Answer
The magnitude of the resultant displacement is
$$
|\vec{r}|=370 \mathrm{m} .
$$
Work Step by Step
$\text{see image below:} \downarrow$
The diagram shows the displacement vectors for the two segments of her walk, labeled $\vec{A}$ and $\vec{B},$ and the total $(\text { "final") displacement vector, labeled } \vec{r} . $
We take east to be the $+x$ direction and north to be the $+y$ direction. We observe that the angle between $\vec{A}$ and the $x$ axis is $60^{\circ} .$
Where the units are not explicitly shown, the distances are understood to be in meters. Thus, the components of
$\vec{A}$ are $A_{x}=250 \cos 60^{\circ}=125$ and $A_{y}$ $=250 \sin 60^{\circ}=216.5 .$
The components of $\vec{B}$ are $B_{x}=175$ and $B_{y}=0 .$
The components of the total displacement are
$r_{x}=A_{x}+B_{x}=125+175=300$
$r_{y}=A_{y}+B_{y}=216.5+0=216.5$
The magnitude of the resultant displacement is
$$
|\vec{r}|=\sqrt{r_{x}^{2}+r_{y}^{2}}=\sqrt{(300 \mathrm{m})^{2}+(216.5 \mathrm{m})^{2}}=370 \mathrm{m} .
$$
