Discussion
A crosswind of \( 15 \ \mathrm{kt} \) at an approach speed of \( 120 \ \mathrm{kt} \) requires a drift angle of approximately
Answer: D
Correct answer: 7.2 degrees
The drift angle is the angle whose sine is the crosswind component over the true airspeed:
\[ \sin\theta = \frac{15}{120} = 0.125 \qquad \theta = 7.18^\circ \approx 7.2^\circ \]
That is the crab angle the aeroplane must hold on approach to track the centreline, and therefore the yaw it must be swung through in the de-crab just before touchdown. The distractor 12.5 degrees is the ratio 15/120 read as a percentage rather than a sine, a mistake worth watching for since the two are numerically close at small angles.
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