Skip to main content

Latitude, longitude and great-circle distances (QCE General Mathematics Unit 3)

Syllabus dot point

“Understand great circles and angles of latitude and longitude, locate and state positions on Earth's surface, and calculate angular distance and distance between two places on the same meridian (D = 111.2 x angular distance) or the same parallel of latitude (D = 111.2 cos θ x angular distance)”

15 min read

Quick answer

Latitude measures degrees north or south of the equator; longitude measures degrees east or west of the prime meridian through Greenwich. Great circles (the equator and meridians) give the shortest paths; other parallels are small circles. For two places on the same meridian, subtract or add the latitudes and use D=111.2×angular distanceD = 111.2 \times \text{angular distance} km. For two places on the same parallel at latitude θ\theta, subtract or add the longitudes and use D=111.2cos⁡θ×angular distanceD = 111.2\cos\theta \times \text{angular distance} km, converting minutes to decimal degrees first.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Exam-style questions
  4. Practice questions

What this dot point is asking

This is the first sub-topic, "Locations on the Earth", of Unit 3 Topic 5: Earth geometry and time zones in the QCAA General Mathematics 2025 syllabus. You must:

  • understand the meaning of great circles;
  • understand angles of latitude and longitude, in decimal degrees and in degrees and minutes;
  • locate and state positions on Earth's surface;
  • calculate angular distance (in degrees and minutes) and distance (in kilometres) between two places on the same meridian, using D=111.2×angular distanceD = 111.2 \times \text{angular distance};
  • calculate angular distance and distance between two places on the same parallel of latitude, using D=111.2cos⁡θ×angular distanceD = 111.2 \cos\theta \times \text{angular distance}, where θ\theta is the latitude of the parallel;
  • solve practical problems involving latitude, longitude, angular distance and distance.

The syllabus builds the constant 111.2 (kilometres per degree along a great circle) into both formulas, so you never need the Earth's radius itself: use 111.2 exactly as written. The second sub-topic, time zones, is on time zones and time differences.

The two syllabus formulas

Same meridian (or the equator): D=111.2×angular distanceD = 111.2 \times \text{angular distance}

Same parallel of latitude θ\theta: D=111.2cos⁡θ×angular distanceD = 111.2\cos\theta \times \text{angular distance}

DD is in kilometres and the angular distance is in degrees (convert minutes to decimal degrees before multiplying).

The answer

Great circles and small circles

A great circle is a circle on the Earth's surface whose centre is the centre of the Earth. It cuts the Earth into two equal halves and has the largest possible radius, the radius of the Earth. The equator and every meridian (a line of longitude running from pole to pole, together with the meridian on the opposite side of the Earth) are great circles.

A small circle is a circle on the surface whose centre is not the centre of the Earth. Every parallel of latitude except the equator is a small circle, and parallels get smaller towards the poles.

The shortest distance between two points on the surface of a sphere lies along the great circle through them. That is why aircraft on long routes follow great-circle paths, which look curved on a flat map.

Latitude and longitude

  • Latitude is the angle north (N) or south (S) of the equator, measured at the centre of the Earth, from 0°0\degree at the equator to 90°90\degree at the poles. Lines of equal latitude are parallels.
  • Longitude is the angle east (E) or west (W) of the prime meridian, which passes through Greenwich in London, from 0°0\degree to 180°180\degree. Lines of equal longitude are meridians. The meridian 180°180\degree is roughly followed by the International Date Line.

A position is stated latitude first, then longitude, each with its direction: Brisbane is at about 27.5°27.5\degreeS, 153.0°153.0\degreeE. To locate a position on a map, globe or GPS app, find the parallel for the latitude, then move along it to the meridian for the longitude. On most maps, latitude values are printed on the left and right edges and longitude values along the top and bottom.

Degrees and minutes. One degree is divided into 60 minutes (1°=60′1\degree = 60').

  • Minutes to decimal degrees: divide by 60. 23°15′=23+1560=23.25°23\degree 15' = 23 + \frac{15}{60} = 23.25\degree.
  • Decimal degrees to minutes: multiply the decimal part by 60. 18.75°=18°+0.75×60′=18°45′18.75\degree = 18\degree + 0.75 \times 60' = 18\degree 45'.
  • When adding or subtracting in degrees and minutes, carry or borrow 60 minutes, just as you would with hours and minutes: 27°28′−16°55′=26°88′−16°55′=10°33′27\degree 28' - 16\degree 55' = 26\degree 88' - 16\degree 55' = 10\degree 33'.

