Answer
$\theta $ lies in the Second Quadrant or Quadrant-II.
Work Step by Step
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r} \\ \cos \theta =\dfrac{x}{r} \\ \tan \theta =\dfrac{y}{x}\\ \csc \theta =\dfrac{r}{y} \\ \sec \theta =\dfrac{r}{x} \\ \cot \theta =\dfrac{x}{y}$
where, $ r=\sqrt {x^2+y^2}$
It has been seen that $ x $ is negative and $ y $ is positive; this implies that the angle $\theta $ lies in the Second Quadrant or Quadrant-II.
