(d)
$Here,\ a = 2,\ \ h = \frac{3}{2},\ \ b = - 5$
$\therefore\ \tan\theta = \left| \frac{2\sqrt{\left( \frac{3}{2} \right)^{2} + 10}}{2 - 5} \right| = \left| \frac{\sqrt{49}}{- 3} \right|$
$\Rightarrow \ \theta = \tan^{- 1}\left| \frac{7}{3} \right|$