Answer
$\theta=135^{\mathrm{o}} \quad $($45^{\mathrm{o}}$ north of due west)
Work Step by Step
We have found $\Delta\vec{r}=(-40.0m)\hat{i}+(40.0m)\hat{j}.$
Eq 4-8 gives us
$ \vec{v}_{avg}$=$\displaystyle \frac{\Delta\vec{r}}{\Delta t}=\frac{(-40.0m)\hat{ i}+(40.0m)\hat{j}}{30.0s}=(\frac{-40.0m}{30.0s})\hat{ i}+(\frac{40.0m}{30.0s})\hat{j}$
Note that the velocity vector points to quadrant II.
Its angle is
$\displaystyle \theta=\tan^{-1}\frac{\frac{40.0m}{30.0s}}{-\frac{40.0m}{30.0s}}=\tan^{-1}(-1),$
which in quadrant II, as in part (b), yields $ 135^{o}$
$\theta=135^{\mathrm{o}} \quad $($45^{\mathrm{o}}$ north of due west)
