What are compass bearings?

A compass bearing uses a primary direction (N or S) followed by an acute angle, then E or W. For example, N30E means start facing north and rotate 30 towards the east; S40W means start facing south and rotate 40 towards the west.

What are back-bearings?

If the bearing of B from A is θ, then the bearing of A from B is θ + 180 (subtract 360 if the result exceeds 360, or equivalently subtract 180 when θ ≥ 180).

What is watch the survey come together, stage by stage?

Suppose a surveyor at station O records three points: P at 040, 70 m; Q at 110, 80 m; R at 205, 60 m. The task is to find the straight-line distance PQ. Here is how the plot and the solution build up.

What are two-digit bearings?

Always use three digits: 075, not 75.

NESA 2022 HSC-style-style practice: A ship leaves port P and sails 40 km on a bearing of 075 to point A, then 55 km on a bearing of 145 to point B. Find the distance and bearing of B from P.

The interior angle at A in triangle PAB is the supplement of the change in direction. From 075 to 145 is a turn of 70, so the interior angle at A is 180 - 70 = 110. Distance PB by the cosine rule: PB^2 = 40^2 + 55^2 - 2 × 40 × 55 × cos 110 = 1600 + 3025 - 4400 × (-0.3420) = 4625 + 1505 = 6130. PB ≈ sqrt(6130) ≈ 78.3 km. For the bearing, find angle APB by the sine rule: sin( APB) = (55 sin 110)/(78.3) ≈ (55 × 0.9397)/(78.3) ≈ 0.660, so APB ≈ 41.3. Bearing of B from P: 075 + 41.3 = 116.3, round to 116. Markers reward the interior angle at A (with the supplement step shown), the cosine rule, the sine rule for the angle at P, and the final bearing.

§-Syllabus dot point

NSWMaths Standard 2Syllabus dot point

How are bearings and radial surveys used to find distances and directions in navigation and surveying?

Use compass and true bearings, and radial surveys, to solve practical navigation and surveying problems

A focused answer to the HSC Maths Standard 2 dot point on bearings and radial surveys. Compass vs true bearings, back-bearings, the structure of a radial survey, and worked Australian navigation examples using the sine and cosine rules, with stage-by-stage diagrams.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-20

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

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What this dot point is asking

NESA wants you to read and write both compass and true bearings, draw radial-survey diagrams, and combine bearings with the sine rule, cosine rule and Pythagoras to find distances and directions in navigation and surveying problems.

Almost every bearings question is a triangle in disguise. The marks are rarely for the trigonometry itself. The sine and cosine rules are on the NESA reference sheet, so you do not need to recall them. The marks are for turning the words and the compass directions into a correctly labelled triangle with the right interior angle. Get that angle right and the rest is routine. Get it wrong and a perfect calculation earns nothing. So the real skill tested here is translation: turning "sails on a bearing of $075\degree$ , then changes course" into "a triangle with these two sides and this enclosed angle".

The answer

True bearings

A true bearing is measured clockwise from north, in three digits, from $000\degree$ to $360\degree$ .

North: $000\degree$
East: $090\degree$
South: $180\degree$
West: $270\degree$

So a bearing of $075\degree$ means $75\degree$ clockwise from north. Always write three digits ( $075\degree$ , not $75\degree$ ). The leading zeros are not decoration. They mark the answer as a bearing and stop a reader misreading $075$ as $750$ . A bearing larger than $180\degree$ points into the western half of the compass. That is a useful first sense-check: $250\degree$ is roughly west-south-west, so if your diagram has that direction pointing east, something is wrong.

Compass bearings

A compass bearing uses a primary direction (N or S) followed by an acute angle, then E or W. For example, N $30\degree$ E means start facing north and rotate $30\degree$ towards the east; S $40\degree$ W means start facing south and rotate $40\degree$ towards the west.

Compass bearings are less common in the HSC than true bearings, but you must be able to convert. The clean way is to anchor on the cardinal the compass bearing names and add or subtract:

N $\theta$ E is $\theta$ true (measured east from north, the same direction as clockwise from north).
N $\theta$ W is $360\degree - \theta$ true.
S $\theta$ E is $180\degree - \theta$ true.
S $\theta$ W is $180\degree + \theta$ true.

