Answer:
379.12 km/h N 9.86° W
Explanation:
You want to know the resultant velocity when a plane's velocity of 373.52 km/h N is modified by a crosswind at 64.90 km/h W.
Vector sum
In (N, E) coordinates, the sum of the velocities is ...
(373.52, 0) +(0, -64.90) = (373.52, -64.90)
The magnitude of this vector is ...
|v| = √(373.52² +64.90²) ≈ 379.12 . . . . km/h
The direction will be west of north by the angle ...
θ = arctan(-64.90/373.52) ≈ -9.86° . . . . . . N = 0°, + = clockwise
The resultant velocity is 379.12 km/h N 9.86° W.
