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MST-12-S2-04 ~55 min ⚡ +95 XP available

Compass Radial Surveys

One person, standing still in the middle of a paddock, can measure the whole thing. A bearing and a distance to each corner cuts an irregular field into triangles, and every one of those triangles arrives ready for the two rules you already have.

Today's hook, A paddock has four corners, no right angles and no parallel sides. A surveyor stands at one point inside it and records four bearings and four distances. She never walks the fence line, never measures a boundary, and never finds a perpendicular height. She still gets the area to the nearest square metre. How?

0/5QUESTS

Orient to radial surveys

Meet the survey from a central point, set the goal and settle the key terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any question in this focus area.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any question

Recall, your gut answer first

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You know the area of a triangle from two sides and the angle between them. You know how to find a side from two sides and the angle between them. A four-sided paddock is not a triangle, and you are standing in the middle of it.

Without calculating write down how you would cut that paddock into shapes you already know how to measure, and what you would need to record from where you are standing.

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What a compass radial survey is

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A compass radial survey is taken from a single point inside the field, usually called $O$. For each corner the surveyor records two things and only two things: the true bearing of that corner from $O$, and the distance to it.

Why it works. Every pair of neighbouring corners forms a triangle with $O$. Both sides of that triangle are distances you measured, and the angle between them is the difference of two bearings. That is SAS, every single time, for every triangle in the field.

The check that comes free. The angles at $O$ go all the way round, so they must add to $360°$. If yours do not, one of your subtractions is wrong, and it is nearly always the wrap-around one. You find that out before you calculate a single area.

angle at $O$ $=$ difference of the two bearings

The ANGLES

Subtract the smaller bearing from the larger. For the last gap, the one that crosses north, add what is left to $360°$ to the first bearing.

The AREA

Area rule on each triangle, $A = \tfrac{1}{2}ab\sin C$, then add them up. The two radii are $a$ and $b$ and the angle at $O$ is $C$.

The BOUNDARIES

Cosine rule on the same triangle, $c^2 = a^2 + b^2 - 2ab\cos C$. That gives a fence length nobody ever walked.

What you'll master

Know

Key facts

A radial survey records a bearing and a distance to each corner from one point
Each pair of neighbouring radial lines makes an SAS triangle
The angles at the central point must total $360°$
Area rule for areas, cosine rule for boundary lengths

Understand

Concepts

Why a radial survey turns an irregular field into triangles you can already solve
Why the angle at $O$ is a difference of bearings and never a bearing itself
Why the last angle, the one crossing north, is calculated differently
Why an obtuse angle at $O$ is no obstacle to the area rule

Can do

Skills

Construct a radial survey diagram from a table of bearings and distances
Find every angle at the central point and check they total $360°$
Find the total area of the field and the length of any boundary
Combine bearings, elevation and both triangle rules in one problem

Key terms

Compass radial surveyA survey of a field taken from one point inside it, recording a true bearing and a distance to each corner. Like this: a four-corner paddock gives four bearings and four distances, and nothing else needs measuring.

Central pointThe single spot the surveyor works from, labelled $O$. Like this: every triangle in the survey has $O$ as one of its vertices, which is why they all share two measured sides.

Radial lineThe straight line from $O$ out to one corner, whose length is the recorded distance. Like this: if corner $A$ is $85\text{ m}$ away on $041°$T, the radial line $OA$ is $85\text{ m}$ long and points on a bearing of $041°$T.

Included angle at OThe angle between two neighbouring radial lines, found by subtracting their bearings. Like this: corners on $041°$T and $128°$T give an angle at $O$ of $128° - 041° = 87°$.

Closing the surveyChecking that the angles you worked out at $O$ add to $360°$. Like this: $87° + 86° + 91° + 96° = 360°$, so every subtraction is right and the calculation can start.

Work the method in order

Turn bearings into angles at the centre, then check they close to 360 degrees.

