In the above figure ∠BAO=60°
Therefore in Tringle ABO, ∠ABO=30°
We have drawn a Right Triangle to use trigonometry to solve this problem.
We need to find AO and BO. We will start with AO:
Cos(BAO)=AO/AB
So, AO = AB x Cos(BAO) = 200 x Cos(60) = 200 x 0.5 = 100NM
AO helps us find the latitude of B. B is on the latitude 100NM sth of A.
i.e. 100NM = 1°40'S
Now, BO:
Cos(ABO)=BO/AB
So, BO = AB x Cos(ABO) = 200 x Cos(30) = 200 x 0.866 = 173.2NM
BO helps us find the longitude on which B lies.
Departure= dLong x Cos(Lat)
dLong= Departure/Cos(Lat) = 173.2/Cos(1.67) = 173.2/0.999 = 173.2' = 2°53'
So B lies 2°53'E of A(100°E). i.e. 100°-2°53'= 97°07'E
Co-ords of B are 1°40'S 97°07'E
Originally Posted by pilotbaba
Psn A 000 N/S 100 E
Psn B True Track 240, 200 Nm
Co-ordinates of B
Even though this question & the one on post # 341; both are more of an elimination questions, still would like to know the tehory behind these.
Thanks in Advance,
Above, I have solved this question. I would be able to tell you the elimination if you provide the options. One noticeable thing is that upto Latitude of 10°N/S, departure is approximately equal to dLong. This is because Cos(10) = 0.98.