Answer
c) $|\textbf{V}|=9.9$, $\theta=13^{\circ}$
Work Step by Step
$|\textbf{V}|=\sqrt {|\textbf{V}_{x}|^{2}+|\textbf{V}_{y}|^{2}}=\sqrt {(9.6)^{2}+(2.3)^{2}}=9.9$
$\theta=\tan^{-1}(\frac{|\textbf{V}_{y}|}{|\textbf{V}_{x}|})=\tan^{-1}(\frac{2.3}{9.6})=13^{\circ}$