If A lies in the third quadrant and 3 tan 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 = \f

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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{	ext{opposite}}{	ext{adjacent}} = \frac{4}{3}$,we find the hypotenuse $r = \sqrt{3^2 + 4^2} = 5$.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$$= 10 \left(-\frac{4}{5}ight) \left(-\frac{3}{5}ight) + 3 \left(-\frac{4}{5}ight) + 4 \left(-\frac{3}{5}ight)$$= 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 find the value of $5 \sin 2A + 3 \sin A + 4 \cos A$.

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