Answer
The magnitude and direction of the fourth displacement is 144m ,41$^{\circ}$ south of west.
Work Step by Step
Let A,B,C and D be the displacement vectors where
A=180m due West
B=210m at 45$^{\circ}$ East of South
C=280m at 30$^{\circ}$ East of North
Since she returns back to her original place, the vector sum of the four displacements equates to zero i.e A+B+C+D=0.
The vector addition diagram is sketched in the figure .
On carelful measurement, the displacement D has a magnitude of 144m in the direction 41$^{\circ}$ South of West.
We can check this measurements as follows:
We Know,
D=-[A+B+C]
=-[($A_{x}+B_{x}+C_{x}$)i+($A_{y}+B_{y}+C_{y}$)j]
=-[(180cos(180$^{\circ}$)+210cos(-45$^{\circ}$)+280cos(60$^{\circ}$)i+(180sin(180$^{\circ}$)+210sin(-45$^{\circ}$)+280sin(60$^{\circ}$)j]
=-108.5i-94.0j ->(1)
Now,the magnitude of D is
mag.D=$\sqrt ((-108.5)^{2}+(-94.0)^{2})$
=144m
And direction of D is
$\theta$=$arctan(\frac{-94.0^{\circ}}{-108.5^{\circ}})$
=41$^{\circ}$
As the displacement vector D lies in the 3rd quadrant ,we have
$\theta=41^{\circ}$ South of West
