Answer
$27^{o}$
Work Step by Step
$\vec{u}=\vec{a}+\vec{b}$
$=[4.0m+6.0m]\hat{i}+[-3.0m+8.0m]\hat{j}$
$=(10m)\hat{i}+(5.0m)\hat{j}$
Using (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)
$\displaystyle \theta=\tan^{-1}\frac{5.0m}{10m}=27^{o}$
(counterclockwise to the +x axis)
