Answer
$\theta=225^{o}\qquad$ or $-135^{o}$
Work Step by Step
(b)
Having found $\Delta\vec{r} =(-10cm)\hat{i}+(-10cm)\hat{j}$ in part (a), we note that it points to the third quadrant.
(We need to know this when we calculate $\tan^{-1}$)
$\displaystyle \theta=\tan^{-1}(\frac{-10cm}{-10cm})=45^{o}$ (the calculator returms the first quadrant angle)
Adding (or subtracting) $180^{o}$ , we get our angle in quadrant III.
$\theta=225^{o}\qquad$ or $-135^{o}$
