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Year 12 Maths Standard 2 (2027) Trigonometry

Radial Surveys

20 practice questions 0 video lessons Theory + worked examples

Master compass radial surveys for NSW Year 12 Mathematics Standard 2. A radial survey fixes every corner of a field by its bearing and distance from one central station \(O\), then splits the field into triangles from \(O\).

You will learn to find the angle at \(O\) from two bearings, use the cosine rule for a boundary length, and find each sub-triangle and the whole-field area with \(A=\tfrac{1}{2}ab\sin C\) — core Standard 2 skills for surveying a paddock, park or building block.

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Theory

Compass radial surveys fix each corner of a field by its bearing and distance from one central station. This Year 12 Standard 2 (NSW) guide shows how to read the field notes, find the angle at \(O\) between two rays, use the cosine rule for a boundary length, and find each sub-triangle and the whole-field area with \(A=\tfrac{1}{2}ab\sin C\).

A compass radial survey is a survey of a field taken from a single central station \(O\). Standing at \(O\), the surveyor records, for every corner, its three-figure bearing (clockwise from north) and its distance from \(O\) — this list of bearings and distances is the field notes.

Joining \(O\) to two adjacent corners forms a triangle, so the whole field is split into sub-triangles that all share the vertex \(O\). Each sub-triangle has two known legs (the distances) and a known included angle at \(O\), so it can be solved with non-right-angled trigonometry.

The angle at \(O\) between two rays is the difference of their bearings; a boundary length comes from the cosine rule; and an area comes from \(A=\tfrac{1}{2}ab\sin C\). This is a core Year 12 Standard 2 (NSW) Trigonometry skill.

Compass radial survey from central point OA compass radial survey taken from central point O, with rays to corner stations A (035 deg, 48 m), B (110 deg, 62 m), C (195 deg, 40 m) and D (295 deg, 55 m); the dashed quadrilateral ABCD is the field boundary. N A 48 m 035° B 62 m 110° C 40 m 195° D 55 m 295° O
A radial survey from \(O\): each corner is fixed by a bearing (red) and a distance (green).
Sub-triangle AOB extracted from the surveyTriangle A-O-B with OA = 48 m, OB = 62 m and the included angle AOB = 75 degrees, with the boundary side AB labelled x. O A B 48 m 62 m x 75°
Pull out one sub-triangle \(AOB\): two legs and the angle at \(O\) give the boundary \(AB\) and the area.

The angle at \(O\) between two rays is the difference of their bearings (adding across north when they lie either side of it):

\[\angle = \text{(larger bearing)} - \text{(smaller bearing)}\]
=β2β1

A boundary length (the side opposite \(O\)) comes from the cosine rule:

\[AB^2 = OA^2 + OB^2 - 2\,OA\cdot OB\cos(\angle AOB)\]
c2=a2+b22abcosC

The area of a sub-triangle (and, added up, the whole field) is:

\[A = \tfrac{1}{2}\,ab\sin C\]
A=12absinC
Whole field. Add the areas of every sub-triangle from \(O\): \(\text{Total}=\triangle AOB+\triangle BOC+\triangle COD+\triangle DOA\).

How to solve a radial survey

  1. Draw the survey: mark north at \(O\), then a ray to each corner at its bearing, labelled with its distance.
  2. Angle at \(O\): subtract the two bearings; if the rays sit either side of north, add the piece up to north to the piece past it.
  3. Boundary length: apply the cosine rule to the sub-triangle, using the two legs and the angle between them.
  4. Area: use \(A=\tfrac{1}{2}ab\sin C\) on each sub-triangle.
  5. Whole field: add the areas of all the sub-triangles.
Example 1 — Angle at O
The radial survey shown was taken from \(O\). Find \(\angle AOB\) and \(\angle DOA\).
Solution

\(A\) and \(B\) are the same side of north; \(OD\) and \(OA\) sit either side of it.

Compass radial survey from central point OA compass radial survey taken from central point O, with rays to corner stations A (035 deg, 48 m), B (110 deg, 62 m), C (195 deg, 40 m) and D (295 deg, 55 m); the dashed quadrilateral ABCD is the field boundary. N A 48 m 035° B 62 m 110° C 40 m 195° D 55 m 295° O
\(\angle AOB\)\(=\)\(110^\circ - 35^\circ = 75^\circ\)
\(OD\text{ to N}\)\(=\)\(360^\circ - 295^\circ = 65^\circ\)
\(\angle DOA\)\(=\)\(65^\circ + 35^\circ = 100^\circ\)
75°,100°
Example 2 — Distance by the cosine rule
Find the boundary length \(AB\), with \(OA=48\) m, \(OB=62\) m and \(\angle AOB=75^\circ\). Give the answer to \(2\) d.p.
Solution

Apply the cosine rule to triangle \(AOB\).

