<s>[INST] $\sin\angle HIO$ given $\triangle ABC$ with $(\overline{AB},\overline{BC},\overline{CA})=(8,13,15)$? In triangle $ABC, AB = 8, BC = 13$ and $CA = 15$.
Let $H, I, O$ be the orthocenter, incenter and circumcenter of triangle $ABC$ respectively. Find $\sin$ of angle $HIO$ .
My attempt: By using law of cosine I can see angle $A = 60^{\circ}$. I also observed that angle $BHC$, $BIC$ and $BOC$ equals $120^{\circ} $. How do I proceed after this?
 [/INST] $\displaystyle \small \angle IAO = \angle IAH = \frac{\angle B - \angle C}{2}$
Now use the identity, $\small AH = 2R \cos \angle A \implies AH = R = AO \ $ (as $\angle A = 60^0$)
That means $\small \triangle HAO$ is an isosceles triangle. As $\small AI$ is angle bisector of $\small \angle HAO, IH = OI$.
Now remember that, $\small OI^2 = R(R-2r), OH^2 = 9R^2 - (a^2+b^2+c^2)$
$\small R = \displaystyle \frac{a}{2 \sin A} = \frac{13}{\sqrt3}$
$\small \displaystyle r = \frac{\Delta}{s} = \frac{(bc \sin A) / 2}{(a+b+c)/2} = \frac{5}{\sqrt3}$
So $\small OI = \sqrt {13} = IH, \ OH = 7$
$\small \displaystyle \cos \angle HIO = - \frac{23}{26} \implies \fbox {$\sin \angle HIO = \frac{7 \sqrt3}{26}$}$

This is just an addendum showing how $\displaystyle \small \angle IAO = \angle IAH = \frac{B - C}{2}$
$\small \displaystyle \angle BAH = 90^0 - B, \angle IAH = \angle IAB - \angle BAH = \frac{A}{2} - 90^0 + B$
As $\small \displaystyle \frac{A+B+C}{2} = 90^0, \angle IAH = \frac{B-C}{2}$
Now $\small \angle AOB = 2C, \angle OAB = \angle OBA = 90^0 - C$
$\small \displaystyle \angle IAO = \angle OAB - \angle IAB = \frac{A+B+C}{2} - C - \frac{A}{2} = \frac{B-C}{2}$
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