If frac{cos A}{3} = frac{cos B}{4} = frac{1}{ | vedclass

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(C) Given $\frac{\cos A}{3} = \frac{1}{5} \implies \cos A = \frac{3}{5}$.Since $-\frac{\pi}{2} < A < 0$, $A$ lies in the fourth quadrant, so $\sin A$

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$-4$

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(C) Given $\frac{\cos A}{3} = \frac{1}{5} \implies \cos A = \frac{3}{5}$.Since $-\frac{\pi}{2} $\sin A = -\sqrt{1 - \cos^2 A} = -\sqrt{1 - (\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.Given $\frac{\cos B}{4} = \frac{1}{5} \implies \cos B = \frac{4}{5}$.Since $-\frac{\pi}{2} $\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{4}{5})^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.Now, calculate $2 \sin A + 4 \sin B = 2(-\frac{4}{5}) + 4(-\frac{3}{5}) = -\frac{8}{5} - \frac{12}{5} = -\frac{20}{5} = -4$.

If $\frac{\cos A}{3} = \frac{\cos B}{4} = \frac{1}{5}$, $-\frac{\pi}{2} < A < 0$, and $-\frac{\pi}{2} < B < 0$, then the value of $2 \sin A + 4 \sin B$ is

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