12 / 20

An aircraft departs points A (50°N, 012°E) on a westerly track and flies for 300 NM to point B. From point B, the aircraft flies a southerly track of 300 NM to point C. From point C, the aircraft finally flies a westerly track of 300 NM to point D. The coordinates of point D are:

• A

64°N, 010°48’W.

• B

45ºN​, 002º51’W

• C

45°N, 010°48’E.

• D
64°N, 033°06’E.

To solve this type of question, it is important to be familiar with:

• Departure (NM) = Change of Longitude (min) x 60 x cos Lat

• One minute of latitude equals one nautical mile and degrees of latitude are 60 nm apart.

(1) From 50ºN 012ºE => Aircraft flies westwards for 300 NM
Start by using the Departure formula to find the change of longitude:
300 NM = Change of longitude (min) x 60 x cos (50º)
Change of longitude (º) = 300 / [cos (50º) x 60] = 7.78 (Convert it to minutes)
Change of longitude (º) = 7.47º westerly

- After the first leg (westwards), the aircraft will be at 012ºE - 007.47º = 4.13º E

• 50ºN 004.13ºE

(2) From 50ºN 003.7ºE => Aircraft flies southwards for 300 NM
1º Latitude = 60 NM
Therefore, 300 NM / 60 NM = 5º southerly

- After the second leg (southwards), the aircraft will be at 50ºN - 5ºS = 45ºN

• 45ºN 004.13ºE

(3) From 45ºN 003.7ºE => Aircraft flies westwards for 300 NM
Start by using the Departure formula to find the change of longitude:
300 NM = Change of longitude (min) x 60 x cos (45º)
Change of longitude (º) = 300 / [cos (45º) x 60] = 7.07 (Convert it to minutes)
Change of longitude (º) = 7.04º westerly

- After the third leg (westwards), the aircraft will be at 004.13ºE – 7.04º = 2.51ºW

• 45ºN 002º51’W

=> Aircraft's final position: 45ºN​ 002º51’W

Your Notes (not visible to others)

This question has appeared on the real examination, you can find the related countries below.