Ans: D
Join $BC$.
Join $BC$.
$\begin{array}{rcl}
\angle BAC & = & \dfrac{1}{2} \times 124^\circ \\
& = & 62^\circ
\end{array}$
Since $AB$ is the angle bisector of $\angle CAE$, then we have
$\begin{array}{rcl}
\angle BAE & = & \angle BAC \\
& = & 62^\circ
\end{array}$
Since $DE$ is the tangent to the circle at $A$, then we have
$\begin{array}{rcl}
\angle ACB & = & \angle BAE \\
& = & 62^\circ
\end{array}$
In $\Delta OBC$,
$\because OC = OB$,
$\therefore \angle OCB = \angle OBC$.
Hence, we have
$\begin{array}{rcl}
\angle OCB & = & \dfrac{1}{2} (180^\circ – 124^\circ) \\
& = & 28^\circ
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\angle ACO & = & \angle ACB – \angle OCB \\
& = & 62^\circ – 28^\circ \\
& = & 34^\circ
\end{array}$
