FAA Commercial Pilot Airplane (CAX)Navigation and Flight PlanningMedium
A pilot is planning a cross-country leg with a true course of 360° and a true airspeed of 150 knots. The forecast wind is from 090° at 30 knots, a direct crosswind. Using the wind triangle, what true heading should the pilot fly to remain on the 360° course (ignoring magnetic variation)?
- A000°
- B012°
- C348°
- D024°
Show answer & explanationAnswer & explanation
Correct answer: B. 012°
With a 90° crosswind, sin(WCA) = wind speed/TAS = 30/150 = 0.20, so WCA = arcsin(0.20) ≈ 11.5°, rounded to 12°. Since the wind is from the right (090°) of the northbound course, it pushes the aircraft left, so the pilot must correct to the right by adding the WCA: 360° + 12° = 012°.
Why the other options are wrong
- A. This ignores the crosswind correction entirely.
- C. This subtracts the correction instead of adding it, which would be wrong for a wind from the right.
- D. This overstates the required correction angle.
Wind Correction Angle (WCA)
The angular difference between true course and true heading needed to counteract wind drift so the aircraft tracks the desired course over the ground.
- sin(WCA) = (wind speed × sin θ)/TAS, where θ is angle between wind and course
- Correct into the wind: turn heading toward the direction the wind is coming from
- A direct crosswind (90° angle) maximizes the WCA for a given wind speed
Memory trick: Turn into the wind that shoves you off course