The method, in the order you should work

core concept

Every radial survey question runs the same four steps, and the order matters because step 2 catches your errors before they spread into everything after it:

1 · Draw $O$, mark north, and draw each radial line at its bearing with its length

2 · Angles at $O$: $\quad$ difference of neighbouring bearings, then check they total $360°$

3 · Area: $\quad A = \tfrac{1}{2}ab\sin C$ on each triangle, then add

4 · Boundaries: $\quad c^2 = a^2 + b^2 - 2ab\cos C$ on whichever triangle you need

Step 2 has one awkward case and it is always the same one. Going clockwise from north, the last angle is the one that crosses north on its way back to the first corner. You cannot subtract those two bearings directly, because the smaller one is on the far side of $0°$. Take what is left of the circle after the last bearing, $360°$ minus it, and add the first bearing to that.

An obtuse angle at $O$ changes nothing here. The area rule uses $\sin C$, and $\sin 140°$ is a perfectly ordinary positive number equal to $\sin 40°$. The awkwardness you met in the sine rule was about finding an unknown angle, and in a radial survey every angle at $O$ is already known.

In a compass radial survey, each corner is recorded as a true bearing and a distance from one central point $O$. The angle at $O$ between two neighbouring corners is the difference of their bearings, and all the angles at $O$ must total $360°$. Each triangle is SAS, so its area is $\tfrac{1}{2}ab\sin C$ and its boundary side comes from $c^2 = a^2 + b^2 - 2ab\cos C$. Total area is the sum of the triangles.

Pause, copy the four steps in order, the $360°$ check, and one line about the wrap-around angle that crosses north, into your book.

Quick check: In a radial survey from $O$, corner $A$ is on a bearing of $041°$T and corner $B$ is on a bearing of $128°$T. What is angle $AOB$?

Work three survey examples

Follow constructing and closing a survey, finding total area, and recovering an unwalked boundary.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CONSTRUCT AND CLOSE THE SURVEY

A paddock $ABCD$ is surveyed from a central point $O$. The record reads: $A$, $041°$T, $85\text{ m}$; $B$, $128°$T, $70\text{ m}$; $C$, $214°$T, $96\text{ m}$; $D$, $305°$T, $60\text{ m}$. Find all four angles at $O$ and check the survey closes.

$\angle AOB = 128° - 041° = 87°$

The bearings are already in clockwise order, so neighbouring pairs subtract directly. Sketch it first, with north drawn up from $O$.

PROBLEM 2 · TOTAL AREA OF THE FIELD

Using the same survey, find the total area of paddock $ABCD$, correct to the nearest square metre.

$[OAB] = \tfrac{1}{2}(85)(70)\sin 87° = 2975(0.998630) = 2970.92$

The two radii are the two sides and the angle at $O$ sits between them, so this is the area rule with no rearranging at all.

PROBLEM 3 · A FENCE LINE NOBODY WALKED

Using the same survey again, find the length of boundary $AB$, correct to 2 decimal places.

$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(\angle AOB)$

Triangle $OAB$ has two known sides and the angle between them, which is the cosine rule situation exactly. $AB$ is the side opposite the angle at $O$.

True or false: In a compass radial survey taken from a point inside the field, the angles at the central point must add to $360°$.

Avoid the three traps

Spot the bearing used as an angle, the wrap-around gap and the dropped half in the area rule.

Common errors · the 3 traps that cost marks

Trap 01

Using a bearing as if it were the angle at O

A bearing is measured from north, not from the neighbouring radial line. Putting $128°$ into the area rule instead of the $87°$ between the two lines gives a confident, wrong area. The angle at $O$ is always a difference of two bearings.

Trap 02

Subtracting straight through the wrap-around gap

For the last angle, the two bearings sit either side of north, so a direct subtraction gives the reflex angle instead. With $305°$ and $041°$, subtracting gives $264°$ when the answer is $96°$. The $360°$ check catches this every time, which is why it is step 2 and not the last thing you do.

