A Classic Circumradius Problem

Given the side lengths, we can find the area of the triangle by Heron's formula, $\sqrt{s(s-a)(s-a)(s-c)}$ . In this case, $s$ is the semiperimeter and $a$ , $b$ , and $c$ are sides of the triangle. Therefore,

$A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{7056} = 84$

We can then relate the area to the circumradius by the area formula $\frac{abc}{4R}$ as well. Thus, equating our previously found area to the circumradius area formula gives us

$84 = \frac{(13)(14)(15)}{4R} \Longrightarrow R = \frac{2730}{336} = \frac{65}{8}$

Therefore, our circumradius is $\frac{65}{8}$ and $\alpha + \beta = 65 + 8 = \boxed{73}$ .