Most scientific calculators have a degrees-minutes-seconds key that performs these conversions; show the conversion in your working anyway.

Where 111.2 and cos⁡θ\cos\theta come from

Why a degree along a parallel is 111.2 cos theta kilometresCross-section of the Earth through the poles. A circle of radius R with its centre O. The equator is a horizontal diameter. A parallel at latitude theta is a horizontal chord above the equator. A radius from O to the end of the parallel makes angle theta with the equator. The horizontal distance from the Earth's axis to that point, labelled R cos theta, is the radius of the parallel. Notes at the right say that a great circle has radius R of about 6371 kilometres so one degree along it is about 111.2 kilometres, and a parallel at latitude theta has radius R cos theta so one degree of longitude along it is 111.2 cos theta kilometres.θNSORR cos θparallelequatorGreat circle: radius R ≈ 6371 km1° along it ≈ 111.2 kmParallel at latitude θ:radius R cos θ1° along it ≈ 111.2 cos θ kmCross-section through the poles (not to scale for labels)

A great circle has the Earth's radius, about 6371 km, so its circumference is 2π×6371≈40 0302\pi \times 6371 \approx 40\,030 km. Dividing by 360°360\degree gives about 111.2111.2 km for each degree of angle at the centre:

D=111.2×angular distance.D = 111.2 \times \text{angular distance}.

A parallel at latitude θ\theta has radius Rcos⁡θR\cos\theta (the figure shows the right-angled triangle that gives this), so its circumference, and the length of each of its degrees, is multiplied by cos⁡θ\cos\theta:

D=111.2cos⁡θ×angular distance.D = 111.2\cos\theta \times \text{angular distance}.

At the equator cos⁡0°=1\cos 0\degree = 1 and the formulas agree. At 60°60\degree, cos⁡60°=0.5\cos 60\degree = 0.5, so a degree of longitude is only 55.6 km. At the poles, cos⁡90°=0\cos 90\degree = 0 and every meridian meets at a point.

Finding the angular distance

Same meridian (north-south). Use the latitudes.

  • Same hemisphere: subtract. 35°35\degreeN and 20°20\degreeN are 15°15\degree apart.
  • Opposite sides of the equator: add. 12°12\degreeN and 25°25\degreeS are 37°37\degree apart.

Same parallel (east-west). Use the longitudes.

  • Both east or both west: subtract. 115°115\degreeE and 153°153\degreeE are 38°38\degree apart.
  • One east and one west: add. 20°20\degreeW and 25°25\degreeE are 45°45\degree apart. If the sum is more than 180°180\degree, the shorter way round crosses the 180°180\degree meridian: use 360°360\degree minus the sum.
Worked examples: distances on meridians and parallels

Example 1: same meridian, opposite hemispheres

Point P is at 12°12\degreeN, 140°140\degreeE and point Q is at 25°25\degreeS, 140°140\degreeE. Find the distance between them.

Angular distance. Opposite sides of the equator, so add: 12+25=37°12 + 25 = 37\degree.

Distance. D=111.2×37=4114.4D = 111.2 \times 37 = 4114.4 km.

Marker's note: say explicitly "same meridian, so D=111.2×D = 111.2 \times angular distance" to show you chose the right formula.

Example 2: degrees and minutes

Two towns on the meridian 145°145\degreeE are at 16°55′16\degree 55'S and 27°28′27\degree 28'S. Find the distance to the nearest kilometre.

Angular distance
27°28′−16°55′27\degree 28' - 16\degree 55': borrow one degree, 26°88′−16°55′=10°33′26\degree 88' - 16\degree 55' = 10\degree 33'.
Convert
33′=3360=0.55°33' = \frac{33}{60} = 0.55\degree, so the angular distance is 10.55°10.55\degree.
Distance
D=111.2×10.55=1173.16D = 111.2 \times 10.55 = 1173.16 km, about 1173 km.

Marker's note: writing 10°33′10\degree 33' as 10.33°10.33\degree gives 1148.7 km and loses the mark.

Example 3: same parallel, either side of Greenwich

Two points lie on the parallel 40°40\degreeN at 20°20\degreeW and 25°25\degreeE. Find the distance along the parallel.

Angular distance. One west and one east, so add: 20+25=45°20 + 25 = 45\degree.

Distance. D=111.2×cos⁡40°×45≈3833.3D = 111.2 \times \cos 40\degree \times 45 \approx 3833.3 km.

