$\begin{array}{rcl}
\angle COE & = & 307^\circ – 127^\circ \\
\angle COE & = & 180^\circ
\end{array}$
Therefore, $COE$ is a straight line.
$\begin{array}{rcl}
\angle COD & = & 217^\circ – 127^\circ \\
\angle COD & = & 90^\circ
\end{array}$
Therefore, $\Delta CDO$ and $\Delta DEO$ are both right-angled triangle.
By applying Pythagoras Theorem to $\Delta CDO$, we have
$\begin{array}{rcl}
CD^2 & = & OC^2 + OD^2 \\
CD^2 & = & 16^2 + 12^2 \\
CD & = & 20
\end{array}$
By applying Pythagoras Theorem to $\Delta DEO$, we have
$\begin{array}{rcl}
DE^2 & = & OD^2 + OE^2 \\
DE^2 & = & 12^2 + 5^2 \\
DE & = & 13
\end{array}$
Therefore, the perimeter of $\Delta CDE$
$\begin{array}{cl}
= & CE + CD + DE \\
= & CO + OE + CD + DE \\
= & 16 + 5 + 20 + 13 \\
= & 54
\end{array}$
