Determine the heading and groundspeed using TRACK = 170 deg, WIND = 330 deg/28 KT, TAS = 137 KTS.

• A. HDG = 174 deg , GS = 163 KTS
• B. HDG = 166 deg , GS = 154 KTS
• C. HDG = 195 deg, GS = 135KTS
• D. HDG = 184 deg , GS = 122 KTS

Answer: (A)

The idea here is to use the wind triangle: the aircraft’s motion relative to the air (airspeed at TAS) plus the wind vector equals the motion over the ground (track with groundspeed). So you set up vector components for both the wind and the airspeed, then solve for the heading that gives the desired ground track, and from that the groundspeed.

Wind from 330 degrees at 28 knots means the wind is blowing toward 150 degrees. Its components are:

• east: 28 × sin(150) = 14

• north: 28 × cos(150) = -24.25

Ground track is 170 degrees with an unknown groundspeed GS, so its components are:

• east: GS × sin(170) ≈ 0.17365 GS

• north: GS × cos(170) ≈ -0.98481 GS

Airspeed is TAS 137 along the unknown heading h, so its components are:

• east: 137 × sin(h)

• north: 137 × cos(h)

Equations from vector addition:

• 137 sin(h) + 14 = 0.17365 GS

• 137 cos(h) - 24.25 = -0.98481 GS

Solving these gives a heading of about 174 degrees and a groundspeed of about 163 knots. This matches the chosen option: heading around 174° with GS near 163 knots.