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A plane is flying at a speed of 330 miles per hour on a bearing of N65°E. Its ground speed is 390 miles per hour and its true course, given by the direction angle of the ground speed vector, is 30°. Find the speed, in miles per hour, and the direction angle, in degrees, of the wind.

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Answer:

R= V + W;

x-axis projection: Rcos15° = Vcos30° + Wx;

y-axis projection: Rsin15° = Vsin30° + Wy.

From here Wx = 330cos15° - 390cos30° = - 19.0 mi/h

Wy = 330sin15° - 390sin30° = - 109.6 mi/h

So W = ((- 19.0)2 + (- 109.6)2)1/2 = 111.2 mi/h;

tanθ = Wy/Wx = 109.6/19.0; θ = tan-1(109.6/19.0) + 180° = 260.2°

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