2011SP-II-22

$\begin{array}{rcl}
\angle AOE & = & 2\times \angle ABE \\
& = & 2\times 30^\circ \\
& = & 60^\circ
\end{array}$

Also,

$\begin{array}{rcl}
\mbox{reflex }\angle COE & = & 2\times\angle CDE\\
& = & 2 \times 105^\circ\\
& = & 210^\circ
\end{array}$

Therefore, we have

$\begin{array}{rcl}
\angle COA & = & \mbox{reflex }\angle COE – \angle AOE \\
& = & 210^\circ-60^\circ \\
& = & 150^\circ
\end{array}$