Answer
The angle of the average velocity is $~~22.5^{\circ}~~$ east of north.
Work Step by Step
We can find the horizontal (east-west) displacement:
$x = (60.0~km/h)(\frac{2}{3}~h)+(60.0~km/h)(\frac{1}{3}~h)~sin~50.0^{\circ}-(60.0~km/h)(\frac{5}{6}~h)$
$x = 5.32~km~~$(east)
We can find the north-south displacement:
$y = (60.0~km/h)(\frac{1}{3}~h)~cos~50.0^{\circ}$
$y = 12.86~km~~$(north)
We can find the magnitude of the displacement:
$r = \sqrt{(5.32~km)^2+(12.86~km)^2}$
$r = 13.917~km$
We can find the angle $\theta$ of the average velocity where $\theta$ is east of north:
$tan ~\theta = \frac{5.32~km}{12.86~km}$
$\theta = tan^{-1}~\frac{5.32~km}{12.86~km}$
$\theta = 22.5^{\circ}$
The angle of the average velocity is $~~22.5^{\circ}~~$ east of north.
