The area (in sq. units) of the part of the circle x^2+y^2=169 which is below the line 5 x-y=13 is (π α)/(2 β)-65/2+(α)/(β) sin^-1(12/13), where α, β…

The area (in sq. units) of the part of the circle $\displaystyle x^2+y^2=169$ which is below the line $\displaystyle 5 x-y=13$ is $\displaystyle \frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$, where $\displaystyle \alpha, \beta$ are coprime numbers. Then $\displaystyle \alpha+\beta$ is equal to $\displaystyle \_\_\_\_$