You should not memorise these as four facts; derive each one on the diagram by walking clockwise from north. N $30\degree$ E lands $30\degree$ clockwise from north, so $030\degree$ . S $40\degree$ W is past south ( $180\degree$ ) by another $40\degree$ , so $220\degree$ .

Back-bearings

If the bearing of $B$ from $A$ is $\theta$ , then the bearing of $A$ from $B$ is $\theta + 180\degree$ (subtract $360\degree$ if the result exceeds $360\degree$ , or equivalently subtract $180\degree$ when $\theta \geq 180\degree$ ).

So if $B$ is at bearing $075\degree$ from $A$ , then $A$ is at bearing $255\degree$ from $B$ . Here is why. The two north arrows (one at $A$ , one at $B$ ) point the same way, so they are parallel. This makes the line out and the line back form a straight angle, that is, a half-turn of $180\degree$ . The bearing therefore reverses by exactly half a turn. This is also why every point in a bearings diagram needs its own north arrow: a bearing is always read from the local north at the point you are standing on.

Radial surveys

A radial survey (also called a plane-table survey) records the distances and true bearings of several points from a single central station. It is the standard way to fix the positions of features around one observation point, and the data is given as a table or a diagram. Once plotted, every point sits at a known distance and bearing from the centre, so any two points and the centre form a triangle you can solve.

To find the interior angle of the triangle at the central station, take the difference of the two bearings to the surveyed points.
To find a distance or angle between two surveyed points (a side not touching the centre), you almost always need the cosine rule (two radii and the included angle is SAS), and then the sine rule for any further angle.

The centre is the key: it turns a scatter of bearings and distances into a set of triangles that all share the station as one vertex.

Watch the survey come together, stage by stage

Suppose a surveyor at station $O$ records three points: $P$ at $040\degree$ , $70$ m; $Q$ at $110\degree$ , $80$ m; $R$ at $205\degree$ , $60$ m. The task is to find the straight-line distance $PQ$ . Here is how the plot and the solution build up.

Stage 1, plot each point from the station. Draw a north arrow at $O$ . For each point, turn clockwise from north by its bearing and step out by its distance to scale. This fixes $P$ , $Q$ and $R$ around the centre. Plotting is worth doing even when the question is numerical: a correct sketch is a marker expectation and it catches gross errors.

Stage 2, find the angle at the station. The wanted side $PQ$ sits in triangle $OPQ$ . Its angle at $O$ is the difference of the two bearings: $110\degree - 040\degree = 70\degree$ . This is the included angle between the two known sides $OP$ and $OQ$ , so the triangle is now a clean SAS setup.

Stage 3, apply the cosine rule. With $OP = 70$ , $OQ = 80$ and the included angle $70\degree$ , the side opposite the angle is $PQ$ :

PQ^2 = 70^2 + 80^2 - 2(70)(80)\cos 70\degree.

Stage 4, evaluate and read off the answer. $PQ^2 = 4900 + 6400 - 11200 \times 0.3420 = 11300 - 3830 = 7470$ , so $PQ \approx \sqrt{7470} \approx 86.4$ m. The plot now carries every piece of information the survey produced: bearings, distances and the computed gap.

The geometry pitfall: change of bearing versus interior angle

The single most common error in this topic is using the change of bearing as the interior angle of the triangle at a bend. They are different. The interior angle at the bend is the supplement of the change of bearing.

The logic: you arrive at $A$ heading on bearing $060\degree$ . You leave heading on bearing $150\degree$ . The change in heading is $90\degree$ . Now picture the original direction carried straight on (the dashed "straight ahead" line). This dashed line and the new leg make that $90\degree$ angle on the outside of the path. The inside angle of the triangle is the one between the leg you came in on and the leg you leave on. That inside angle is the supplement (the rest of the straight angle), $180\degree - 90\degree = 90\degree$ .

A reliable instinct: if the change is small (you barely turn), the interior angle is large (close to $180\degree$ , an almost-straight path). If the change is large (you almost double back), the interior angle is small (a tight, narrow triangle). When the result of your supplement step contradicts that picture, recheck.

