Answer
$4.3^{\mathrm{o}}$ south of due west.
Work Step by Step
From part (a),
$\vec{b}$=$(-5.01 \mathrm{k}\mathrm{m}) \hat{\mathrm{i}}-(0.38 \mathrm{k}\mathrm{m}) \hat{\mathrm{j}}$,
(pointing to quadrant III).
Using (3-6) for the angle
$\displaystyle \tan^{-1}[\frac{-0.38 \mathrm{k}\mathrm{m}}{-5.01 \mathrm{k}\mathrm{m}}]=4.3^{\mathrm{o}} $(calc. gives angle in q. I)
In q.III, $\theta=4.3^{\mathrm{o}}+180.0^{\mathrm{o}}=184.3^{\mathrm{o}},$ or
$4.3^{\mathrm{o}}$ south of due west.
