If A lies in the third quadrant and 3tan A - | vedclass

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(A) Given $3	an A - 4 = 0$,so $	an A = \frac{4}{3}$.Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.Using $	an A = \fr

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(A) Given $3	an A - 4 = 0$,so $	an A = \frac{4}{3}$.Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.Using $	an A = \frac{4}{3}$,we can represent this as a right triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.Thus,$\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.Now,substitute these values into the expression $5\sin 2A + 3\sin A + 4\cos A$:$5(2\sin A \cos A) + 3\sin A + 4\cos A$$= 10\sin A \cos A + 3\sin A + 4\cos A$$= 10\left(-\frac{4}{5}ight)\left(-\frac{3}{5}ight) + 3\left(-\frac{4}{5}ight) + 4\left(-\frac{3}{5}ight)$$= 10\left(\frac{12}{25}ight) - \frac{12}{5} - \frac{12}{5}$$= \frac{120}{25} - \frac{24}{5} = \frac{24}{5} - \frac{24}{5} = 0$.

If $A$ lies in the third quadrant and $3\tan A - 4 = 0,$ then $5\sin 2A + 3\sin A + 4\cos A = $

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