<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] 
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}
</s>