Answer
$225^\circ.$
Work Step by Step
If $v=ai+bj$ then by definition $tan(\alpha)=\frac{b}{a}$, hence $\alpha=tan^{-1}(\frac{b}{a})$. Also $0^\circ\lt\alpha\lt360^\circ.$
Therefore here $\alpha=tan^{-1}(\frac{-5}{-5})=tan^{-1}(1)=225^\circ.$ (because it must be in Quadrant III).