Contrast this with a radial survey, where the two legs both start at the central station (no bend), so the angle between them is the plain difference of bearings, not the supplement. Knowing which situation you are in (bend in a path, or two radii from a centre) is what decides whether you subtract or take the supplement.

How exam questions ask about bearings

"Sails / drives / walks ... then changes course / turns ..." A bend in a path. Find the change of bearing, take its supplement for the interior angle, then use the cosine rule (SAS) for the distance back to the start.
"From the lighthouse / tower, two ships / landmarks are observed ..." A radial survey from one station. The angle at the station is the difference of the two bearings; use the cosine rule for the gap between the two observed points.
"Find the bearing of $B$ from $A$ ... hence the bearing of $A$ from $B$ ." A back-bearing: add or subtract $180\degree$ .
"Find the distance and bearing of $C$ from $A$ ." Two-part: cosine rule for the distance, then the sine rule for an interior angle, then add or subtract that angle from a known bearing to get the final bearing. State the bearing in three digits.
"Convert N $\theta$ E (or similar) to a true bearing." Walk clockwise from north on a sketch.

Worked example

Distance across a two-leg walk

A walker leaves base camp and walks $5$ km on bearing $060\degree$ to point $A$ , then $7$ km on bearing $150\degree$ to point $B$ . Find the straight-line distance from base camp to $B$ .

Set up: Label the triangle: base camp $C$ , the bend $A$ , the end $B$ . The two known sides are $CA = 5$ km and $AB = 7$ km, meeting at the bend $A$ .

Find the interior angle at the bend: The bearing changes from $060\degree$ to $150\degree$ , a change of $90\degree$ . The interior angle of the triangle at $A$ is the supplement of this change:

180\degree - 90\degree = 90\degree

The interior angle happens to be exactly $90\degree$ , so the triangle is right-angled at $A$ and Pythagoras applies directly (the cosine rule would give the same thing, since $\cos 90\degree = 0$ ).

Apply Pythagoras:

\text{dist}^2 = 5^2 + 7^2 = 25 + 49 = 74

Take the square root:

\text{dist} = \sqrt{74} \approx 8.6 \text{ km}

Answer: the straight-line distance from base camp to $B$ is approximately $8.6$ km.

Distance between two ships from Norah Head lighthouse

From the lighthouse at Norah Head on the NSW Central Coast, two ships are observed: ship $X$ at bearing $060\degree$ true and distance $4$ km, and ship $Y$ at bearing $135\degree$ true and distance $6$ km. Find the distance between the ships.

Find the angle at the station: Both ships are seen from the same station (the lighthouse), so this is a radial survey. The angle at the lighthouse is the difference of the two bearings:

135\degree - 060\degree = 75\degree

This $75\degree$ is the included angle of triangle (lighthouse) $XY$ between the two radii.

Apply the cosine rule: With sides $4$ km and $6$ km and included angle $75\degree$ :

XY^2 = 4^2 + 6^2 - 2 \times 4 \times 6 \times \cos 75\degree = 16 + 36 - 48 \times 0.2588 = 52 - 12.42 = 39.58

XY \approx \sqrt{39.58} \approx 6.29 \text{ km}

Answer: the distance between the two ships is approximately $6.29$ km.

Distance and bearing of B from port P

A yacht sails $40$ km from port $P$ on a bearing of $075\degree$ to $A$ , then $55$ km on a bearing of $145\degree$ to $B$ . Find the distance and bearing of $B$ from $P$ .

Find the distance: The course turns by $145\degree - 075\degree = 70\degree$ , so the interior angle at the bend $A$ is the supplement, $180\degree - 70\degree = 110\degree$ . By the cosine rule:

PB^2 = 40^2 + 55^2 - 2(40)(55)\cos 110\degree = 1600 + 3025 - 4400(-0.3420) = 6130

PB \approx \sqrt{6130} \approx 78.3 \text{ km}

Use the sine rule for the bearing: To turn the distance into a bearing, find the angle $APB$ ... the angle the line $PB$ makes at $P$ , away from the first leg. By the sine rule in triangle $PAB$ :

\sin(\angle APB) = \frac{55 \sin 110\degree}{78.3} \approx \frac{55 \times 0.9397}{78.3} \approx 0.660

so $\angle APB \approx 41.3\degree$ . The first leg left $P$ on bearing $075\degree$ , and $B$ lies a further $41.3\degree$ clockwise from it, so the bearing of $B$ from $P$ is:

075\degree + 41.3\degree \approx 116\degree

Answer: $B$ is about $78.3$ km from $P$ on a bearing of $116\degree$ .

