Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let the internal angle bisector of $\angle BAC$ intersect $BC$ and $\Gamma$ at $E$ and $N$, respectively. Let $A'$ be the antipode of $A$ on $\Gamma$ and let $V$ be the point where $AA'$ intersects $BC$. Given that $EV=6$, $VA'=7$, and $A'N=9$, compute the radius of $\Gamma$.

[i]Proposed by James Lin.[/i]
**
   Since $R = 5.1$ is too small, we select $R = 7.5$.

The final answer is $\boxed{\frac{15}{2}}$