CPL/ATPL Question
 

Hi,
I have a question that has appeared in DGCA CPL exam, I know the answer is B. 16 o Left – 156Kt. I know it is easy to get this answer using the CX-2.
Can anyone please tell me how to get the answer using a scientific calculator, I know there is a formula but can’t seem to remember it.
Given: TAS = 140Kt, True HDG = 302, W/V 045(T)/45kt, calculate drift angle & GS
Thanks in advance.

Crosswind Component = W x Sin (diff between Track and W/V)
= 45 x Sin (103)
= 43.8kts
Head wind Component = W x Cos (diff between Track and W/V)
= 45 x Cos (103)
= -`10 kts.......- stands for Tail wind
so Gs will be TAS +(Headwind comp)= Ground speed

140+(10)=150Kts

drift angle
Tan(theta)= Crosswind component/TAS
= 43.8/140=17.13deg

Wind is from around 5 o clock...therefore ur drifting left i.e. 17Left

i don't care what are the options..these are the accurate figures:ok::ok::ok::ok:

Thanks Buddy.

How we get 103...do explain.
Ty

360+45 -302

Could anyone explain to me how does 43.8/140=17.13deg, can't figure it out whatever way I look at it

Nevermind, finally figured it out, tan−1(43.8/140)=17.13 I hadn't noticed the tan function there

Can someone explain how to solve this

Aircraft present Hdg 170 (M), ADF 345 relative. Make 450 intercept of the 355 track out bound. What shall be the Hdg to intercept and ADF indication at the time of intercept
– ADF on intercept (Relative) Intercept Hdg (M)
a) 310 175 b) 040 135 c) 045 130
Ans. c

General nav question
 

Guys i have a doubt. What is chlat btw these two points....A 71DEG 20MIN north and point B 86deg 45min north (over the north pole). In normal questions i was to just subtract point B from point A but in this question i need to subtract both from 90deg and then add to get ans. Plz explain why so