Equation of a tangent to a parabola $y^{2}=4 x$ is
$y=m x+\frac{1}{m}$
Since, it passes through point $(1,4)$. So,
$4=m+\frac{1}{m}$
or $4 m=m^{2}+1$
$\Rightarrow m^{2}-4 m+1$
$\therefore m=\frac{4 \pm \sqrt{16-4}}{2} =\frac{4 \pm \sqrt{12}}{2} $
$m =2 \pm \sqrt{3}$
$\therefore m_{1}=2+\sqrt{3}$ and $m_{2}=2-\sqrt{3}$
Hence, angle between two slopes is
$ \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
$=\left|\frac{2 \sqrt{3}}{1+1}\right| $
$\Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \tan \theta =\tan \frac{\pi}{3} $
$ \Rightarrow \theta =\frac{\pi}{3}$