Convert a compass bearing and take a back-bearing

A yacht observes a headland at compass bearing S $70\degree$ W. What is this as a true bearing, and what is the bearing of the yacht from the headland?

Convert and reverse: S $70\degree$ W is past south ( $180\degree$ ) by another $70\degree$ towards the west, so the true bearing is $180\degree + 70\degree = 250\degree$ . The back-bearing (yacht from headland) reverses by $180\degree$ ; since $250\degree \geq 180\degree$ , subtract:

250\degree - 180\degree = 070\degree

Answer: the headland is at $250\degree$ true from the yacht, and the yacht is at $070\degree$ from the headland.

Common traps

Confusing change of direction with interior angle: At a bend in a path the interior angle of the triangle is the supplement of the bearing change, not the change itself. (At a central station between two radii, it is the difference of the bearings.)
Two-digit bearings: Always use three digits: $075\degree$ , not $75\degree$ .
Mixing up compass and true: N $30\degree$ E happens to equal $030\degree$ true, but N $30\degree$ W is $330\degree$ and S $30\degree$ W is $210\degree$ . Walk clockwise from north on a sketch every time.
Forgetting the second part of the answer: "Distance and bearing" needs both; the bearing usually takes a further sine-rule step after the cosine rule gives the distance.
Wrong calculator mode: Always degrees, not radians.
No diagram, or one shared north: Mark a north arrow at every point. Bearings are reckoned from each point's local north, not a single global one. The marker expectation is a labelled diagram before any algebra.

Exam-style practice questions

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

2022 HSC-style5 marksA ship leaves port

P

and sails

40

km on a bearing of

075\degree

to point

A

, then

55

km on a bearing of

145\degree

to point

B

. Find the distance and bearing of

B

from

P

Show worked answer →

The interior angle at $A$ in triangle $PAB$ is the supplement of the change in direction. From $075\degree$ to $145\degree$ is a turn of $70\degree$ , so the interior angle at $A$ is $180\degree - 70\degree = 110\degree$ .

Distance $PB$ by the cosine rule:

$PB^2 = 40^2 + 55^2 - 2 \times 40 \times 55 \times \cos 110\degree = 1600 + 3025 - 4400 \times (-0.3420) = 4625 + 1505 = 6130$ .

$PB \approx \sqrt{6130} \approx 78.3$ km.

For the bearing, find angle $APB$ by the sine rule: $\sin(\angle APB) = \frac{55 \sin 110\degree}{78.3} \approx \frac{55 \times 0.9397}{78.3} \approx 0.660$ , so $\angle APB \approx 41.3\degree$ .

Bearing of $B$ from $P$ : $075\degree + 41.3\degree = 116.3\degree$ , round to $116\degree$ .

Markers reward the interior angle at $A$ (with the supplement step shown), the cosine rule, the sine rule for the angle at $P$ , and the final bearing.

2021 HSC-style4 marksFrom point

A

, point

B

is at a bearing of

030\degree

and a distance of

200

m. From point

A

, point

C

is at a bearing of

120\degree

and a distance of

150

m. Find the distance

BC

Show worked answer →

In triangle $ABC$ , the angle at $A$ is the difference in bearings: $120\degree - 030\degree = 90\degree$ .

Use Pythagoras (or the cosine rule with $\cos 90\degree = 0$ ):

$BC^2 = 200^2 + 150^2 = 40000 + 22500 = 62500$ .

$BC = \sqrt{62500} = 250$ m.

Markers reward the angle at $A$ identified as the difference of bearings, recognising it as a right angle, and the distance. A radial-survey diagram is expected.

Practice questions

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

foundation1 marksA surveyor at a central station records a water tank on a true bearing of fifty degrees and a flagpole on a true bearing of one hundred and thirty degrees. Find the size of the angle at the station between the line to the tank and the line to the flagpole.

