Answer
Magnitude $=0.9\space km$
Direction $=56^{\circ}$ north of west
Work Step by Step
Here we use the component method to find the magnitude and the direction of R.
Let's take +x &+y on the directions of east & north respectively.
We can write,
$R=A+B+C+D$
x component of $R =R_{x}= [2+0+(-2.5)+0]\space km=-0.5\space km$
y component of $R =R_{y}= [0+3.75+0+-3]\space km=0.75\space km$
By using the Pythagorean theorem, we can get
$R=\sqrt {R_{x}^{2}+R_{y}^{2}}=\sqrt {(-0.5\space km)^{2}+(0.75\space km)^{2}}=0.9\space km$
Let's take the angle = $\theta$ which R makes with the direction due west. So, using trigonometry we can get
$tan\theta=\frac{0.75\space km}{0.5\space km}=\gt \theta=tan^{-1}(1.5)=56^{\circ}$ north of west
