Question

A jet plane is traveling with a velocity of 720 kilometers/hour due north. There is a crosswind of 40 kilometers/hour from the east. Find the resultant velocity and the direction of the jet plane.

Answer 1

\[Tan\Theta=\frac{opposite}{adjacent}\]
From the diagram, the triangle formed from the velocities makes the "opposite" 40 and "adjacent" 720. Thus,
\[\Theta=3.17^{o} from\:the\:North\]
Not that when reporting angles in bearing, the North is 0 and it sweeps clockwise.

The magnitude of the resultant belocity is simply:
\[\sqrt{720^{2}+40^{2}}=721km/h\]

You can check if the angles are correct if you apply the relation:
\[\frac{40}{720}=sin\Theta\]
If the values are the same, it shows there there aren't any mistakes!