Marker's note: θ\theta is the latitude, 40°40\degree, not the angular distance.

Example 4: across the 180°180\degree meridian

Two islands lie on the parallel 20°20\degreeS at 150°150\degreeE and 170°170\degreeW. Find the shortest distance along the parallel.

Angular distance. 150+170=320°150 + 170 = 320\degree, more than 180°180\degree, so the shorter way is 360−320=40°360 - 320 = 40\degree.

Distance. D=111.2×cos⁡20°×40≈4179.8D = 111.2 \times \cos 20\degree \times 40 \approx 4179.8 km.

Marker's note: a sketch of the two longitudes on a circle makes the "shorter way round" obvious.

Example 5: working backwards

Two places on the parallel 35°35\degreeS are 1500 km apart along the parallel. Find their angular distance.

angular distance=D111.2cos⁡θ=1500111.2×cos⁡35°≈16.47°.\text{angular distance} = \frac{D}{111.2\cos\theta} = \frac{1500}{111.2 \times \cos 35\degree} \approx 16.47\degree.

In degrees and minutes: 0.47×60≈28′0.47 \times 60 \approx 28', so about 16°28′16\degree 28'.

Marker's note: rearrange the formula symbolically first, then substitute; it shows your method even if you slip on the calculator.

Comparing the two formulas

Angular distance of 10°10\degree Distance
Along a meridian or the equator 111.2×10=1112111.2 \times 10 = 1112 km
Along the parallel 30°30\degree 111.2×cos⁡30°×10≈963.0111.2 \times \cos 30\degree \times 10 \approx 963.0 km
Along the parallel 60°60\degree 111.2×cos⁡60°×10=556111.2 \times \cos 60\degree \times 10 = 556 km

The further a parallel is from the equator, the shorter each degree of longitude becomes, reaching zero at the poles.

A parallel is not the shortest route

Two stations on 60°60\degreeS at longitudes 0°0\degree and 180°180\degree are 111.2cos⁡60°×180=10 008111.2 \cos 60\degree \times 180 = 10\,008 km apart along the parallel. But the meridians 0°0\degree and 180°180\degree together form a single great circle through the South Pole, and along it the stations are only 30°+30°=60°30\degree + 30\degree = 60\degree apart: 111.2×60=6672111.2 \times 60 = 6672 km. The great-circle route saves 3336 km. The syllabus only asks you to calculate distances along a meridian or a parallel, but this example shows why "great circle" and "shortest distance" go together.

Key fact

Great circles (the equator and all meridians) have the Earth's radius; parallels other than the equator are small circles. Same meridian: D=111.2×D = 111.2 \times angular distance. Same parallel: D=111.2cos⁡θ×D = 111.2\cos\theta \times angular distance, with θ\theta the latitude of the parallel. Subtract angles on the same side of the equator or prime meridian, add them on opposite sides, and use 360°360\degree minus the sum if it exceeds 180°180\degree.

Background only: nautical miles

Navigators use the nautical mile, defined as exactly 1852 m, which is about one minute of arc along a great circle. It is not part of the QCAA General Mathematics syllabus. In assessment, give distances in kilometres using the 111.2 formulas unless a question defines another unit.

Common traps
Adding latitudes on the same side of the equator
35°35\degreeN and 20°20\degreeN are 15°15\degree apart, not 55°55\degree.
Leaving out cos⁡θ\cos\theta on a parallel
Only the equator is a great circle among the parallels.
Using the longitude instead of the latitude as θ\theta
In D=111.2cos⁡θ×angular distanceD = 111.2\cos\theta \times \text{angular distance}, θ\theta is the latitude of the parallel.
Treating minutes as decimals
10°33′10\degree 33' is 10.55°10.55\degree, not 10.33°10.33\degree.
Calculator in radians
Set the calculator to degrees before finding cos⁡θ\cos\theta. cos⁡30\cos 30 in radians is about 0.1540.154, not 0.8660.866.
Going the long way round
If the longitudes add to more than 180°180\degree, subtract the sum from 360°360\degree.
Using a radius instead of 111.2
Some textbooks use arc-length formulas with a rounded radius such as 6400 km, which give slightly different answers. In QCE General Mathematics, use the syllabus constant 111.2.
Exam technique

Write one line for each step: "Same meridian/parallel, so ...", "Angular distance =…= \ldots", "D=…D = \ldots", and a final sentence with units and the requested rounding. If a map is given, read the coordinates carefully and state them before calculating; a small reading error only costs one mark if your method is clear.

