$\text{ Angle between the asymptotes of a hyperbola }{\mathrm{x}}^2-3{\mathrm{y}}^2=1\text{ is }$

$$\begin{gathered} Given hyperbola x^2-3y^2=1 where a^2=1,b^2=\frac{1}{3} \\ If \theta isangle between asymptotes then \\ \theta =2{\tan }^{-1}\frac{\mathrm{b}}{\mathrm{a}} \\ =2{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =2\left({30}^0\right) \\ ={60}^0 \end{gathered}$$