The area (in sq. units) of the part of 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} (\frac{12}{13}) where \alpha, \beta are coprime numbers. Then \alpha+\beta is equal to [29-Jan-2024 Shift 1]

JEE Main 29-Jan-2024 Shift 1 Solved Paper

Section: Mathematics

Question : 28 of 90

Marks: +1, -0

The area (in sq. units) of the part of circle x2+y2=169 which is below the line 5x−y=13 is ‌

πα

2β

−‌

+‌

sin‌−1(‌

)‌‌ where α,β‌‌ are coprime numbers. Then α+β is equal to

[29-Jan-2024 Shift 1]

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