Trap 03

Rounding each triangle before adding them

Four areas each rounded to the nearest square metre can shift the total by several square metres, and in a question marked to the nearest square metre that is the difference between right and wrong. Keep the decimals in your calculator and round only the final total.

Fill the gaps: With corners on $041°$T, $128°$T, $214°$T and $305°$T: $\angle COD =$ degrees, the wrap-around $\angle DOA =$ degrees, and the four angles total degrees.

Drill radial surveys

Run the five drills across one survey until the order of work is automatic.

Quick-fire practice · 5 calculations

All five drills use one radial survey from a central point $O$: $P$, $035°$T, $40\text{ m}$; $Q$, $110°$T, $55\text{ m}$; $R$, $250°$T, $48\text{ m}$; $S$, $310°$T, $62\text{ m}$.

Find angle $POQ$ and the wrap-around angle $SOP$, then check the survey closes.

Find the area of triangle $POQ$, correct to 2 decimal places.

Find the length of boundary $PQ$, correct to 2 decimal places.

Find the area of triangle $QOR$, correct to 2 decimal places. Its angle at $O$ is obtuse, so say what difference that makes.

Find the total area of the field $PQRS$, correct to the nearest square metre.

Match each part of a radial survey question to the tool that does it:

The angle between two radial lines
The area of one triangle in the survey
The length of one boundary of the field
Checking your four angles before you go on

Cosine rule
Subtract the two bearings
They must total 360 degrees
Area rule, one half ab sin C

Revisit your thinking

Return to your opening answer and name what has changed.

Revisit your thinking

Back to the surveyor standing in the paddock. Four bearings and four distances, taken without moving, and here is everything they bought her:

Four angles at $O$ of $87°$, $86°$, $91°$ and $96°$, which total $360°$ and so confirm the bearings were ordered and subtracted consistently. The total does not prove the readings themselves are right: gaps around a point always sum to $360°$, so a mistyped bearing can still pass this check. Four triangle areas by the area rule, totalling $\mathbf{11738\text{ m}^2}$. Any fence length she wants by the cosine rule, such as $AB = 107.25\text{ m}$.

She never walked a boundary, never measured a perpendicular height, and never needed a right angle anywhere in the field. Eight numbers, taken from one spot, describe the whole paddock.

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Practise radial surveys

Answer the question bank, then write full short-answer responses.

Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 33 marks

Q1. In a radial survey from a central point $O$, corner $X$ is $44\text{ m}$ away on a bearing of $052°$T and corner $Y$ is $61\text{ m}$ away on a bearing of $137°$T. Find the area of triangle $XOY$, correct to 2 decimal places. (3 marks)

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ApplyBand 45 marks

Q2. A paddock $ABCD$ is surveyed from a central point $O$: $A$, $041°$T, $85\text{ m}$; $B$, $128°$T, $70\text{ m}$; $C$, $214°$T, $96\text{ m}$; $D$, $305°$T, $60\text{ m}$. Find the total area of the paddock to the nearest square metre, and the length of boundary $CD$ correct to 2 decimal places. (5 marks)

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AnalyseBand 55 marks

Q3. From a point $P$ the bearing of a communications tower $T$ is $040°$T. A second point $Q$ lies $500\text{ m}$ due east of $P$, and from $Q$ the bearing of $T$ is $325°$T.
(a) Find the distance $PT$, correct to the nearest $0.1\text{ m}$. (3 marks)
(b) From $P$, the angle of elevation to the top of the tower is $8°$. Find the height of the tower, correct to the nearest $0.1\text{ m}$. (2 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $\angle POQ = 110° - 035° = 75°$ and $\angle SOP = (360° - 310°) + 035° = 85°$. The other two are $\angle QOR = 140°$ and $\angle ROS = 60°$, and $75° + 140° + 60° + 85° = 360°$, so the survey closes.

