QLDGeneral MathematicsSyllabus dot point

Topic 3: Earth geometry - how do we locate points on the Earth and measure distances along its surface?

Locate a position on the Earth using latitude and longitude, define great circles and small circles, calculate the distance along a meridian or the equator using the angular separation, and convert between nautical miles, minutes of arc and kilometres

A focused answer to the QCE General Mathematics Unit 3 dot point on Earth geometry positions and distances. Covers latitude and longitude coordinates, great and small circles, distance along meridians and the equator using angular separation, and conversions between minutes of arc, nautical miles and kilometres, with arithmetic-verified worked examples for IA2 and the external assessment.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to treat the Earth as a sphere, give any point a position using latitude and longitude, and measure how far apart two points are along the surface. The key idea is that distance along a great circle is found from the angle the arc makes at the centre of the Earth, and that one minute of that angle is one nautical mile. This is the first half of Unit 3 Topic 3 (Earth geometry) and is examined in IA2 and the external assessment.

The answer

Latitude and longitude

A point on the Earth is located by two angles measured from the centre.

Latitude is the angle north or south of the equator, from $0$ degrees at the equator to $90$ degrees at each pole, labelled N or S.
Longitude is the angle east or west of the prime meridian (through Greenwich), from $0$ to $180$ degrees, labelled E or W.

A position is written as latitude then longitude, for example $27\degree$ S, $153\degree$ E for Brisbane.

Great circles and small circles

A great circle is any circle on the sphere whose plane passes through the centre of the Earth; it has the largest possible radius, equal to the Earth's radius. The equator and every meridian (line of longitude) are great circles. The shortest surface distance between two points always lies on a great circle.

A small circle is any circle on the sphere whose plane does not pass through the centre. Every parallel of latitude except the equator is a small circle, with a radius smaller than the Earth's.

Distance along a great circle

Because a great circle has the Earth's radius, the arc length between two points depends only on the angle between them at the centre. The standard tool for this in General Mathematics is the nautical mile.

One minute of arc ( $1'$ , which is $1/60$ of a degree) along a great circle is defined as one nautical mile.
So the distance in nautical miles equals the angular separation measured in minutes.

To find the angular separation along a meridian, the two points share a longitude, so you combine their latitudes: subtract if they are on the same side of the equator, add if on opposite sides. Along the equator you combine longitudes the same way.

Converting to kilometres

One nautical mile is approximately $1.852$ kilometres, so

\text{distance in km} = \text{nautical miles} \times 1.852.

A useful check: each degree of arc is $60$ nautical miles, which is about $111.1$ kilometres.

Worked example

Distance between two points on the same meridian

Two ships lie on the same meridian: ship A at $20\degree$ N and ship B at $35\degree$ N. Find the distance between them in nautical miles and in kilometres.

Step 1: Angular separation. Both are north of the equator on the same meridian, so subtract the latitudes.

35\degree - 20\degree = 15\degree.

Step 2: Convert degrees to minutes of arc. One degree is $60$ minutes.

15 \times 60 = 900 \text{ minutes}.

Verify: $15 \times 60 = 900$ .

Step 3: Distance in nautical miles. One minute of arc along a meridian is one nautical mile.

900 \text{ minutes} = 900 \text{ nautical miles}.

Step 4: Convert to kilometres.

900 \times 1.852 = 1666.8 \text{ km}.

Verify: $900 \times 1.852 = 1666.8$ . The ships are $900$ nautical miles, about $1666.8$ km, apart.

Common traps

Adding latitudes on the same side of the equator: Two points both north of the equator have an angular gap equal to the difference of their latitudes. Add only when the points straddle the equator.
Using a parallel of latitude as a great circle: Only the equator is a great circle among the parallels. Distances along other parallels are shorter than the angular formula suggests because the small-circle radius is reduced.
Confusing minutes of arc with minutes of time: One minute of arc is $1/60$ of a degree of angle; it is a distance unit here, not a unit of time.
Forgetting to convert degrees to minutes: The nautical-mile rule uses minutes of arc, so multiply the degree gap by $60$ before reading it as nautical miles.
Using the wrong conversion factor: One nautical mile is about $1.852$ km, not $1.6$ km (that is a statute mile). Mixing them up throws the kilometre answer out.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 QCAA4 marksMarovoay and Iakora are located on the same meridian at 46.6 degrees E, as shown on the map of Madagascar (latitude grid from 15 degrees S to 25 degrees S). a) Determine the latitudes of Marovoay and Iakora. [1 mark] b) Use the result from part a) to determine the shortest distance between Marovoay and Iakora. [3 marks]

Show worked answer →

a) Read the latitudes (1 mark). From the map, Marovoay is at about 16.1 degrees S and Iakora is at about 23.1 degrees S, both on the meridian 46.6 degrees E (answers within about 0.2 degrees are accepted).

b) Shortest distance along the meridian (3 marks). Because both points share a meridian, the shortest path runs north-south along that great circle.

Step 1 - angular separation (1 mark): both are south of the equator, so subtract: 23.1 - 16.1 = 7 degrees.

Step 2 - convert to distance (1 mark): each degree of latitude is about 111.2 km, so distance = 111.2 x 7.

Step 3 - evaluate and state with units (1 mark): distance = 778.4 km, so the towns are about 778 km apart.

Since both points are in the same hemisphere you subtract the latitudes; if one were north and one south you would add them.

2021 QCAA6 marksThe map shows regional federal electorates in Victoria. a) In which federal electorate is the position N (about 36 degrees S 146 degrees E)? [1 mark] b) Determine the distance between point A in the electorate of Mallee and point B in the electorate of Gippsland along the same parallel of latitude, to the nearest 100 km. [5 marks] (Point A is at 37.25 degrees S 141.75 degrees E and point B is at 37.25 degrees S 148.5 degrees E.)

Show worked answer →

a) Locate the point (1 mark): the position N lies in the electorate of Indi.

b) Distance along a parallel of latitude (5 marks). Both points sit on the same parallel (37.25 degrees S), so the path runs east-west along a small circle.

Step 1 - read the latitudes (1 mark): A and B are both at about 37.25 degrees S.

Step 2 - read the longitudes (1 mark): A is at about 141.75 degrees E and B at about 148.5 degrees E.

Step 3 - angular separation in longitude (1 mark): 148.5 - 141.75 = 6.75 degrees.

Step 4 - apply the small-circle (parallel) rule (1 mark): along a parallel the radius shrinks by cos(latitude), so distance = 111.2 x cos(37.25 degrees) x angular separation = 111.2 x cos(37.25 degrees) x 6.75.

Step 5 - evaluate and round (1 mark): = 597.48, so to the nearest 100 km the points are about 600 km apart.

The east-west case needs the extra cos(latitude) factor; without it you would wrongly treat the parallel like a great circle and overstate the distance.