## University Physics with Modern Physics (14th Edition)

The plane should fly 60.9 km at an angle of $33.0^{\circ}$ south of west.
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.