QJEEM24Maf05a: JEE Main 2024 Mathematics - Class 12 JEEM24Maf05a Solution: The area (in sq. units) of the part of the circle which is…

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 $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$ , where $\alpha, \beta$ are coprime numbers. Then $\alpha+\beta$ is equal to __________.