$\sin\angle HIO$ 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?
 A: Recall that $B,I,C$ lie on a circle centered at $D$, midpoint of arc $BC$ not containing $A$ ($AD$ is the angle bisector of $\angle A$). We have $O,H$ lying on this circle too.

Since $O,H$ are isogonal conjugates, $\angle OCI=\angle ICH$. It follows, $IH=IO$ and $\angle OCI=\angle HCI=\angle OHI=\angle HOI$.
Hence $\angle HIO=180-2\angle OCI$.
But $\angle OCI=\angle ACI-\angle ACO=C/2-(90-B)$. Therefore, $$2\angle OCI=C-180+2B=C-(A+B+C)+2B=B-A$$
Finally,
$$\sin \angle HIO =\sin 2\angle OCI=\sin (B-A)=\ldots$$
A: $\displaystyle \small \angle IAO = \angle IAH = \frac{\angle B - \angle C}{2}$
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 $\small \angle AOB = 2C, \angle OAB = \angle OBA = 90^0 - C$
$\small \displaystyle \angle IAO = \angle OAB - \angle IAB = \frac{A+B+C}{2} - C - \frac{A}{2} = \frac{B-C}{2}$
A: 
Given $a=13$, $b=15$, $c=8$, we
can find semiperimeter, area, inradius and circumradius
ot the triangle:
\begin{align} 
\rho&=\tfrac12(a+b+c)=18
,\\
S_{ABC}&= 
\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
=
30\sqrt3
,\\
r&=\frac{S_{ABC}}\rho=\tfrac53\sqrt3
,\\
R&=\frac{abc}{4S_{ABC}}=
\tfrac{1}3\sqrt3
\end{align}
And use known expressions for $|OI|$, $|OH|$, and $|IH|$:
\begin{align} 
|OI|&=\sqrt{R(R-2r)}=\sqrt{13}
,\\
|OH|&=\sqrt{R^2+2((r+2R)^2-\rho^2)}
=7
,\\
|IH|&=
\sqrt{(r+2R)^2+2r^2-\rho^2}
=\sqrt{13}
,\\
S_{OIH}&= 
\tfrac14\sqrt{4|OI|^2|OH|^2-(|OI|^2+|OH|^2-|IH|^2)^2}
=\tfrac74\sqrt3
,\\
S_{OIH}&=\tfrac12|OI|\cdot|IH|\sin OIH
=\tfrac{13}2\sin OIH
,\\
\sin OIH&=
\frac{\tfrac74\sqrt3}{\tfrac{13}2}
=
\tfrac7{26}\sqrt3
.
\end{align}
