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Determine the aircraft's final position, given:

An aircraft takes off from position 05°52’N, 000°00’E/W and flies westwards for 704 NM, then southwards for 704 NM, then eastwards for 1 304 km and finally northwards for 5°52’.

  • A
    00º00' N/S 000º00' E/W
  • B
    02º00'N 000ºE/W
  • C
    00º00' N/S 010º02'E
  • D
    05º52'S 000º00' E/W
To solve this type of question, it is important to be familiar with:
  • Departure (NM) = Change of Longitude (min) x cos Lat
  • One minute of latitude equals one nautical mile and degrees of latitude are 60 nm apart.

(1) From 05º52' N 000º00' E/W => Aircraft flies westwards for 704 NM

Start by using the Departure formula to find the change of longitude:

704 NM = Change of longitude (min) x cos (5º52')
Change of longitude (min) = 704 / cos (5º52')
Change of longitude (min) = 708 min

Change of longitude (º) = 708 / 60 = 11º47' W

- After the first leg (westwards), the aircraft will be at 000º00' + 11º47' W = 11º47'W

  • 05º52'N 11º47'W

(2) From 05º52' N 11º47' W => Aircraft flies southwards for 704 NM

1º Latitude = 60 NM
Therefore, 704 NM / 60 NM = 11º44' S

- After the second leg (southwards), the aircraft will be at 11º44'S - 05º52'N = 05º52' S

  • 05º52'S 11º47'W

(3) From 05º52' S 11º47' W => Aircraft flies eastwards for 1304 KM

Start by converting km into NM so that we can use the Departure equation:

1 NM = 1.852 km

1304 / 1.852 = 704 NM

704 NM = Change of longitude (min) x cos (5º52')
Change of longitude (min) = 704 / cos (5º52')
Change of longitude (min) = 708 min
Change of longitude (º) = 708 / 60 = 11º47' E

- After the third leg (eastwards), the aircraft will be at 11º47'W + 11º47'E = 000ºE/W

  • 05º52'S 000ºE/W

(4) From 05º52' S 000ºE/W => Aircraft flies northwards for 5º52'

- After this final leg, the aircraft will be at 05º52'S + 05º52'N = 00ºN/S


=> Aircraft's final position: 00ºN/S 000ºE/W

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