Show worked solution →

Identify the situation. Both points are sighted from the same central station, so this is a radial survey and the two lines are radii from the station. The angle between two radii at the station is the difference of their true bearings, not a supplement.

Subtract the bearings. Take the larger bearing minus the smaller.

130\degree - 50\degree = 80\degree

State the answer. The angle at the station between the two lines is $80\degree$ . Check: $80\degree$ is less than $180\degree$ and the flagpole is the more clockwise of the two sightings, so a positive acute-to-obtuse opening between them is exactly what the plot shows.

foundation1 marksFrom a fishing boat, a jetty is on a true bearing of two hundred and forty-eight degrees. Find the true bearing of the boat from the jetty (the back-bearing).

Show worked solution →

Recognise a back-bearing. The two north arrows, one at the boat and one at the jetty, are parallel, so the line out and the line back make a straight angle. The return bearing therefore differs from the forward bearing by exactly $180\degree$ .

Decide add or subtract. The forward bearing is $248\degree$ , which is already at least $180\degree$ , so adding $180\degree$ would push past $360\degree$ . Subtract instead.

248\degree - 180\degree = 68\degree

State the answer. The boat is on a true bearing of $068\degree$ from the jetty. Check: $068\degree$ and $248\degree$ differ by exactly $180\degree$ , and one bearing falls in the north-east quadrant while the other falls in the south-west quadrant, as opposite directions must.

foundation2 marksA bushwalker records a saddle at the compass bearing N sixty-eight degrees W. Convert this compass bearing to a three-figure true bearing.

Show worked solution →

Picture the compass bearing. N $68\degree$ W means start facing north and rotate $68\degree$ towards the west. Turning from north towards the west is an anticlockwise turn, so the true bearing (measured clockwise from north) is the rest of the way round.

Walk clockwise from north. A full turn is $360\degree$ , and the direction sits $68\degree$ short of north on the western side, so subtract from $360\degree$ .

360\degree - 68\degree = 292\degree

State the answer. The saddle is on a true bearing of $292\degree$ . Check: N $68\degree$ W points into the north-west quadrant, and $292\degree$ lies between $270\degree$ (west) and $360\degree$ (north), exactly the north-west quadrant, so the conversion is sound.

foundation3 marksFrom a yacht, a headland is observed at the compass bearing S twenty-five degrees W. First convert this compass bearing to a true bearing. Then, given that a buoy is on a true bearing of one hundred and ten degrees from the yacht, find the true bearing of the yacht from the buoy.

Show worked solution →

Convert the compass bearing to true. S $25\degree$ W means start facing south and turn $25\degree$ towards the west. South is $180\degree$ true, and turning further towards the west increases the bearing, so add.

180\degree + 25\degree = 205\degree

Set up the back-bearing. The bearing of the buoy from the yacht is $110\degree$ . The bearing of the yacht from the buoy is the back-bearing, found by adding $180\degree$ (the two north arrows are parallel, so the return line reverses by a straight angle).

110\degree + 180\degree = 290\degree

State the answer. The headland is at $205\degree$ true from the yacht, and the yacht is at $290\degree$ from the buoy. Check: $205\degree$ lies in the south-west quadrant ( $180\degree$ to $270\degree$ ), matching S $25\degree$ W; and $290\degree$ exceeds $110\degree$ by exactly $180\degree$ , as a back-bearing must.

core3 marksFrom a lighthouse, a radial survey fixes two fishing boats. The first boat is on a true bearing of thirty-five degrees at a distance of eight kilometres, and the second boat is on a true bearing of one hundred and ten degrees at a distance of twelve kilometres. Find the distance between the two boats, correct to two decimal places.

Show worked solution →

Find the angle at the lighthouse. Both boats are sighted from the one station, so the angle between the two radii is the difference of the bearings.

110\degree - 35\degree = 75\degree

Recognise the SAS triangle. In the triangle formed by the lighthouse and the two boats, the two known sides are the radii ( $8$ km and $12$ km) and the angle between them is $75\degree$ , so this is a side-angle-side setup and the cosine rule gives the opposite side, the gap between the boats.

