If $\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}$ , where A and B are angles in the second quadrant, the find the value of $4 cos A + 3 cos B$

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Solution

$\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}$
$4 cos A + 3 cos B$
$sin A = \frac{3}{5} sin B = \frac{4}{5}$
$cos A = \frac{4}{2} cos B = \frac{3}{5}$
$\Rightarrow \frac{16}{5} + \frac{9}{5} = \frac{25}{5} = 5$

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