An airplane travels north with an airspeed of 300 km/h. There is a wind from the Southwest at 50 km/h. What is the resulting course of the plane? (magnitude and direction)

Answer 1

If you break the wind vector into two components, one blowing north and the other blowing east then this is easy. The wind velocity blowing north, using trig ratios, is 50sin(45°) = 50(√2/2) = 25√2 km/hr. The wind velocity blowing east, using trig ratios, is 50cos(45°) = 50(√2/2) = 25√2 km/hr (approx 35.36 km/hr). You can add the two vectors that are blowing north: 300 + 25√2 (approx 335.36 km/hr). So the resultant vector is made of one component blowing north at 335.36 and one blowing east at 35.36. To put them back together into a single vector, use the Pythagorean Theorem. True velocity magnitude = √(335.36² + 35.36²) = 337.2 km/hr. The direction is found using inverse tangent (tan^-1):

θ = tan^-1(35.36/335.36) = 6.02° east of north.

Answer 2

May I offer an alternative to William W.'s vector analysis?

Using Law of Cosines

c² = 300² + 50² - 2(300)(50) cos 135°

c² = 92500 - 30000 cos 135°

c² = 92500 + 21213.20

c² = 113713.20

c = 337.21

Speed is 337.21 kph

Using Law of Sines

sin θ / 50 = sin 135º / 337.21

sin θ / 50 = 0.0020969

sin θ = 0.1048

θ = 6.018º

Bearing is 6.018º