Answer:
True course = 49.4° = N40.6°E
Ground speed = 145.32 km/h
Step-by-step explanation:
Velocity of the wind = (vₓî + vᵧj) km/h
Magnitude = 50 km/h
Direction = N45°W
Velocity of the wind = 50 [(-sin 45°)î + (cos 45°)j] = 50 (-0.7071î + 0.7071j) = (-35.36î + 35.36j) km/h
Velocity of the plane in still air = (vₓî + vᵧj) km/h
Magnitude = 150 km/h
Direction = N60°E
Velocity of the plane in still air = 150 [(sin 60°)î + (cos 60°)j] = 150 (0.866î + 0.5j) = (129.9î + 75j) km/h
Resultant velocity = vector sum of the two velocities
R = (-35.36î + 35.36j) + (129.9î + 75j)
R = (94.54î + 110.36j) km/h
True course = direction of the resultant
True course = tan⁻¹ (110.36/94.54)
True course = 49.4° = N40.6°E
Ground speed = magnitude of the resultant velocity
Ground speed = √(94.54² + 110.36²) = 145.32 km/h