Let ABC be a triangle with $\mathrm{A}\left(-3, 1\right)$ and $\mathrm{\angle }\mathrm{ACB}=\mathrm{\theta }, 0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2}$. If the equation of the median through B is $2\mathrm{x}+\mathrm{y}-3=0$ and the equation of angle bisector of C is $7\mathrm{x}-4\mathrm{y}-1=0$, then $\tan \mathrm{\theta }$ is equal to

Given, the equation of median through B i.e. BE : $2\mathrm{x}+\mathrm{y}-3=0$

Equation, of angle bisector of C i.e. $\mathrm{CD} : 7\mathrm{x}-4\mathrm{y}=1$

Since, E satisfies the equation of BE.

$\begin{array}{l}\therefore 2\left(\frac{\mathrm{a}-3}{2}\right)+\left(\frac{\mathrm{b}+1}{2}\right)-3=0\\ 2\mathrm{a}-6+\mathrm{b}+1-6=0\\ 2\mathrm{a}+\mathrm{b}=11 ...\left(\mathrm{i}\right)\end{array}$

Since, C satisfies CD.

$\therefore 7\mathrm{a}-4\mathrm{b}=1 ...\left(\mathrm{ii}\right)$

On solving Eqs. (i) and (ii), we get

$\mathrm{a}=3, \mathrm{b}=5$

Slope of $\mathrm{AC}=\frac{2}{3}$

Slope of $\mathrm{CD}=\frac{7}{4}$

$\begin{array}{l}\therefore \tan \left(\frac{\mathrm{\theta }}{2}\right)=\left|\frac{\frac{2}{3}-\frac{7}{4}}{1+\frac{14}{12}}\right|=\frac{1}{2}\\ \mathrm{Now}, \tan \mathrm{\theta }=\frac{2\tan \left(\frac{\mathrm{\theta }}{2}\right)}{1-{\tan }^2\left(\frac{\mathrm{\theta }}{2}\right)}=\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}\end{array}$