Apply the cosine rule. Let $d$ be the distance between the boats.

d^2 = 8^2 + 12^2 - 2 \times 8 \times 12 \times \cos 75\degree = 208 - 192 \times 0.2588 = 208 - 49.69 = 158.31

Take the square root.

d = \sqrt{158.31} \approx 12.58 \text{ km}

State the answer. The boats are about $12.58$ km apart. Check: the longer radius is $12$ km, and $12.58$ km is a little more than that, which is sensible because the $75\degree$ opening between the radii spreads the boats wider than either single distance from the lighthouse.

core2 marksA plane-table survey from a central station fixes a marker tree on a true bearing of twenty degrees at a distance of thirty metres, and a boundary rock on a true bearing of ninety-five degrees at a distance of forty-five metres. Find the area of the triangle joining the station, the tree and the rock, correct to the nearest square metre.

Show worked solution →

Find the included angle at the station. The two sightings are radii from the one station, so the angle between them is the difference of the bearings.

95\degree - 20\degree = 75\degree

Choose the area rule. Two sides (the radii $30$ m and $45$ m) and the included angle $75\degree$ are known, so use the area formula $\text{Area} = \frac{1}{2}ab\sin C$ from the reference sheet.

Substitute and evaluate.

\text{Area} = \frac{1}{2} \times 30 \times 45 \times \sin 75\degree = 675 \times 0.9659 \approx 652 \text{ m}^2

State the answer. The area of the triangle is about $652$ m $^2$ . Check: half the product of the radii is $675$ m $^2$ , and multiplying by $\sin 75\degree$ (just under $1$ ) trims it slightly to $652$ m $^2$ , exactly as expected for an angle near a right angle.

core3 marksA fire tower acts as the central station for a radial survey. A hut is recorded on a true bearing of three hundred and ten degrees at a distance of five kilometres, and a cabin is recorded on a true bearing of forty degrees at a distance of seven kilometres. Find the straight-line distance between the hut and the cabin, correct to one decimal place.

Show worked solution →

Find the angle at the tower. The two bearings straddle north, so subtracting directly ( $310\degree - 40\degree = 270\degree$ ) gives the reflex angle on the wrong side. The angle inside the triangle is the rest of the full turn.

360\degree - 270\degree = 90\degree

Equivalently, swinging clockwise from $310\degree$ round through north ( $000\degree$ ) to $40\degree$ is $50\degree + 40\degree = 90\degree$ .

Recognise the triangle. The two radii ( $5$ km and $7$ km) enclose this $90\degree$ angle, so the triangle is right-angled at the tower and Pythagoras applies (the cosine rule gives the same result since $\cos 90\degree = 0$ ).

Apply Pythagoras. Let $d$ be the distance between the hut and the cabin.

d^2 = 5^2 + 7^2 = 25 + 49 = 74

d = \sqrt{74} \approx 8.6 \text{ km}

State the answer. The hut and the cabin are about $8.6$ km apart. Check: $8.6$ km is the hypotenuse, larger than either radius ( $5$ km and $7$ km) but smaller than their sum ( $12$ km), which the triangle inequality demands.

core4 marksFrom a road junction, a town is on a true bearing of thirty degrees at a distance of eight kilometres, and a silo is on a true bearing of one hundred degrees at a distance of eleven kilometres. Find the true bearing of the silo from the town, to the nearest degree.

Show worked solution →

Find the angle at the junction. The town and the silo are two radii from the one junction, so the angle between them is the difference of the bearings.

100\degree - 30\degree = 70\degree

Find the distance between them by the cosine rule. With radii $8$ km and $11$ km enclosing $70\degree$ , the side opposite (the town-to-silo distance $d$ ) is

d^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos 70\degree = 185 - 176 \times 0.3420 = 185 - 60.20 = 124.80

d = \sqrt{124.80} \approx 11.2 \text{ km}

Find the angle at the town by the sine rule. The side opposite the angle at the town is the radius to the silo ( $11$ km); the side opposite the $70\degree$ angle is $d \approx 11.2$ km.

\sin(\text{angle at town}) = \frac{11 \times \sin 70\degree}{11.2} \approx \frac{10.34}{11.2} \approx 0.9253

so the angle at the town is about $67.7\degree$ .

