G15Geometry & MeasuresFoundation & HigherGrades 3–7

Bearings

Q: Why do bearings use three figures?

So they cannot be misread or altered — "45" could become "145" by accident, but "045" is unambiguous. It is a navigation convention.

Q: Can a bearing be 360^?

Due north is written 000^; 360^ is the same direction but is not normally used.

Q: Are the sine and cosine rules needed for bearings?

Only at Higher, when the triangle formed has no right angle.

A bearing gives a direction as an angle measured clockwise from north, written with three figures: east is $090^\circ$ , south is $180^\circ$ , and a direction just past north-west is around $320^\circ$ . Bearings combine angle facts, scale drawings and, at Higher, trigonometry.

40 min AQA GCSE Maths

The big picture

Three rules define a bearing: measure from $\textbf{north}$ , turn $\textbf{clockwise}$ , and write $\textbf{three figures}$ . "The bearing of $B$ from $A$ " means stand at $A$ , face north, and turn clockwise until you face $B$ .

North lines at different points are parallel. That single fact turns every "find the bearing back" or "find the angle in the triangle" question into a parallel-lines problem — co-interior angles add up to $180^\circ$ .

What you'll learn

Measure and draw three-figure bearings
Find the bearing of $A$ from $B$ given the bearing of $B$ from $A$
Use bearings with scale drawings
Calculate distances and bearings using angle facts and trigonometry

Measuring and drawing bearings

Draw a north line at the starting point. Place your protractor on it and measure clockwise. For bearings over $180^\circ$ , measure the anticlockwise angle and subtract it from $360^\circ$ .

The bearing of

B

from

A

062^\circ

. The bearing of

A

from

B

062^\circ + 180^\circ = 242^\circ

Example 1Write each direction as a three-figure bearing: (a) east (b) south-west (c) north-west.

1(a) Quarter turn clockwise from north:

090^\circ

2(b) South is

180^\circ

; south-west is

45^\circ

further:

225^\circ

3(c)

360 - 45 = 315^\circ

Answer

(a)

090^\circ

(b)

225^\circ

(c)

315^\circ

Your turnSame method, new numbers — try it before moving on.

What is the three-figure bearing of south-east?

Your answer

Show worked solution

Solution

1East is $090^\circ$ and south is $180^\circ$ .
2South-east is halfway: $135^\circ$ .

Answer: $135^\circ$

Back bearings

If the bearing of $B$ from $A$ is $\theta$ , the bearing of $A$ from $B$ differs by exactly $180^\circ$ : add $180^\circ$ if $\theta < 180^\circ$ , otherwise subtract $180^\circ$ .

Example 1The bearing of a lighthouse from a ship is

295^\circ

. What is the bearing of the ship from the lighthouse?

295 > 180

, so subtract:

295 - 180 = 115

Answer

115^\circ

Your turnSame method, new numbers — try it before moving on.

The bearing of $Q$ from $P$ is $047^\circ$ . Find the bearing of $P$ from $Q$ , in degrees.

Your answer

Show worked solution

Solution

1 $47 < 180$ , so add $180$ : $227^\circ$ .

Answer: $227^\circ$

Why ±180°

The two north lines are parallel, and the path between the points is a transversal. The bearing at $A$ and the angle measured at $B$ back to the north line are co-interior angles — they add to $180^\circ$ — which is why the back bearing is always $180^\circ$ away.

Scale drawings

A scale such as $1$ cm $: 5$ km means every centimetre on the drawing is 5 km in real life. Draw the north line, measure the bearing, then measure the scaled distance along it.

Example 1On a map with scale

1

: 4

km, a harbour is 6.5 cm from a buoy on a bearing of

140^\circ

. How far apart are they in real life?

6.5 \times 4 = 26

Answer

26

Your turnSame method, new numbers — try it before moving on.

A map has scale $1 : 50{,}000$ . Two villages are 7 cm apart on the map. How far apart are they in real life, in km?

Your answer

Show hint

$1$ cm on the map is $50{,}000$ cm in real life.

Show worked solution

Solution

1 $7 \times 50{,}000 = 350{,}000$ cm.
2 $350{,}000$ cm $= 3500$ m $= 3.5$ km.

Answer: $3.5$ km

Calculating with bearings

Sketch the situation with north lines at each point. Use angle facts (co-interior angles, angles on a straight line) to find angles inside the triangle, then use Pythagoras or trigonometry — or, at Higher, the sine and cosine rules.

Example 1A walker goes 5 km due east, then 3 km due north. Find the bearing of her finishing point from her start, to the nearest degree.

1The angle at the start between east and the direct line:

\tan^{-1}\left(\frac{3}{5}\right) = 30.96^\circ

2Bearing is measured from north:

90 - 30.96 = 59.04

Answer

059^\circ

Your turnSame method, new numbers — try it before moving on.

A boat sails 8 km due south then 6 km due west. How far is it from its starting point, in km?

Your answer

Show worked solution

Solution

1The legs are perpendicular: $d = \sqrt{8^2 + 6^2} = \sqrt{100}$ .
2 $d = 10$ km.

Answer: $10$ km

Example 2From that boat's finishing point, what is the bearing back to the start, to the nearest degree?

1The start is north-east of the finish: 8 km north and 6 km east.

2Angle from north:

\tan^{-1}\left(\frac{6}{8}\right) = 36.87^\circ

Answer

037^\circ

Tip — Label each point, draw its north line, and mark every bearing you are given before calculating.

Think like an examiner

Always write three figures: " $45^\circ$ " loses the mark where " $045^\circ$ " gains it.
Accurate drawings are marked within $\pm 2^\circ$ and $\pm 2$ mm — use a sharp pencil.
Read the wording "from" carefully: the bearing of $B$ from $A$ starts at $A$ .

