Actually the answer IS D, as VC10 says.
Track is 310o T
Heading is 296o T
Therefore drift angle (DA) is 14o
TAS is 225 Kt
GS is 250 Kt
Using wind triangle and cosine rule wind velocity is:-
WS2 = TAS2 + GS2 – 2 x TAS x GS x Cos DA
WS2 = 2252 + 2502 – 2 x 225 x 250 x Cos 14o
WS2 = 3966.73
Therefore WS = 62.98 = 63 Kt
Now having wind velocity, we can calculate wind direction:-
WV/Sin DA = TAS/Sin X
Therefore Sin X = (TAS x Sin DA) / WV
Sin X = (225 x Sin 14o ) / 63
Sin X = 0.864007
Therefore X = 59.77o
Finally Wind Direction = (Track – 180) + X
Wind Direction = 310 – 180 + 59.77 = 189.77 = 190o
Therefore answer is 190/63 = D
PS Sorry, could not figure out how to do degrees and squared as suffixes!
Last edited by Groundloop; 19th December 2006 at 11:55.