Angle of intersection of the circles $S_1:x^2+y^2-2x-6y-39=0$ and $S_2:x^2+y^2+10x-4y+20=0$ is

 

Centre of $S_1$ is $C_1(1,3)$ and of $S_2$ is $C_2(-5,2)$ radius

of $S_1$ is $r_1=7$ and of $S_2$ is $r_2=3$

If $\theta$ is the required angle, then

$\begin{array}{l}\cos \theta =\frac{r_1^2+r_2^2-{\left(c_1{c}_2\right)}^2}{2r_1{r}_2}\\ =\frac{49+9-(36+1)}{2\left(7\right)\left(3\right)}=\frac{1}{2}\\ \Rightarrow \theta =\frac{\pi }{3}\end{array}$