Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.
SOLUTION
Since she got back from where she started then her total displacement, $\vec{R}$ is zero but not her total distance. Displacement is the shortest path traveled while distance is total path traveled. We're given with the following displacements,
$\vec{A}=180 \ m, west$
$\vec{B}=210 \ m, 45^o$ east of south
$\vec{C}=280 \ m, 30^o$ east of north
$\vec{D}=?$
Then, $\vec{R}=0=\vec{A}+\vec{B}+\vec{C}+\vec{D}$. Let's solve for the components of the given vectors, where the angles are measured from the +x-axis and rotates toward +y-axis,
$A_x=(180 \ m) \ cos \ 180^o=-180 \ m$
$A_y=(180 \ m) \ sin\ 180^o=0$
$B_x=(210 \ m) \ cos \ (270^o+45^o)=148.5 \ m$
$B_y=(210 \ m) \ sin \ (270^o+45^o)=-148.5 \ m$
$C_x=(280 \ m) \ cos \ (90^o-30^o)=140.0 \ m$
$C_y=(280 \ m) \ sin \ (90^o-30^o)=242.5 \ m$
Vector $\vec{D}=-\vec{A}-\vec{B}-\vec{C}$, then adding by components
$\vec{D}=-(A_x+A_y)-(B_x+B_y)-(C_x+C_y)$
$=(-A_x-B_x-C_x)+(-A_y-B_y-C_y)$
$=D_x + D_y$
$=(180 \ m-148.5 \ m-140.0 \ m)\hat{i}+(0+148.5 \ m-242.5 \ m)\hat{j}$
$=-108.5 \ m \ \hat{i} -94.0 \ m \ \hat{j}$
Then solving for the magnitude and direction using Pythagorean and trigonometry,
$D=\sqrt{(-108.5 \ m)^2+(-94.0 \ m)^2}=144 \ m$
$tan \ \theta=\frac{D_y}{D_x}=\frac{-94.0 \ m}{-108.5 \ m}$
$\theta=41^o$ south of west.
A vector-addition diagram is shown below,
