Answer
$135^{\circ},\space -45^{\circ}(315^{\circ})$
Work Step by Step
Please see the image first.
Given that, $\space V_{y}=-V_{x}$; Let's plug known values from the image as follows.
$Vsin\theta=-Vcos\theta$
$\frac{sin\theta}{cos\theta}=-1$
$tan\theta= -1$
$\theta=tan^{-1}(-1)$
If $tan\theta$ has a negative value we know that the angle is in the second or fourth quadrant.
$tan\theta=-tan(45^{\circ})$
$tan\theta=tan(180^{\circ}-45^{\circ})\space or\space tan\theta=tan(360^{\circ}-45^{\circ})$
$tan\theta=tan(135^{\circ})\space or\space tan\theta=tan(-45^{\circ})$
So we can get $\theta=135^{\circ}\space or\space \theta=-45^{\circ}(315^{\circ})$