Bearing rules

\text{from North, clockwise, 3 figures}

e.g.

045^\circ

270^\circ

\theta_{\text{back}} = \theta \pm 180^\circ

Back bearing.

\text{co-interior angles on north lines} = 180^\circ

North lines are parallel.

Watch out for these

Measuring anticlockwise from north.

Bearings are always clockwise.

Measuring from the line joining the points instead of north.

Draw the north line at the starting point first.

Writing a bearing as

62^\circ

Use three figures:

062^\circ

Mixing up "the bearing of

A

from

B

" and "of

B

from

A

Start at the point after "from".

Stretch yourself

Ship $S$ is 12 km from port $P$ on a bearing of $070^\circ$ . Lighthouse $L$ is 12 km from $P$ on a bearing of $130^\circ$ . Find the distance $SL$ and the bearing of $L$ from $S$ .

Hint — Angle $SPL = 130 - 70 = 60^\circ$ , and $PS = PL$ .

Reveal solution

SolutionTriangle

PSL

is isosceles with angle

P = 60^\circ

, so the base angles are also

60^\circ

— it is equilateral, and

SL = 12

km. At

S

, the bearing back to

P

070^\circ + 180^\circ = 250^\circ

. Seen from

S

L

lies on the south side of the line

SP

, so it is

60^\circ

anticlockwise of that direction:

250^\circ - 60^\circ = 190^\circ

. (A coordinate check with east and north axes gives

S \approx (11.28, 4.10)

and

L \approx (9.19, -7.71)

, a direction just west of south ✓.)

SL = 12

km on a bearing of

190^\circ

\blacksquare

Questions students ask

Why do bearings use three figures?

So they cannot be misread or altered — "45" could become "145" by accident, but "045" is unambiguous. It is a navigation convention.

Can a bearing be

360^\circ

Due north is written

000^\circ

;

360^\circ

is the same direction but is not normally used.

Are the sine and cosine rules needed for bearings?

Only at Higher, when the triangle formed has no right angle.

Key takeaways

Measure from north, clockwise, and write three figures.
Back bearing: add or subtract $180^\circ$ .
North lines are parallel — use co-interior and alternate angles.
Scale drawings: measure the bearing, then scale the distance.
Sketch, find angles inside the triangle, then use Pythagoras or trigonometry.

Exam-style questions

Practise it the way it is examined

3 original questions · 10 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 32 marks

(a)Write down the three-figure bearing of west.[1]
Show mark scheme (a)
- B1 $270^\circ$
Answer: $270^\circ$
(b)The bearing of $B$ from $A$ is $108^\circ$ . Work out the bearing of $A$ from $B$ .[1]
Show mark scheme (b)
- B1 $288^\circ$
Answer: $288^\circ$

Question 2Grade 54 marks

A plane flies 120 km due north from airport $X$ , then 50 km due east to $Y$ .

Work out the distance $XY$ and the bearing of $Y$ from $X$ , to the nearest degree.[4]
Show mark scheme
- M1 $\sqrt{120^2 + 50^2}$
- A1 $130$ km
- M1 $\tan^{-1}\left(\frac{50}{120}\right)$
- A1 $023^\circ$ (22.6…)
Answer: $130$ km, $023^\circ$

Question 3Grade 74 marks

Town $B$ is 40 km from town $A$ on a bearing of $055^\circ$ . Town $C$ is 25 km from $B$ on a bearing of $145^\circ$ .

(a)Show that angle $ABC = 90^\circ$ .[2]
Show mark scheme (a)
- M1Bearing of $A$ from $B$ is $235^\circ$
- A1Angle $ABC = 235 - 145 = 90^\circ$
(b)Work out the distance $AC$ , to 3 significant figures.[2]
Show mark scheme (b)
- M1 $\sqrt{40^2 + 25^2}$
- A1 $47.2$ km (47.16…)
Answer: $47.2$ km

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

The three-figure bearing of south is…
- $180^\circ$
- $090^\circ$
- $270^\circ$
- $000^\circ$
Show answer and explanation
Answer: $180^\circ$
A half turn clockwise from north.
Bearings are measured…
- clockwise from north
- anticlockwise from north
- clockwise from east
- from the horizontal
Show answer and explanation
Answer: clockwise from north
The definition of a bearing.
Write a bearing of $8^\circ$ correctly.
- $008^\circ$
- $8^\circ$
- $080^\circ$
- $800^\circ$
Show answer and explanation
Answer: $008^\circ$
Three figures.
Bearing of $B$ from $A$ is $070^\circ$ . Bearing of $A$ from $B$ ?
- $250^\circ$
- $110^\circ$
- $290^\circ$
- $070^\circ$
Show answer and explanation
Answer: $250^\circ$
$70 + 180$ .
Bearing of $D$ from $C$ is $310^\circ$ . Bearing of $C$ from $D$ ?
- $130^\circ$
- $490^\circ$
- $050^\circ$
- $230^\circ$
Show answer and explanation
Answer: $130^\circ$
$310 - 180$ .
Scale $1$ cm : $2$ km. A 4.5 cm line represents…
- 9 km
- 2.25 km
- 6.5 km
- 90 km
Show answer and explanation
Answer: 9 km
$4.5 \times 2$ .
3 km north then 3 km east. Bearing of the end from the start?
- $045^\circ$
- $135^\circ$
- $090^\circ$
- $315^\circ$
Show answer and explanation
Answer: $045^\circ$
Equal legs give $45^\circ$ from north.
North-west as a bearing?
- $315^\circ$
- $045^\circ$
- $225^\circ$
- $135^\circ$
Show answer and explanation
Answer: $315^\circ$
$360 - 45$ .

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 14 lessons