Find the angle between the asymptotes of the hyperbola $\;$ $24x^2 - 8y^2 = 27$
Equation of hyperbola is $\;$ $24x^2 - 8y^2 = 27$
i.e. $\;$ $\dfrac{x^2}{27 / 24} - \dfrac{y^2}{27 / 8} = 1$
i.e. $\;$ $\dfrac{x^2}{9 / 8} - \dfrac{y^2}{27 / 8} = 1$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = \dfrac{9}{8}$ $\implies$ $a = \dfrac{3}{2 \sqrt{2}}$
and $b^2 = \dfrac{27}{8}$ $\implies$ $b = \dfrac{3 \sqrt{3}}{2 \sqrt{2}}$
The angle between the asymptotes is
$\begin{aligned}
2 \alpha & = 2 \tan^{-1} \left(\dfrac{b}{a}\right) \\\\
& = 2 \tan^{-1} \left(\dfrac{3 \sqrt{3} / 2 \sqrt{2}}{3 / 2 \sqrt{2}}\right) \\\\
& = 2 \tan^{-1} \left(\sqrt{3}\right)
\end{aligned}$
i.e. $\;$ $2 \alpha = \dfrac{2 \pi}{3}$