Answer
5.0 km
Work Step by Step
Selecting $\hat{\mathrm{i}}$ as "east", $\hat{\mathrm{j}}$ as "north",
the explorer wanted his displacement to be $\vec{r}=(5.6$ km $)\hat{\mathrm{j}}$.
His real displacement is
$\vec{a}=(7.8\mathrm{k}\mathrm{m})(\mathrm{c}\mathrm{o}\mathrm{s}50^{\mathrm{o}}\hat{\mathrm{i}}+\sin 50^{\mathrm{o}}\hat{\mathrm{j}})\\=(5.01\mathrm{k}\mathrm{m})\hat{\mathrm{i}}+(5.98\mathrm{k}\mathrm{m})\hat{\mathrm{j}}$
Now, we need a displacement $\vec{b}$ such that $\vec{r}=\vec{a}+ \vec{b}.$
$\vec{b}=\vec{r}-\vec{a} =(-5.01 \mathrm{k}\mathrm{m}) \hat{\mathrm{i}}-(0.38 \mathrm{k}\mathrm{m}) \hat{\mathrm{j}}$
----
(a)
Using (3-6) for the magnitude,
$b= \sqrt{(-5.01\mathrm{k}\mathrm{m})^{2}+(-0.38\mathrm{k}\mathrm{m})^{2}}=5.0$ km.