Convert to a bearing. Stand at the town. The junction lies on the back-bearing $30\degree + 180\degree = 210\degree$ . The silo sits $67.7\degree$ anticlockwise from the line back to the junction, so subtract.

210\degree - 67.7\degree = 142.3\degree

State the answer. The silo is on a true bearing of about $142\degree$ from the town. Check: the silo is the more easterly and southerly feature from the junction, so a bearing near $142\degree$ (south-east quadrant) from the town is the sensible direction.

exam5 marksA radial survey from a central station fixes two features. A pylon is on a true bearing of twenty-five degrees at a distance of sixty metres, and a gate is on a true bearing of one hundred degrees at a distance of ninety metres. Find, correct to one decimal place where rounding is needed, the distance from the pylon to the gate and the true bearing of the gate from the pylon.

Show worked solution →

Find the angle at the station. Both features are radii from the station, so the angle between them is the difference of the bearings.

100\degree - 25\degree = 75\degree

Find the distance by the cosine rule. With radii $60$ m and $90$ m enclosing $75\degree$ , the side opposite is the pylon-to-gate distance $d$ .

d^2 = 60^2 + 90^2 - 2 \times 60 \times 90 \times \cos 75\degree = 11700 - 10800 \times 0.2588 = 11700 - 2795.25 = 8904.75

d = \sqrt{8904.75} \approx 94.4 \text{ m}

Find an interior angle by the sine rule. To turn the distance into a bearing, find the angle of the triangle at the pylon. The side opposite that angle is the radius to the gate ( $90$ m), and the side opposite the $75\degree$ angle is $d \approx 94.4$ m.

\sin(\text{angle at pylon}) = \frac{90 \times \sin 75\degree}{94.4} \approx \frac{86.93}{94.4} \approx 0.9212

so the angle at the pylon is about $67.1\degree$ .

Convert to a bearing. Stand at the pylon. The station lies on the back-bearing of $025\degree$ , that is $25\degree + 180\degree = 205\degree$ . The gate sits $67.1\degree$ anticlockwise from the line back to the station, so subtract.

205\degree - 67.1\degree = 137.9\degree

State the answer. The gate is about $94.4$ m from the pylon on a true bearing of about $138\degree$ . Check: the gate is the more easterly feature from the station, so a bearing of $138\degree$ (south-east quadrant) from the pylon is a sensible direction, and rounding $137.9\degree$ gives $138\degree$ .

exam4 marksA council survey of a triangular reserve is taken from a single central station. Corner A is on a true bearing of zero degrees at a distance of fifty metres, corner B is on a true bearing of seventy degrees at a distance of eighty metres, and corner C is on a true bearing of one hundred and fifty degrees at a distance of sixty metres. Find the total area of the reserve enclosed by the three radii, correct to the nearest square metre.

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Split the region into triangles from the station. Taking the corners in bearing order, the region is the fan of two triangles sharing the station: one joining the station to A and B, the other joining the station to B and C. Find each area with $\frac{1}{2}ab\sin C$ and add.

First triangle (station, A, B). The included angle is the bearing difference $70\degree - 0\degree = 70\degree$ , between radii $50$ m and $80$ m.

\text{Area}_1 = \frac{1}{2} \times 50 \times 80 \times \sin 70\degree = 2000 \times 0.9397 \approx 1879.4 \text{ m}^2

Second triangle (station, B, C). The included angle is $150\degree - 70\degree = 80\degree$ , between radii $80$ m and $60$ m.

\text{Area}_2 = \frac{1}{2} \times 80 \times 60 \times \sin 80\degree = 2400 \times 0.9848 \approx 2363.5 \text{ m}^2

Add the two areas.

\text{Total} = 1879.4 + 2363.5 = 4242.9 \approx 4243 \text{ m}^2

State the answer. The reserve has an area of about $4243$ m $^2$ . Check: the two angles at the station total $80\degree$ short of a half-turn, fanning the corners across the eastern side, and each triangle area is a believable fraction of half its two radii multiplied together ( $2000$ m $^2$ and $2400$ m $^2$ ), so the total near $4243$ m $^2$ is reasonable.

exam5 marksA boat leaves a marina and motors nine kilometres on a true bearing of sixty-five degrees to a buoy, then turns and motors twelve kilometres on a true bearing of one hundred and fifty-five degrees to an island. Find the straight-line distance from the marina to the island, and the true bearing of the island from the marina, rounding to one decimal place where needed.

