8. Given $I$ is the incenter of $\triangle A B C$, and $5 \overrightarrow{I A}=4(\overrightarrow{B I}+\overrightarrow{C I})$. Let $R, r$ be the circumradius and inradius of $\triangle A B C$, respectively. If $r=15$, then $R=$ $\qquad$ .
Let $D$ be the midpoint of $BC$, then $5 \overrightarrow{A I}=4(\overrightarrow{I B}+\overrightarrow{I C})=8 \overrightarrow{I D}$
$\Rightarrow A, I, D$ are collinear and $\frac{A I}{I D}=\frac{8}{5}=\frac{A B}{B D}$.
Suppose the three sides of $\triangle ABC$ are $8x, 8x, 10x$, then
$S_{\triangle ABC}=5 \sqrt{39} x^{2}=13 r x \Rightarrow x=\frac{13 r}{5 \sqrt{39}}=\sqrt{39}$.
Also, $\sin \angle ABC=\frac{\sqrt{39}}{8}$, so we have $2 R=\frac{8 \sqrt{39}}{\frac{\sqrt{39}}{8}}=64 \Rightarrow R=32$.