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Distances on Earth Surface — DGCA CPL Air Navigation
Chapter GN07 of the Air Navigation paper. 50 practice questions in the Avielevate bank, every one with its answer key, filterable into a timed set on this chapter alone.
What this chapter covers
The learning objectives we file Distances on Earth Surface questions under, and how many sit behind each. These come straight out of the bank and are regenerated whenever it changes — they are what is there today, not a claim.
| Learning objective | In the bank |
|---|
| __GEN__ General | 50 questions |
Sample Distances on Earth Surface questions, answered and explained
Real questions from this chapter, with the answer marked and the reasoning written out — the same explanations you get inside the app.
0215059 · Distances on Earth Surface
One nautical mile is equal to:
- A0.869 statute mile.
- B1609 meters.
- C6076 feet.
Why that is the answer
The **nautical mile** started life as the length of one minute of latitude along a meridian. Because the Earth is slightly flattened, that length changes a little from place to place, so an exact international value was agreed: **1 NM = 1,852 m**.
Converting to feet: 1 m ≈ 3.2808 ft, so 1,852 × 3.2808 ≈ 6,076 ft.
So one nautical mile is equal to **6,076 feet**.
**Tip:** in rough working 6,080 ft is often used; other handy links are 1 NM = 1.852 km and 1 NM ≈ 1.15 statute miles.
0215090 · Distances on Earth Surface
Position A is on the equator at longitude 130°00'E. Position B is 100 nm from A on a true bearing of 225°. What are the coordinates of B?
- A01°11'S 128°49'E
- B01°11'N 131°11'E
- C01°11'S 131°11'E
Why that is the answer
To find a position from a bearing and distance, split the movement into a **north-south part** and an **east-west part**, like the two sides of a right-angled triangle.
A true bearing of 225° is exactly south-west, halfway between south (180°) and west (270°), so the two parts are equal.
South part = 100 × cos 45° = 100 × 0.7071 ≈ 70.7 NM.
West part = 100 × sin 45° ≈ 70.7 NM.
North-south: 1′ of latitude = 1 NM, so 70.7 NM ≈ 71′ = 1°11′. Starting on the equator and going south: **01°11′S**.
East-west: so close to the equator cos lat is almost exactly 1, so 1′ of longitude ≈ 1 NM. 70.7 NM ≈ 71′ = 1°11′ towards the West: 130°00′E − 1°11′ = **128°49′E**.
So B is at **01°11′S 128°49′E**.
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Frequently asked
How many Distances on Earth Surface questions are there for the DGCA CPL Air Navigation paper?
Avielevate's bank carries 50 published Distances on Earth Surface questions for Air Navigation, mapped across 1 learning objective. Every one has its answer key, and explanations are being written across the whole bank.
How much of the Air Navigation paper is Distances on Earth Surface?
The DGCA does not publish a per-chapter breakdown of its papers, so nobody can honestly tell you — and we will not guess. What we can tell you is what students who have just sat a paper reported meeting, which is measured rather than rumoured: that is the exam recall report inside the app.
What is the best way to revise Distances on Earth Surface?
Sit a short timed set on this chapter alone, tag how sure you were of each answer, then drill the two that actually leak marks: the ones you got right while guessing, and the ones you got wrong while confident. Avielevate does that tagging for you and puts anything you got wrong back in front of you on a spaced schedule.
The rest of the Air Navigation paper
One account covers every chapter of every paper.
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