Answer
As a result the woman is moving in the direction $8^\circ$ west of north with a speed $|\boldsymbol{v_f}|=\sqrt{493}=22.2\frac{mi}{h}$.
Work Step by Step
Visualize this problem with a bird's eye view, with the woman's velocity vector $\boldsymbol{v_w}=\langle x_w,y_w\rangle$ and the ship's velocity vector $\boldsymbol{v_s}=\langle x_s,y_s\rangle$. If we think of north as the direction of the positive y axis we can rewrite $\boldsymbol{v_w}=\langle-3,0\rangle$ and $\boldsymbol{v_s}=\langle0,22\rangle$. In order to find the woman's velocity vector with respect to the water we simply add $\boldsymbol{v_w}$ to $\boldsymbol{v_s}$.
$\boldsymbol{v_f}=\boldsymbol{v_w}+\boldsymbol{v_s}=\langle-3,0\rangle+\langle0,22\rangle=\langle-3,22\rangle$
In order to find the speed of the woman relative to the water we just calculate $|\boldsymbol{v_f}|$.
$|\boldsymbol{v_f}|=\sqrt{(-3)^2+(22)^2}=\sqrt{493}\frac{mi}{h}$
Finding the direction the woman is traveling realitve to the water is simply the direction that $\boldsymbol{v_f}$ points in. In order to put this in terms of north and south we can simply calculate $\theta$ using trigonometry.
$\theta=\arctan\frac{3}{22}\approx8^\circ$
