A bearing is a way of giving a direction as an angle. This note shows how to measure and draw bearings, find a back bearing, and combine bearings with Pythagoras and right-angled trigonometry to solve journey problems. Navigation, surveying and flight planning all depend on these ideas.
Key ideas
A bearing is an angle measured clockwise from north. It is always written with three figures, so east is 090°, not 90°, and a direction just east of north is 005°. Bearings run from 000° up to 360°.
Always follow the same three steps when you measure or draw a bearing:
- Put your pencil (or imagine yourself) at the point you are going from.
- Draw a north line through that point.
- Measure clockwise from the north line to the line that joins the two points.
The main compass directions are: north 000°, east 090°, south 180°, west 270°. North-east is 045°, south-east is 135°, south-west is 225° and north-west is 315°.
Reading the wording. "The bearing of B from A" means you stand at A and look towards B. The north line is drawn at A. The wording "from" tells you where to put the north line.
Back bearings. If you know the bearing of B from A, the bearing of A from B points the opposite way. North lines at A and B are parallel, so co-interior angles add to 180°, or equivalently alternate angles are equal. The result is simple:
- If the bearing is less than 180°, add 180°.
- If the bearing is 180° or more, subtract 180°.
Solving problems. Most bearing questions become a triangle. Draw a sketch, draw every north line, and work out the angles inside the triangle using parallel-line rules and the fact that angles at a point add to 360°. If the triangle has a right angle, use Pythagoras or SOH CAH TOA. Remember that a bearing is measured from north, so the angle you use in the triangle is often not the bearing itself.
A useful shortcut for a right-angled triangle with a northerly side: for a journey of distance \(d\) on bearing \(\theta\) (where \(\theta\) is between 000° and 090°), the distance travelled north is \(d\cos\theta\) and the distance travelled east is \(d\sin\theta\).
Scale drawing. You may be asked to draw a bearing problem accurately. Choose a sensible scale (for example 1 cm to 2 km), use a protractor from a north line, and measure the answer back off the diagram, converting with the scale. Give an answer within a sensible tolerance, as drawings are never perfect.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Bearing | Direction as a three-figure angle, measured clockwise from north. |
| Back bearing | The bearing of the starting point from the end point. Add or subtract 180°. |
| \(a^2 + b^2 = c^2\) | Pythagoras: find a side of a right-angled triangle, where \(c\) is the hypotenuse. |
| \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\) | Right-angled triangle, opposite and hypotenuse involved. |
| \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\) | Right-angled triangle, adjacent and hypotenuse involved. |
| \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\) | Right-angled triangle, opposite and adjacent involved. |
| Angles at a point | Add up to 360°. Angles on a straight line add up to 180°. |
Worked examples
Example 1 A hiker walks 8.0 km from camp on a bearing of 050°. How far north and how far east of camp is she?
Sketch the north line at camp. The path makes an angle of 50° with north, so the right-angled triangle has hypotenuse 8.0 km and an angle of 50° next to the north side.
North distance is the side adjacent to 50°: \(8.0\cos 50^\circ = 5.1423\ldots\)
East distance is the side opposite 50°: \(8.0\sin 50^\circ = 6.1283\ldots\)
Answer: 5.14 km north and 6.13 km east.
Example 2 A ferry leaves harbour H and sails 15 km on a bearing of 070° to point B. It then turns and sails 20 km on a bearing of 160° to point C. (a) Find the distance HC. (b) Find the bearing of C from H. (c) Find the bearing of H from C.
(a) The two bearings differ by \(160^\circ - 70^\circ = 90^\circ\). The ferry turns through a right angle at B, so triangle HBC has a right angle at B.
\(HC^2 = 15^2 + 20^2 = 225 + 400 = 625\), so \(HC = 25\) km.
(b) Let \(x\) be angle BHC. Then \(\tan x = \dfrac{BC}{HB} = \dfrac{20}{15}\), so \(x = \tan^{-1}\!\left(\dfrac{20}{15}\right) = 53.13^\circ\ldots\)
The bearing of B from H is 070°, and C lies further clockwise from B, so the bearing of C from H is \(70^\circ + 53.13^\circ = 123.13^\circ\ldots\), which is 123° to the nearest degree.
(c) The bearing of C from H is less than 180°, so add 180°: \(123.13^\circ + 180^\circ = 303.13^\circ\ldots\)
Answer: (a) 25 km. (b) 123°. (c) 303°.
Example 3 Town B is 12 km north and 9 km east of town A. Find the distance AB and the bearing of B from A. Then give the bearing of A from B.
Sketch a right-angled triangle with the north line at A. The north side is 12 km and the east side is 9 km.
Distance: \(AB^2 = 12^2 + 9^2 = 144 + 81 = 225\), so \(AB = 15\) km.
The angle \(\theta\) at A between north and AB is opposite the 9 km side and adjacent to the 12 km side, so \(\tan\theta = \dfrac{9}{12}\) and \(\theta = 36.87^\circ\ldots\)
B is to the north and east of A, so the bearing is measured clockwise from north and equals \(\theta\), written with three figures: 037°.
Back bearing: \(36.87^\circ + 180^\circ = 216.87^\circ\ldots\), which is 217°.
Answer: AB = 15 km, bearing of B from A is 037°, bearing of A from B is 217°.
Common errors
- Common error: measuring anticlockwise, or measuring from east or from the line itself instead of from north.
- Common error: putting the north line at the wrong point. "Bearing of B from A" needs the north line at A.
- Common error: writing 50° instead of 050°. A bearing must have three figures.
- Common error: forgetting to add or subtract 180° for a back bearing, or adding when the bearing is already above 180° and ending up over 360°.
- Common error: giving the angle inside the triangle as the bearing, without checking where north is.
Exam tip: Sketch the situation first and draw a north line at every point where a bearing is given or asked for. A labelled sketch shows you which angles are equal or add up to 180°.
Check yourself
- Write down the bearing of south-west.
Answer
South is 180° and west is 270°, so south-west is halfway: 225°. - The bearing of Q from P is 125°. Find the bearing of P from Q.
Answer
125° is less than 180°, so add 180°: 125° + 180° = 305°. - A plane flies 40 km on a bearing of 030°. How far east has it travelled?
Answer
East distance is opposite the 30° angle: \(40\sin 30^\circ = 20\) km. - A yacht sails 6 km due south and then 8 km due west. Find its distance from the start and the bearing of the start from the yacht.
Answer
Distance: \(\sqrt{6^2 + 8^2} = 10\) km. The yacht is south-west of the start, so the start is north-east of the yacht. The angle between north and the line back to the start, measured at the yacht, is \(\tan^{-1}(8/6) = 53.13^\circ\ldots\) east of north, so the bearing is 053°.