2019-I-13

$\begin{array}{rcll}
\text{reflex } \angle AOD & = & 2 \times \angle AED & \text{($\angle$ at centre twice $\angle$ at $\unicode{x2299}^{ce}$)} \\
\text{reflex } \angle AOD & = & 2 \times 115^\circ \\
\text{reflex } \angle AOD & = & 230^\circ
\end{array}$

Then, we have

$\begin{array}{rcll}
\angle AOD & = & 360^\circ – \text{reflex } \angle AOD & \text{(vert. opp. $\angle$s)} \\
\angle AOD & = & 360^\circ – 230^\circ \\
\angle AOD & = & 130^\circ
\end{array}$

Since $AC$ is a straight line, we have

$\begin{array}{rcll}
\angle COD & = & 180^\circ – \angle AOD & \text{(adj. $\angle$s on a st. line)} \\
\angle COD & = & 180^\circ – 130^\circ \\
\angle COD & = & 50^\circ
\end{array}$

Hence, we have

$\begin{array}{rcll}
\angle CBF & = & \dfrac{1}{2} \times \angle COD & \text{($\angle$ at centre twice $\angle$ at $\unicode{x2299}^{ce}$)} \\
\angle CBF & = & \dfrac{1}{2} \times 50^\circ \\
\angle CBF & = & 25^\circ
\end{array}$

$\begin{array}{rcll}
\angle ODB & = & \angle CBD & \text{(alt. $\angle$s, $BC//OD$)} \\
\angle ODB & = & 25^\circ \\
OB & = & OD & \text{(radii)} \\
\angle OBD & = & \angle ODB & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle OBD & = & 25^\circ \\
\angle BOD & = & 180^\circ – \angle OBD – \angle ODB & \text{($\angle$ sum of $\Delta$)} \\
\angle BOD & = & 180^\circ – 25^\circ – 25^\circ \\
\angle BOD & = & 130^\circ \\
\angle BOC & = & \angle BOD – \angle COD \\
\angle BOC & = & 130^\circ – 50^\circ \\
\angle BOC & = & 80^\circ
\end{array}$

Therefore, the perimeter of the sector $OBC$

$\begin{array}{cl}
= & 2 \pi \times OB \times \dfrac{\angle BOC}{360^\circ} + 2 OB \\
= & 2 \pi \times 18 \times \dfrac{80^\circ}{360^\circ} + 2 \times 18\\
= & 61.132\ 741\ 23 \text{ cm} \\
> & 60 \text{ cm}
\end{array}$

Therefore, the perimeter of the sector $OBC$ is longer than $60\text{ cm}$.