Answer
$(a)\space 2.7\space km$
$(b)\space60^{\circ}\space from\space east\space to\space north$
Work Step by Step
Here we use the component method to find the displacement vector required to bring the team to the research station.
Let's get the directions due east and due north to be the positive directions.
We can get, the desired displacement A has components,
$A_{east}=4.8\space km\times cos42^{\circ}=3.57\space km$
$A_{north}=4.8\space km\times sin42^{\circ}=3.21\space km$
We can get, the actual displacement B has components,
$B_{east}=2.4\space km\times cos22^{\circ}=2.23\space km$
$B_{north}=2.4\space km\times sin22^{\circ}=0.9\space km$
So, to reach the research station, the research team must go,
$3.57\space km-2.23\space km=1.34\space km-East$
$3.21\space km-0.9\space km=2.31\space km-North$
(a) Magnitude of the displacement vector required to bring the team to the research station $R$ can get by using the Pythagorean theorem,
$R=\sqrt {(1.34\space km)^{2}+(2.31\space km)^{2}}=2.67\space km$
(b) the angle of $R$ with east direction $\theta$ can get by using the Trigonometry.
$tan\theta=\frac{2.31\space km}{1.34\space km}=\gt \theta=tan^{-1}(1.72)\approx60^{\circ}$
