q,D$, and $C$ lie on the circle shown and the measures of arcs $\widehat{BQ}$ and $\widehat{QD}$ 
are $42^\circ$ and $38^\circ$ respectively. The sum of the measures of angles $P$ and $Q$ is
$\textbf{(A) }80^\circ\qquad \textbf{(B) }62^\circ\qquad \textbf{(C) }40^\circ\qquad \textbf{(D) }46^\circ\qquad  \textbf{(E) }\text{None of these}$
We see that the measure of $P$ equals $(\widehat{BD}-\widehat{AC})/2$, and that the measure of $Q$ equals $\widehat{AC}/2$.
Since $\widehat{BD} = \widehat{BQ} + \widehat{QD} = 42^{\circ} + 38^{\circ} = 80^{\circ}$, the sum of the measures of $P$ and $Q$ is $\widehat{BD}/2 = 80^{\circ}/2 = 40^{\circ} \Longrightarrow \textbf{(C) }$.