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

$(a)\;0\qquad(b)\;\large\frac{-24}{5}$$\qquad(c)\;\large\frac{24}{5}$$\qquad(d)\;\large\frac{48}{5}$

$3\tan A-4=0$
$3\tan A=4$
$\tan A=\large\frac{4}{3}$
$\cos A=-\large\frac{3}{5}$
$\sin A=-\large\frac{4}{5}$
$180^{\circ}\;<\;A\;<\;270^{\circ}$
$\sin 2A=\large\frac{2\tan A}{1+\tan^2A}$
$\qquad\;\;\;=\large\frac{24}{25}$
$\therefore 5\sin 2A+3\sin A+4\cos A=5\times\large\frac{24}{25}$$+3\times\large\frac{-4}{5}+$$4\times\large\frac{-3}{5}$
$\qquad\qquad\qquad\qquad\quad\quad\;\;\;=\large\frac{24}{5}-\frac{12}{5}-\frac{12}{5}$
$\qquad\qquad\qquad\qquad\quad\quad\;\;\;=0$
Hence (a) is the correct answer.