The angle between lines joining the origin to the points of the line $\sqrt{3}\mathrm{x}+\mathrm{y}-2=0$ and the hyperbola ${\mathrm{y}}^2-{\mathrm{x}}^2=4$ is

$$\begin{gathered} From the given line\frac{ \sqrt{3}x+y}{2}=1 \\ apply homogenisation then \\ {x}^2-y^2+4{\left(\frac{\sqrt{3}x+y}{2}\right)}^2=0 \\ \Rightarrow 4x^2+2\sqrt{3}xy=0 where a=4,2h=2\sqrt{3},b=0 \end{gathered}$$ 

 Now $$\begin{gathered} \tan \mathrm{\theta }=\frac{2\sqrt{{\mathrm{h}}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}} \\ =\frac{2\sqrt{3}}{4} \\ \Rightarrow \tan \theta =\frac{\sqrt{3}}{2} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right) \end{gathered}$$