Answer
$(a)\space 48^{\circ}$ east of south
$(b)\space 48^{\circ}$ west of south
Work Step by Step
Please see the attached image first.
The drawing shows the two vectors and the resultant vector. By using the Pythagorean theorem, we can get,
(a) $R^{2}=A^{2}+B^{2}=\gt B=\sqrt {R^{2}-A^{2}}$ ; Let's plug known values into this equation.
$B=\sqrt {(3.7\space km)^{2}-(2.5\space km)^{2}}$
$B=2.8\space km$
We can write,
$tan\theta=\frac{B}{A}=\frac{2.8\space km}{2.5\space km}=\gt\theta=tan^{-1}(\frac{2.8\space km}{2.5\space km})=48^{\circ}$ east of south
(b) $R_{1}^{2}=A^{2}+(-B)^{2}=\gt B=\sqrt {R_{1}^{2}-A^{2}}$ ; Let's plug known values into this equation.
$B=\sqrt {(3.7\space km)^{2}-(2.5\space km)^{2}}$
$B=2.8\space km$
We can write,
$tan\theta=\frac{B}{A}=\frac{2.8\space km}{2.5\space km}=\gt\theta=tan^{-1}(\frac{2.8\space km}{2.5\space km})=48^{\circ}$ west of south
