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KEAM 2010 Math Paper

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Question : 101 of 120

Marks: +1, -0

If

\alpha, \beta \in \left(0, \frac{\pi}{2}\right), \sin \alpha = \frac{4}{5}

\cos (\alpha+\beta) = -\frac{12}{13},

\sin \beta

is equal to

$\frac{63}{65}$
$\frac{61}{65}$
$\frac{3}{5}$
$\frac{5}{13}$
$\frac{8}{65}$

Solution:

Given that,

\alpha, \beta \in \left(0, \frac{\pi}{2}\right)

\sin \alpha = \frac{4}{5}

\Rightarrow \cos \alpha = \frac{3}{5}

Now,

\cos (\alpha+\beta) = -\frac{12}{13},

\Rightarrow \sin (\alpha+\beta) = \frac{5}{13}

and

\cos \alpha \cos \beta - \sin \alpha \sin \beta = -\frac{12}{13}

\Rightarrow \frac{3}{5} \cos \beta - \frac{4}{5} \sin \beta = -\frac{12}{13}

.....(i)

Now,

\sin (\alpha+\beta) = \frac{5}{13}

\Rightarrow \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{5}{13}

\Rightarrow \frac{4}{5} \cos \beta + \frac{3}{5} \sin \beta = \frac{5}{13}

....(ii)

Solving Eqs. (i) and (ii) simultaneously, we get

\sin \beta = \frac{63}{65}

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