Note

Think of the Earth as an orange. Cutting it straight through the middle gives the biggest possible circle, a great circle, and the equator and the lines running from pole to pole are all like that. Lines of latitude further north or south are smaller circles, like thin slices near the top of the orange. Every degree around a big circle is about 111 km, but around a smaller slice a degree covers less ground, so you shrink the distance by multiplying by the cosine of the latitude.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Original4 marks
Marovoay and Iakora in Madagascar lie on the same meridian, 46.6 degrees E. Marovoay is at 16.1 degrees S and Iakora is at 23.1 degrees S. (a) State the angular distance between the towns. [1 mark] (b) Determine the shortest distance between Marovoay and Iakora, correct to the nearest kilometre. [3 marks]
Show worked answer →

(a) Angular distance (1 mark): both towns are south of the equator on the same meridian, so subtract the latitudes: 23.1 - 16.1 = 7 degrees.

(b) Shortest distance along the meridian (3 marks). Both points share a meridian, which is a great circle, so the shortest path runs north-south along it.

Formula (1 mark): D = 111.2 x angular distance = 111.2 x 7.

Evaluate (1 mark): D = 778.4 km.

State with units and rounding (1 mark): the towns are about 778 km apart.

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

Original5 marks
Point A in western Victoria is at 37.25 degrees S 141.75 degrees E and point B in eastern Victoria is at 37.25 degrees S 148.5 degrees E. Determine the distance between A and B along the parallel of latitude, to the nearest 100 km.
Show worked answer →

Distance along a parallel of latitude (5 marks). Both points sit on the parallel 37.25 degrees S, a small circle.

Step 1 - identify the common latitude (1 mark): theta = 37.25 degrees.

Step 2 - angular distance in longitude (1 mark): both are east, so subtract: 148.5 - 141.75 = 6.75 degrees.

Step 3 - apply the parallel formula (1 mark): D = 111.2 cos(37.25 degrees) x 6.75.

Step 4 - evaluate (1 mark): D = 597.48 km.

Step 5 - round as instructed (1 mark): about 600 km to the nearest 100 km.

The east-west case needs the cos theta factor; without it you would treat the parallel like a great circle and overstate the distance (111.2 x 6.75 = 750.6 km).

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks
Convert (a) 23°15′23\degree 15' to decimal degrees and (b) 18.75°18.75\degree to degrees and minutes.
Show worked solution →

(a) 15′=1560=0.25°15' = \frac{15}{60} = 0.25\degree, so 23°15′=23.25°23\degree 15' = 23.25\degree. (1 mark)

(b) 0.75×60=45′0.75 \times 60 = 45', so 18.75°=18°45′18.75\degree = 18\degree 45'. (1 mark)

foundation3 marks
Two towns lie on the meridian 145°145\degreeE. Town P is at 16°55′16\degree 55'S and town Q is at 27°28′27\degree 28'S. Calculate the distance between the towns along the meridian, correct to the nearest kilometre.
Show worked solution →

Angular distance (1 mark): both are south, so subtract: 27°28′−16°55′=10°33′27\degree 28' - 16\degree 55' = 10\degree 33', and 33′=3360=0.55°33' = \frac{33}{60} = 0.55\degree, giving 10.55°10.55\degree.

Formula (1 mark): D=111.2×10.55D = 111.2 \times 10.55.

Answer (1 mark): D=1173.16D = 1173.16 km, about 1173 km.

foundation2 marks
Point A is at 12°27′12\degree 27'N, 30°30\degreeE and point B is at 42°33′42\degree 33'S, 30°30\degreeE. Find the distance between them along the meridian, to the nearest kilometre.
Show worked solution →

Angular distance (1 mark): opposite sides of the equator, so add: 12°27′+42°33′=54°60′=55°12\degree 27' + 42\degree 33' = 54\degree 60' = 55\degree.

Distance (1 mark): D=111.2×55=6116D = 111.2 \times 55 = 6116 km.