Sub-triangle AOB extracted from the surveyTriangle A-O-B with OA = 48 m, OB = 62 m and the included angle AOB = 75 degrees, with the boundary side AB labelled x. O A B 48 m 62 m x 75°
\(AB^2\)\(=\)\(OA^2 + OB^2 - 2\,OA\cdot OB\cos(\angle AOB)\)
\(AB^2\)\(=\)\(48^2 + 62^2 - 2(48)(62)\cos 75^\circ\)
\(AB^2\)\(=\)\(6148 - 1540.49 = 4607.51\)
\(\therefore\ AB\)\(\approx\)\(67.88\text{ m}\)
AB67.88
Example 3 — Area of a sub-triangle
Find the area of sub-triangle \(AOB\), with \(OA=48\) m, \(OB=62\) m and \(\angle AOB=75^\circ\). Give the answer to \(2\) d.p.
Solution

Use the area rule \(A=\tfrac{1}{2}ab\sin C\) on the two legs and the angle between them.

Sub-triangle AOB for the area ruleTriangle A-O-B with OA = 48 m, OB = 62 m and the included angle AOB = 75 degrees, used with the area rule. O A B 48 m 62 m 75°
\(A\)\(=\)\(\tfrac{1}{2}\times OA \times OB \times \sin(\angle AOB)\)
\(A\)\(=\)\(\tfrac{1}{2}\times 48 \times 62 \times \sin 75^\circ\)
\(A\)\(=\)\(1488 \times 0.96593\)
\(\therefore\ A\)\(\approx\)\(1437.30\text{ m}^2\)
A1437.30
Example 4 — Area of the whole field
Find the total area of field \(ABCD\), given \(\angle AOB=75^\circ\), \(\angle BOC=85^\circ\), \(\angle COD=100^\circ\), \(\angle DOA=100^\circ\). Give the answer to \(2\) d.p.
Solution

Find each sub-triangle with \(\tfrac{1}{2}ab\sin C\), then add the four.

Compass radial survey from central point OA compass radial survey taken from central point O, with rays to corner stations A (035 deg, 48 m), B (110 deg, 62 m), C (195 deg, 40 m) and D (295 deg, 55 m); the dashed quadrilateral ABCD is the field boundary. N A 48 m 035° B 62 m 110° C 40 m 195° D 55 m 295° O
\(\triangle AOB\)\(=\)\(\tfrac{1}{2}(48)(62)\sin 75^\circ = 1437.30\)
\(\triangle BOC\)\(=\)\(\tfrac{1}{2}(62)(40)\sin 85^\circ = 1235.28\)
\(\triangle COD\)\(=\)\(\tfrac{1}{2}(40)(55)\sin 100^\circ = 1083.29\)
\(\triangle DOA\)\(=\)\(\tfrac{1}{2}(55)(48)\sin 100^\circ = 1299.95\)
\(\therefore\ \text{Total}\)\(\approx\)\(5055.82\text{ m}^2\)
5055.82

Common pitfalls

Add across north. When two rays lie either side of north, do not just subtract the bearings; add the angle up to north to the angle past it — for example \(295^\circ\) and \(035^\circ\) give \(65^\circ+35^\circ=100^\circ\).
Use the angle at \(O\). The cosine rule and the area rule both use the angle between the two legs. If it is obtuse, its cosine is negative — keep the sign or the length comes out wrong.
Add every sub-triangle. The whole-field area is the sum of all the triangles from \(O\); the boundary side \(AB\) is not a leg from \(O\), so it is never used in \(A=\tfrac{1}{2}ab\sin C\).

Frequently asked questions

What is a compass radial survey?

It is a survey of a field taken from one central point, called the central station or O. From that point the surveyor measures the bearing and the distance to each corner of the field. Those bearings and distances are the field notes, and the field is then split into triangles from O to find lengths and areas.

How do you find the angle between two rays in a radial survey?

Subtract the two bearings. If the two rays are on the same side of north the angle is simply the larger bearing minus the smaller. If they lie either side of north, add the angle from one ray up to north to the angle from north on to the other, for example 295 degrees and 035 degrees give 65 plus 35, which is 100 degrees.

How do you find the distance between two corners?

Use the cosine rule on the triangle made by the central point and the two corners. The distance is the side opposite the central point, so it equals the square root of OA squared plus OB squared minus two times OA times OB times the cosine of the angle at O.

How do you find the area of a field from a radial survey?

Find the area of each sub-triangle from the central point using A equals one half a b sine C, where a and b are the two distances and C is the angle between them. Then add the areas of all the sub-triangles to get the total area of the field.

Why is a radial survey split into triangles?

Because every ray from the central station O is a known distance and every pair of adjacent rays has a known angle between them. That gives each triangle two sides and the included angle, which is exactly what the cosine rule and the area rule need, so the whole field can be solved without any right angles.

What is the difference between a radial survey and using the trapezoidal rule?

A radial survey works from a central point using bearings, distances and the cosine and area rules for triangles. The trapezoidal rule instead estimates an area from offset measurements taken along a straight baseline. In Standard 2 the trapezoidal-rule area of a survey belongs to the Ratios and rates topic, while the triangle methods here belong to Trigonometry.