I understood the 85 deg part but Shabez how did you get 212.132 kts..

This is the diagram I came up with

http://img411.imageshack.us/img411/2498/diagramy.jpg

Now since Triangle ABD is congruent to Triangle ADC,
BD = DC

cos 35 = BD/50 => BD = cos35*x 50 = 40.95 nm
=> DC =40.95
=> BC which is the distance travelled by the A/C in 20 minutes is 40.95 + 40.95 = 81.9 nm

Therefore speed is 81.9/0:20 = 245.74


I think you mistook the Angle BAC as 90 deg when it is actually 110 (60 + 50)

You're right DJ.

I made the mistake of applying pythagoras to it, when this should only be applied to 90 degree triangles.

I should have taken more care in constructing the question.

:D

ure right

For this question use chart E(LO)1
What is the average track (° T) and distance between WTD NDB (N5211.3 W00705.0) and FOY NDB (N5234.0 W00911.7)?
A) 277° - 83 NM.
B) 075° - 81 NM.
C) 286° - 81 NM.
D) 294° - 80 NM.
ans b
any clue

@flying machine
 

do you have the chart?

no i dont have but does this type of questions come in indigo exam

Given: Magnetic heading 311° Drift angle 10° left Relative bearing of NDB 270° What is the magnetic bearing of the NDB measured from the aircraft?
A) 221° .
B) 211° .
C) 180° .
D) 208° .
any light on this plz

For this question use chart E(LO)1
What is the average track (° T) and distance between WTD NDB (N5211.3 W00705.0) and FOY NDB (N5234.0 W00911.7)?
A) 277° - 83 NM.
B) 075° - 81 NM.
C) 286° - 81 NM.
D) 294° - 80 NM.
ans b
any clue

try visualizing the points on a map...only one direction fits

Given: Magnetic heading 311° Drift angle 10° left Relative bearing of NDB 270° What is the magnetic bearing of the NDB measured from the aircraft?
A) 221° .
B) 211° .
C) 180° .
D) 208° .

ignore drift angle given it is to confuse u

mh+rb = mb

311+270 =581
581-360 = 221

Shabez
 

thanks mate:D

The rhumb line track between position A (45° 00N, 010° 00W) and position B (48° 30N, 015° 00W) is approximately:
A) 330
B) 315
C) 345
D) 300

@flying machine
 

Rhumb line track calculation [Archive] - PPRuNe Forums


If you work out the N/S distance, it comes to 210nm; if you work out the East/West Departure distance at 45N it comes to 212nm-ish. You therefore have an approximate right angled isosceles triangle where the right angle is at 48.5N 15W.

tan(angle) = opposite/adjacent

opp= 210nm and adj = 212nm

Now invert the tan function, you get...

(angle) =tan^(-1) (210/212) = 45 degrees

Add this to the 270 degees and you get 315 degrees precisely!

Shabez
 

thanks mate ru :D ur also giving exam on 20

Given: Course 040° (T), TAS is 120 kt, Wind speed 30 kt. Maximum drift angle will be obtained for a wind direction of:
A) 130°
B) 115°
C) 145°
D) 120°

YEAH I AM

Given: Course 040° (T), TAS is 120 kt, Wind speed 30 kt. Maximum drift angle will be obtained for a wind direction of:
A) 130°
B) 115°
C) 145°
D) 120°

90 Deg perpendicular to course

Constant of cone?
 

Hi! Aviator mates
I have a doubt if we are jist give two standard of parralels lets say 38N and 50N and were asked to find constant of cone.
How do we find it,is there any formula or method.
:(

The parallel of origin is halfway between the standard parallels (actually very slightly poleward) in your example 44 deg N, take sin 44, that's the answer.

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

Rk bali Air navigation
 

Guys do anyone has RK bali for air navigation.i am unable to get the soft copy as its not available so checking in case anyone has pdf copy