DepartureAir Navigation — DGCA CPL practice questions

Question 1 of 5

Aircraft starts 0410S 17822W, heads true north 2950 NM, then 90° left for 314 km RL. Final position?

A.5500N 17422W

B.4500N 17422W

C.5500N 17738E

D.4500N 17738E

All 5 questions — Departure

Air Navigation · DGCA CPL. The correct option is marked on each.

Q1. Aircraft starts 0410S 17822W, heads true north 2950 NM, then 90° left for 314 km RL. Final position?
- A.5500N 17422W
- B.4500N 17422W
- C.5500N 17738E
- D.4500N 17738E✓
Why: 2950 NM north from 04°10′S: 2950/60 = 49°10′ change of latitude, so final latitude = 49°10′ − 4°10′ = 45°00′N. 90° left of a northerly heading = due west. 314 km = 169.5 NM. Ch.long = 169.5/cos 45° = 239.7 min ≈ 4°00′ west. Going 4° west of 178°22′W crosses the Date Line: final longitude = 177°38′E. Final position 4500N 17738E.
Q2. Aircraft at 50°N 006°E: 300 NM South, 300 NM East, 300 NM North, 300 NM West. Final position relative to start?
- A.North
- B.East
- C.West✓
- D.South
Why: S leg: 50°N → 45°N. E leg at 45°N: ch.long = 300/cos45° = 424 min = 7°4′ E. N leg: back to 50°N. W leg at 50°N: ch.long = 300/cos50° = 467 min = 7°47′ W. Net: went 7°4′ E then 7°47′ W → net = 0°43′ W. Final position is west of start. Answer: c
Q3. Aircraft departs 04°00′N 170°00′W: 600 NM south, 600 NM east, 600 NM north, 600 NM west. Final position?
- A.04°00′N 170°00′W
- B.06°00′S 170°00′W
- C.04°00′N 169°58.1′W✓
- D.04°00′N 170°01.8′W
Why: S: 04°N → 06°S. E at 06°S: ch.long = 600/cos6° = 600/0.9945 = 603.3 min = 10°3.3′ E. N: 06°S → 04°N. W at 04°N: ch.long = 600/cos4° = 600/0.9976 = 601.4 min = 10°1.4′ W. Net: 10°3.3′ E − 10°1.4′ W = 1.9′ E. Start 170°00′W − 1.9′ = 169°58.1′W . Answer: c
Q4. Aircraft flying eastwards at 60°N at 240 kts goes once round Earth. At what speed must another fly eastwards along the Equator to take the same time?
- A.600 kts
- B.240 kts
- C.480 kts✓
- D.120 kts
Why: Circumference at 60°N = full circumference × cos(60°) = 21 600 × 0.5 = 10 800 NM. Time = 10 800 / 240 = 45 hours. Equatorial circumference = 360 × 60 = 21 600 NM. Required speed = 21 600 / 45 = 480 kts . Answer: c
Q5. Position 58°33′N 174°00′W. Fly exactly 6 NM eastwards. New position?
- A.58°33′N 174°11.5′W
- B.58°33′N 173°55′W
- C.58°33′N 173°40′W
- D.58°33′N 173°48.5′W✓
Why: At 58°33′N: cos(58.55°) ≈ 0.5217. ch.long = 6 / 0.5217 = 11.5 min of longitude. Going east from 174°00′W: 174°00′ − 11.5′ = 173°48.5′W. Answer: d — 58°33′N 173°48.5′W