Question

An airplane flies with a velocity of 55.0 m/s [ 35o N of W] with respect to the air (this is known as air speed). If the velocity of the airplane according to an observer on the ground is 40.0 m/s [53o N of W], what was the wind velocity?

Answer 1

The magnitude of the wind velocity can be found using the law of cosines. 

c^2 = a^2 + b^2 - 2*a*b*cos(C) 

C = angle between the two vectors = 53 - 35 = 18 degrees = 0.31459265 radians 
c^2 = 55^2 + 40^2 - 2*55*40*cos(0.314159265) = 951.8194212 
c = wind speed = 30.85 m/s 

We can find the wind direction by using the difference in the X components of the velocities along with the wind speed and the cosine. This angle will be the angle that wind vector makes with the X axis (the west-east direction). We can put it into proper context once we have the value. 

Vx = 55 X component = 55*cos(35) = 45.053 m/s 
Ux = 40 X component = 40*cos(53) = 24.07260093 

Vx - Ux = 20.98076151 m/s 

Angle = acos((Vx - Ux)/(wind speed)) = acos(20.98076151/30.85157081) = acos(0.680054887) 
Angle = 47.152 degrees 

If you draw the vectors you cn see how the wind is blowing, it is North of East 
So the wind speed is 30.85 m/s at an angle of 47.152 degrees N of E.