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Find the interior angle at the bend. This is a bend in a path, not two radii, so the interior angle of the triangle at the buoy is the supplement of the change of course. The course changes from $65\degree$ to $155\degree$ , a turn of $90\degree$ .

180\degree - 90\degree = 90\degree

Find the distance. The two legs ( $9$ km and $12$ km) enclose this $90\degree$ angle at the buoy, so the marina-to-island distance $d$ follows from Pythagoras (the cosine rule agrees, since $\cos 90\degree = 0$ ).

d^2 = 9^2 + 12^2 = 81 + 144 = 225

d = \sqrt{225} = 15 \text{ km}

Find an interior angle by the sine rule. To get the final bearing, find the angle of the triangle at the marina. The side opposite it is the second leg ( $12$ km); the side opposite the $90\degree$ angle is $d = 15$ km.

\sin(\text{angle at marina}) = \frac{12 \times \sin 90\degree}{15} = \frac{12}{15} = 0.8

so the angle at the marina is about $53.1\degree$ .

Convert to a bearing. The first leg left the marina on bearing $065\degree$ , and the island lies a further $53.1\degree$ clockwise from that leg, so add.

65\degree + 53.1\degree = 118.1\degree

State the answer. The island is $15$ km from the marina on a true bearing of about $118\degree$ . Check: the boat first heads east-north-east then swings south, so the island ending up south-east of the marina (a bearing near $118\degree$ ) is consistent, and $15$ km is a tidy hypotenuse of the $9$ , $12$ , $15$ right triangle.

exam6 marksA surveyor sets up at a trig station and records two features in a radial survey. A dam is on a true bearing of fifty degrees at a distance of one hundred and thirty metres, and a shed is on a true bearing of one hundred and fifty-five degrees at a distance of one hundred and eighty metres. (a) Find the straight-line distance from the dam to the shed, correct to one decimal place. (b) Hence find the true bearing of the shed from the dam, to the nearest degree. (c) Find the area of the triangle joining the station, the dam and the shed, to the nearest square metre.

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(a) Find the angle at the station. The dam and the shed are two radii from the one station, so the angle between them is the difference of the bearings.

155\degree - 50\degree = 105\degree

Apply the cosine rule. With radii $130$ m and $180$ m enclosing $105\degree$ , the side opposite (the dam-to-shed distance $d$ ) is

d^2 = 130^2 + 180^2 - 2 \times 130 \times 180 \times \cos 105\degree = 49300 - 46800 \times (-0.2588) = 49300 + 12112.7 = 61412.7

d = \sqrt{61412.7} \approx 247.8 \text{ m}

(b) Find the angle at the dam by the sine rule. The side opposite the angle at the dam is the radius to the shed ( $180$ m); the side opposite the $105\degree$ angle is $d \approx 247.8$ m.

\sin(\text{angle at dam}) = \frac{180 \times \sin 105\degree}{247.8} \approx \frac{173.86}{247.8} \approx 0.7016

so the angle at the dam is about $44.6\degree$ .

Convert to a bearing. Stand at the dam. The station lies on the back-bearing $50\degree + 180\degree = 230\degree$ . The shed sits $44.6\degree$ anticlockwise from the line back to the station, so subtract.

230\degree - 44.6\degree = 185.4\degree

so the shed is on a true bearing of about $185\degree$ from the dam.

(c) Find the area. Two sides (the radii $130$ m and $180$ m) and the included angle $105\degree$ are known, so use $\text{Area} = \frac{1}{2}ab\sin C$ .

\text{Area} = \frac{1}{2} \times 130 \times 180 \times \sin 105\degree = 11700 \times 0.9659 \approx 11301 \text{ m}^2

Answer: the dam and the shed are about $247.8$ m apart, the shed is on a true bearing of about $185\degree$ from the dam, and the triangle has an area of about $11301$ m $^2$ . Check: $247.8$ m exceeds the longer radius ( $180$ m) because the wide $105\degree$ opening spreads the features apart, and a bearing near $185\degree$ (just past due south) suits the shed being the more southerly feature.