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{4}{3}$,we have $\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.Now,substitute these values into the expression $5 \sin 2A + 3 \sin A + 4 \cos A$.Recall $\sin 2A = 2 \sin A \cos A = 2 	imes (-\frac{4}{5}) 	imes (-\frac{3}{5}) = \frac{24}{25}$.So,$5 \sin 2A = 5 	imes \frac{24}{25} = \frac{24}{5}$.Then,$3 \sin A = 3 	imes (-\frac{4}{5}) = -\frac{12}{5}$.And $4 \cos A = 4 	imes (-\frac{3}{5}) = -\frac{12}{5}$.Adding them together: $\frac{24}{5} - \frac{12}{5} - \frac{12}{5} = \frac{24 - 12 - 12}{5} = 0$.

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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