FAA Flight Instructor Airplane (FIA)Navigation and AirspaceMedium
A pilot plans a cross-country leg with a true course of 090° and a forecast wind of 360° at 30 knots. The airplane's true airspeed is 120 knots. Using the wind triangle, what true heading should the pilot fly to maintain the desired course?
- A060°
- B105°
- C090°
- D075°
Show answer & explanationAnswer & explanation
Correct answer: D. 075°
The wind is blowing from the north (360°) directly across the eastbound course (090°), a 90° relative angle, so all 30 knots act as crosswind component. WCA = arcsin(30/120) = arcsin(0.25) ≈ 14.5°, rounded to 15°. Since the wind is from the left (north) of the course, the pilot must crab into the wind by turning the nose toward the north, giving a true heading of 090° − 15° = 075°.
Why the other options are wrong
- A. Overcorrects by using too large a wind correction angle.
- B. Applies the correction in the wrong direction (away from the wind instead of into it).
- C. Ignores the wind correction angle entirely, which would result in drift south of course.
Wind Correction Angle (WCA)
The angular difference between true course and true heading needed to counteract wind drift, calculated using the wind triangle or an E6B flight computer.
- WCA = arcsin(wind speed × sin(relative wind angle) / TAS)
- Correct into the wind: turn heading toward the direction the wind is coming from
- True heading = True course ± WCA depending on which side the wind is from
Memory trick: Turn your nose into the wind to stay on the line