Q.The angle between the pair of straight lines $x^2 - 7xy + 4y^2 = 0$ is :

Using $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{a+b}$ with $a=1,\,h=-\tfrac72,\,b=4$ gives $\tan\theta = \dfrac{\sqrt{33}}{5}$, so the answer is option (a).

For a pair of straight lines through the origin $ax^2 + 2hxy + by^2 = 0$, the angle $\theta$ between them is

$\tan\theta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right|.$

Compare $x^2 - 7xy + 4y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$:

$a = 1,\quad 2h = -7 \Rightarrow h = -\tfrac{7}{2},\quad b = 4.$

Compute $h^2 - ab$:

$h^2 - ab = \frac{49}{4} - (1)(4) = \frac{49}{4} - \frac{16}{4} = \frac{33}{4}.$

So …