Answer
$53^{o}$
Work Step by Step
The angle, relative to the +x axis is found from (3-6)
$\vec{a}=a_{x}\hat{\mathrm{i}}+a_{y}\hat{\mathrm{j}} ,$
$a=\sqrt{a_{x}^{2}+a_{y}^{2}}$ and $\displaystyle \tan\theta=\frac{a_{y}}{a_{x}}$ (3-6)
$b_{x}=6.0m,\quad b_{y}=8.0m$
$\displaystyle \theta=\tan^{-1}\frac{8.0m}{6.0m}=53^{o}$
(counterclockwise, pointing to the first quadrant)
