Find the acute angle between the pair of straight lines $\;$ $x^2 - xy - 6y^2 = 0$
Equation of given pair of lines: $\;$ $x^2 - xy - 6y^2 = 0$
Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ 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. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-1}{2}\right)^2 - 1 \times \left(-6\right)}}{1 - 6}\right|$
i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{1}{4} + 6}}{-5}\right| = \left|\dfrac{5}{-5}\right| = \left|-1\right| = 1$
$\implies$ $\theta = \tan^{-1} \left(1\right) = 45^\circ$