Answer
a. $QII , QIII$
Work Step by Step
$\cos{\theta} = \dfrac{x}{r}$
$\because r = \sqrt{x^2+y^2} \hspace{20pt} \therefore$ It is always positive.
$\therefore$ For $\cos{\theta}$ to be negative, $x$ must be negative.
$x$ is negative in $QII$ and $QIII$
$\therefore \theta $ could terminate in $\boxed{QII , QIII }.$
