Asked by John

An airplane is traveling at 650 mph on a heading of 255°. The wind is blowing from a bearing of 320° at 28 mph. What is the ground speed and the actual bearing of the airplane?


P.S. I prefer the law of cosine/sine method
Answered by scott
resolve the two vectors into x and y components

add the respective components

find the resultant vector
Answered by Henry
All angles are measured CW from +Y-axis.

Vp = 650[255o] + 28[320o].

X = 650*sin255 + 28*sin320 =
Y = 650*Cos255 + 28*Cos320 =

Vp = sqrt(X^2+Y^2). = Velocity of the plane(gnd. speed).

Direction: Tan A = X/Y.

















t

Answered by Henry
Vp = 650[255o]
























































Correction: Since the wind is blowing FROM 320o, the heading is 140o.

Vp = 650mi/h[255o] + 28mi/h[140o].

X = 650*sin255 + 28*sin140 = -609.9 mi/h.
Y = 650*Cos255 + 28*Cos140 = -189.7 mi/h.

Vp = -609.9 - 189.7i = 639mi/h[72.7o] W. of S. = 639mi/h[252.7o] CW.

Tan A = X/Y.








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