Find the acute angle between the pair of straight lines $\;$ $6y^2 - xy - x^2 + 30y + 36 = 0$
Equation of given pair of lines: $\;$ $6y^2 - xy - x^2 + 30y + 36 = 0$
Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 + 2 gx + 2 fy + c = 0$ $\;$ gives
$a = -1, \;\; h = \dfrac{-1}{2}, \;\; b = 6$
Let $\theta$ be the angle between the given pair of lines.
Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$
i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-1}{2}\right)^2 - \left(-1\right) \times 6}}{-1 + 6}\right|$
i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{1}{4} + 6}}{5}\right| = \dfrac{5}{5} = 1$
$\implies$ $\theta = \tan^{-1} \left(1\right) = 45^\circ$