13. $\frac{7 \sqrt{3}}{26}$
13. Since $\cos A=\frac{8^{2}+15^{2}-13^{2}}{2(8)(15)}=\frac{1}{2}$, we have $\angle A=60^{\circ}$.
As a result we have $\angle B H C=180^{\circ}-60^{\circ}=120^{\circ}$, $\angle B I C=90^{\circ}+\frac{60^{\circ}}{2}=120^{\circ}$ and $\angle B O C=60^{\circ} \times 2=120^{\circ}$.
Hence $B, H, I, O, C$ are concyclic. Let $D$ be a point on $A C$ such that $\triangle A B D$ is equilateral. Then we have
$$
\angle H I O=180^{\circ}-\angle H B O=60^{\circ}-\angle C=\angle D B C .
$$

Applying cosine law in $\triangle B C D$ (with $B D=8$ and $D C=7$ ), we have
$$
\cos \angle D B C=\frac{8^{2}+13^{2}-7^{2}}{2(8)(13)}=\frac{23}{26}
$$
and so $\sin \angle H I O=\sin \angle D B C=\sqrt{1-\left(\frac{23}{26}\right)^{2}}=\frac{7 \sqrt{3}}{26}$
13. In $\triangle A B C, A B=8, B C=13$ and $C A=15$. Let $H, I, O$ be the orthocentre, incentre and circumcentre of $\triangle A B C$ respectively. Find $\sin \angle H I O$.
(2 marks)
在 $\triangle A B C$ 中, $A B=8 、 B C=13 、 C A=15$ 。設 $H 、 I 、 O$ 分別為 $\triangle A B C$ 的垂
心、內心和外心。求 $\sin \angle H I O$ 。
(2 分)