Answer
The plane should fly 60.9 km at an angle of $33.0^{\circ}$ south of west.
Work Step by Step
We can find the west component $d_x$ of the direction $d$.
$d_x - 23.0~cos(34.0^{\circ}) = 32.0~km$
$d_x = 32.0~km + 23.0~cos(34.0^{\circ})$
$d_x = 51.07~km$
We can find the south component $d_y$ of the direction $d$.
$d_y + 23.0~sin(34.0^{\circ}) - 46.0 = 0$
$d_y = - 23.0~sin(34.0^{\circ})+46.0$
$d_y = 33.14~km$
We can use $d_x$ and $d_y$ to find the magnitude of the distance $d$.
$d = \sqrt{(51.07~km)^2+(33.14~km)^2}$
$d = 60.9~km$
We can find the angle south of west.
$tan(\theta) = \frac{33.14}{51.07}$
$\theta = tan^{-1}(\frac{33.14}{51.07}) = 33.0^{\circ}$
The plane should fly 60.9 km at an angle of $33.0^{\circ}$ south of west.
