Gen Nav

[dojeevs]

A course of 120° (T) is drawn between X (61° 30N) and Y (58° 30N) on a Lambert Conformal conic chart with a scale of 1: 1 000 000 at 60° N. The chart distance between X and Y is: A) 66.7 cm B) 33.4 cm C) 38.5 cm D) 36.0 cm

[Alex Whittingham]

Just confirm, please, that the 61 and the 58 are longitudes East or West and the latitude in both cases is meant to be 30N? or is latitude only given with one point 61°30'N and the other 58°30'N?

[B2N2]

Again a glorious example of how these exam questions having no practical use or whatsoever.

[dojeevs]

The question is printed as it has been given in the paper. I am sorry i do not have any further information available on this

[Alex Whittingham]

I think we are assuming that we have a right angle triangle with the angle between the track of 120° and the meridian being 60°, then cos60° = A/H, then H = A ÷ cos60° = 180/cos60° = 360 NM. Chart distance = 360*6080÷3280*1000*100 / 1,000,000 = 66.7 cm. Not an EASA Q I am aware of, Indian DGCA?

[340drvr]

Originally Posted by B2N2

Again a glorious example of how these exam questions having no practical use or whatsoever.

Maybe trip-planning for a small bug walking across the chart?

[sshd]

Some of gnav questions are absolutely ridiculous, just pass the exam and then you are allowed to forget everything about it (so 90% of the whole bank)