Ans: A
Join $AB$. Since $A$, $B$, $C$ and $D$ are four points on the circumference, we have
$\begin{array}{rcl}
\angle ADC + \angle ABC & = & 180^\circ \\
84^\circ + \angle ABO + 38^\circ & = & 180^\circ \\
\angle ABO & = & 58^\circ
\end{array}$
Consider $\Delta OAB$, since $OA$ and $OB$ are radii of the circle, we have
$\begin{array}{rcl}
\angle OAB & = & \angle OBA \\
& = & 58^\circ
\end{array}$
Hence, we have
$\begin{array}{rcl}
\angle AOB & = & 180^\circ – \angle OAB – \angle OBA \\
& = &180^\circ – 58^\circ – 58^\circ \\
& = & 64^\circ
\end{array}$
