Answer
$\displaystyle \theta = 42.6^{o}$
Work Step by Step
(b)
With the work done in part (a),
The angle of $ \vec{v}=(11.7m/s)\hat{i}+(10.7m/s)\hat{j},$
keeping in mind that it points to quadrant I, is
$\displaystyle \theta=\tan^{-1}\frac{10.7}{11.7}=42.6^{o}$
