2019-I-13

$\begin{array}{rcll}
\text{反角 } AOD & = & 2 \times \angle AED & \text{(圓心角兩倍於圓周角)} \\
\text{反角 } AOD & = & 2 \times 115^\circ \\
\text{反角 } AOD & = & 230^\circ
\end{array}$

所以，可得

$\begin{array}{rcll}
\angle AOD & = & 360^\circ – \text{反角 } AOD & \text{(對頂角)} \\
\angle AOD & = & 360^\circ – 230^\circ \\
\angle AOD & = & 130^\circ
\end{array}$

由於，$AC$ 為一直線，可得

$\begin{array}{rcll}
\angle COD & = & 180^\circ – \angle AOD & \text{(直線上的鄰角)} \\
\angle COD & = & 180^\circ – 130^\circ \\
\angle COD & = & 50^\circ
\end{array}$

由此，可得

$\begin{array}{rcll}
\angle CBF & = & \dfrac{1}{2} \times \angle COD & \text{(圓心角兩倍於圓周角)} \\
\angle CBF & = & \dfrac{1}{2} \times 50^\circ \\
\angle CBF & = & 25^\circ
\end{array}$

$\begin{array}{rcll}
\angle ODB & = & \angle CBD & \text{(錯角，$BC//OD$)} \\
\angle ODB & = & 25^\circ \\
OB & = & OD & \text{(半徑)} \\
\angle OBD & = & \angle ODB & \text{(等腰 $\Delta$ 的底角)} \\
\angle OBD & = & 25^\circ \\
\angle BOD & = & 180^\circ – \angle OBD – \angle ODB & \text{($\Delta$ 的內角和)} \\
\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}$

所以，扇形 $OBC$ 的周界

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

所以，扇形 $OBC$ 的周界大於 $60\text{ cm}$。