Drill 2: $[POQ] = \tfrac{1}{2}(40)(55)\sin 75° = 1100 \times 0.965926 = 1062.52\text{ m}^2$

Drill 3: $PQ^2 = 40^2 + 55^2 - 2(40)(55)\cos 75° = 4625 - 1138.80 = 3486.20$, so $PQ = 59.04\text{ m}$

Drill 4: $\angle QOR = 250° - 110° = 140°$, so $[QOR] = \tfrac{1}{2}(55)(48)\sin 140° = 1320 \times 0.642788 = 848.48\text{ m}^2$. The obtuse angle makes no difference at all, because $\sin 140° = \sin 40°$ is a perfectly ordinary positive number. Obtuseness only caused trouble in the sine rule when the angle was unknown, and here every angle at $O$ is known.

Drill 5: $[ROS] = \tfrac{1}{2}(48)(62)\sin 60° = 1288.65$ and $[SOP] = \tfrac{1}{2}(62)(40)\sin 85° = 1235.28$. Total $= 1062.52 + 848.48 + 1288.65 + 1235.28 = 4434.93$, which is $\mathbf{4435\text{ m}^2}$ to the nearest square metre.

Q1 (3 marks): The angle at $O$ is the difference of the bearings, $\angle XOY = 137° - 052° = 85°$ [1]. That angle is included between the two radial lines, so the area rule applies directly [1]: $A = \tfrac{1}{2}(44)(61)\sin 85° = 1342 \times 0.996195 = \mathbf{1336.89\text{ m}^2}$ [1].

Q2 (5 marks): Angles at $O$: $87°$, $86°$, $91°$ and $(360° - 305°) + 041° = 96°$, totalling $360°$ so the survey closes [1]. Areas: $\tfrac{1}{2}(85)(70)\sin 87° = 2970.92$, $\tfrac{1}{2}(70)(96)\sin 86° = 3351.82$, $\tfrac{1}{2}(96)(60)\sin 91° = 2879.56$, $\tfrac{1}{2}(60)(85)\sin 96° = 2536.03$ [2]. Total $= 11738.33$, which is $\mathbf{11738\text{ m}^2}$ [1]. For the boundary, $CD^2 = 96^2 + 60^2 - 2(96)(60)\cos 91° = 12816 + 201.05 = 13017.05$, so $CD = \mathbf{114.09\text{ m}}$ [1]. Note that $\cos 91°$ is negative, so that correction term is added and $CD$ comes out longer than the Pythagoras value of $113.21\text{ m}$, the opposite way round from $AB$ in the worked example.

Q3 (5 marks): (a) $Q$ is due east of $P$, so the bearing of $Q$ from $P$ is $090°$ and $\angle TPQ = 090° - 040° = 50°$ [1]. The bearing of $P$ from $Q$ is $270°$, so $\angle TQP = 325° - 270° = 55°$ [1]. Then $\angle PTQ = 180° - 50° - 55° = 75°$, and the sine rule gives $PT = \dfrac{500\sin 55°}{\sin 75°} = \dfrac{500 \times 0.819152}{0.965926} = \mathbf{424.0\text{ m}}$ [1]. (b) The tower is vertical, so its height sits in a right-angled triangle with the horizontal distance $PT$: height $= PT\tan 8° = 424.0237 \times 0.140541 = \mathbf{59.6\text{ m}}$ [2]. Sense check, $\tan 8°$ is close to $\tfrac{1}{7}$, and $424 \div 7 \approx 61$.

Finish the lesson

Take the boss battle, mark the lesson complete and move on.

Boss battle · Radial Surveys and Synthesis

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The full focus area in one challenge: Pythagoras, right-angled trigonometry, elevation and depression, bearings, the sine rule, the cosine rule and the area rule.

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Focus area overview · Year 12 · Maths Standard