Answer
$\theta=135^{\mathrm{o}} \quad $($45^{\mathrm{o}}$ north of due west.)
Work Step by Step
In part (a), we found: $\Delta\vec{r}=(-40.0m)\hat{i}+(40.0m)\hat{j}$.
Note that this veector points to quadrant II.
The direction (angle) of $\Delta\vec{r}$ is
$\displaystyle \theta=\tan^{-1}(\frac{\Delta r_{y}}{\Delta r_{x}})=\tan^{-1}(\frac{40.0\mathrm{m}}{-40.0\mathrm{m}})=-45.0^{\mathrm{o}}$
( our calculator gives quadrant IV angle)
We add (or subtract) $180^{o}$ to get the angle in q.II
$\theta=135^{\mathrm{o}}$
We can describe this angle as
$45^{\mathrm{o}}$ west of due north, or
$45^{\mathrm{o}}$ north of due west.
