Answer
$260^{o}$
Work Step by Step
$\vec{w}=\vec{a}-\vec{b}$
$=[4.0m-6.0m]\hat{i}+[-3.0m-8.0m]\hat{j}$
$=(-2.0m)\hat{i}-(11m)\hat{j}$
Note that the vector points to the 3rd quadrant, where $180^{o} < \theta < 270^{o}.$
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)
$\displaystyle \theta=\tan^{-1}\frac{-11m}{-2.0m}=80^{o}$, as given by the calculator, in quadrant 1,
but we know that the angle is in quadrant 3,
which is why we add (or subtract) $180^{o}$ to the obtained angle
$\theta=80^{o}+180^{o}=260^{o}$
(counterclockwise to the +x axis)