Adding the minutes, 27′+33′=60′27' + 33' = 60', gives exactly one extra degree.

core4 marks
Two points lie on the parallel 33.9°33.9\degreeS, at 115.9°115.9\degreeE and 151.2°151.2\degreeE. (a) Find the angular distance between them. (b) Calculate the distance between them along the parallel, correct to one decimal place.
Show worked solution →

(a) Both are east, so subtract: 151.2−115.9=35.3°151.2 - 115.9 = 35.3\degree. (1 mark)

(b) Formula with θ=33.9°\theta = 33.9\degree (1 mark): D=111.2cos⁡33.9°×35.3D = 111.2 \cos 33.9\degree \times 35.3. Substitution (1 mark) and answer (1 mark): D≈3258.1D \approx 3258.1 km.

core3 marks
Two ships on the parallel 27.5°27.5\degreeS are 800 km apart along the parallel. One is at 153.0°153.0\degreeE. Find the longitude of the other ship, which is further east, to one decimal place.
Show worked solution →

Rearrange (1 mark): angular distance =D111.2cos⁡θ=800111.2×cos⁡27.5°= \dfrac{D}{111.2 \cos\theta} = \dfrac{800}{111.2 \times \cos 27.5\degree}.

Evaluate (1 mark): angular distance ≈8.1°\approx 8.1\degree (8.11).

Longitude (1 mark): 153.0+8.1=161.1°153.0 + 8.1 = 161.1\degreeE.

core3 marks
Two places on the equator are at 150°150\degreeE and 180°180\degree (the International Date Line meridian). (a) Find the distance between them. (b) Explain why the formula D=111.2×D = 111.2 \times angular distance can be used here even though they are on the same parallel.
Show worked solution →

(a) Angular distance =180−150=30°= 180 - 150 = 30\degree (1 mark). D=111.2×30=3336D = 111.2 \times 30 = 3336 km (1 mark).

(b) The equator is a parallel of latitude 0°0\degree and cos⁡0°=1\cos 0\degree = 1, so 111.2cos⁡θ=111.2111.2\cos\theta = 111.2: the equator is a great circle, and both formulas agree. (1 mark)

exam6 marks
A research vessel leaves point A (35°35\degreeS, 140°140\degreeE), sails due north along the meridian to point B (12°12\degreeS, 140°140\degreeE), then sails due east along the parallel to point C (12°12\degreeS, 170°170\degreeE). (a) Calculate the distance from A to B. (b) Calculate the distance from B to C. (c) Find the total distance travelled, correct to the nearest kilometre. (d) Explain why the distance from B to C is less than 111.2×30111.2 \times 30 km.
Show worked solution →

(a) Same meridian, same hemisphere: 35−12=23°35 - 12 = 23\degree, so D=111.2×23=2557.6D = 111.2 \times 23 = 2557.6 km. (1 mark)

(b) Same parallel, both east: 170−140=30°170 - 140 = 30\degree, so D=111.2cos⁡12°×30≈3263.1D = 111.2 \cos 12\degree \times 30 \approx 3263.1 km. (2 marks: formula with θ=12°\theta = 12\degree, answer)

(c) Total =2557.6+3263.1=5820.7= 2557.6 + 3263.1 = 5820.7 km, about 5821 km. (1 mark)

(d) The parallel 12°12\degreeS is a small circle whose radius is the Earth's radius multiplied by cos⁡12°\cos 12\degree, so each degree of longitude along it is shorter than a degree along a great circle (111.2cos⁡12°≈108.8111.2 \cos 12\degree \approx 108.8 km instead of 111.2 km). (2 marks)

exam5 marks
Two research stations X and Y are both on the parallel 60°60\degreeS, at longitudes 0°0\degree and 180°180\degree. (a) Calculate the distance from X to Y travelling along the parallel. (b) Calculate the distance from X to Y travelling along meridians through the South Pole. (c) Use your answers to explain which route is shorter and why.
Show worked solution →

(a) Angular distance along the parallel =180°= 180\degree. D=111.2cos⁡60°×180=111.2×0.5×180=10 008D = 111.2 \cos 60\degree \times 180 = 111.2 \times 0.5 \times 180 = 10\,008 km. (2 marks)

(b) From 60°60\degreeS to the pole is 30°30\degree along one meridian, and from the pole back to 60°60\degreeS is another 30°30\degree along the opposite meridian. These two meridians form one great circle, so the angular distance is 60°60\degree and D=111.2×60=6672D = 111.2 \times 60 = 6672 km. (2 marks)

(c) The route through the pole is shorter by 10 008−6672=333610\,008 - 6672 = 3336 km. It follows a great circle, and the shortest path between two points on a sphere lies along the great circle through them; the parallel is a small circle. (1 mark)

Practise this

Sources & how we know this

ExamExplained