## Fundamentals of Physics Extended (10th Edition)

$4.3^{\mathrm{o}}$ south of due west.
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.