The angle between the pair of tangents drawn from 1 , 3 to the circle x 2 + y 2 - 2 x + 4 y - 11 = 0 is

$x^2+y^2-2x+4y-11=0$

Let the angle between tangents$=2\theta$

Now, $\tan \theta =\frac{r}{L_T}$ 

where $r$ is the radius of the circle and $L_T$ is the length of tangent from the given point.

Here $L_T=\sqrt{S_1}=\sqrt{1^2+3^2-2+12-11}$

$=3$

$r=\sqrt{1^2+2^2+11}=4$

i.e. $\tan \theta =\frac{4}{3}$

$\Rightarrow \tan 2\theta =\frac{2\times {\displaystyle \frac{4}{3}}}{1-{\displaystyle \frac{16}{9}}}=\frac{{\displaystyle \frac{8}{3}}}{-{\displaystyle \frac{7}{9}}}=\frac{-24}{7}$

$\sin 2\theta =\frac{24}{25}$

$\Rightarrow 2\theta ={\sin }^{-1}\left(\frac{24}{25}\right)$