Bearings
A bearing gives a direction. A true bearing is a three-figure angle measured clockwise from north, such as \(070^\circ\), and a compass bearing, such as \(\text{S}\,30^\circ\,\text{E}\), is measured from north or south. This NSW Year 9 Advanced maths subtopic uses both (NESA outcome MA5‑TRG‑C‑02).
You will convert between the two kinds, find the reverse bearing, and use trigonometry to find how far north, south, east or west a boat, hiker or plane has travelled. You will find distances and bearings when two legs of a journey meet at right angles, locate a fire seen from two towers, and find how far apart two planes are after half an hour. It is part of Year 9 Advanced Trigonometry.
Theory
A bearing gives a direction as an angle from north. This Year 9 Advanced maths page covers true bearings and compass bearings, back bearings, finding distances north, south, east and west, journeys with a right-angled turn, and journeys that need to be split into east and north parts.
A true bearing is measured clockwise from north and written with three figures: east is \(090^\circ\), south is \(180^\circ\), west is \(270^\circ\). A bearing of \(45^\circ\) is written \(045^\circ\).
A compass bearing starts at N or S and turns towards E or W by an angle from \(0^\circ\) to \(90^\circ\): N\(\,40^\circ\,\)E means \(40^\circ\) from north towards east.
"The bearing of \(B\) from \(A\)" means stand at \(A\), face north, and turn clockwise until you face \(B\). Always draw the north line at the point you are measuring from.
The back bearing (the bearing of \(A\) from \(B\)) points the opposite way, so it differs by \(180^\circ\). The north lines at \(A\) and \(B\) are parallel, which is why angles carry across between them.
A journey on a bearing is the hypotenuse of a right-angled triangle whose other sides run north–south and east–west. Sine and cosine give how far north or south and how far east or west the journey goes.
| True bearing \(\beta\) | Compass bearing | Back bearing |
|---|---|---|
| \(0^\circ < \beta < 90^\circ\) | N\(\,\beta\,\)E | \(\beta + 180^\circ\) |
| \(90^\circ < \beta < 180^\circ\) | S\(\,(180^\circ - \beta)\,\)E | \(\beta + 180^\circ\) |
| \(180^\circ < \beta < 270^\circ\) | S\(\,(\beta - 180^\circ)\,\)W | \(\beta - 180^\circ\) |
| \(270^\circ < \beta < 360^\circ\) | N\(\,(360^\circ - \beta)\,\)W | \(\beta - 180^\circ\) |
How to solve a bearings problem
- Draw a north line at every point where a bearing is measured.
- Mark each bearing clockwise from its north line. Turn it into an angle from the nearest north–south line (for example \(236^\circ \to 56^\circ\) west of south).
- Find a right angle: a north–south line meets an east–west line at \(90^\circ\); two legs whose bearings differ by \(90^\circ\) meet at \(90^\circ\).
- Solve with Pythagoras' theorem or a trig ratio. Use the inverse ratio for an angle.
- Convert back to a three-figure true bearing (or a compass bearing if asked).
Common set-ups
- Bearing from a diagram: add or subtract the marked angle from \(090^\circ\), \(180^\circ\), \(270^\circ\) or \(360^\circ\).
- Speed and time: distance \(=\) speed \(\times\) time. From 6 pm to 8:30 pm is \(2.5\) hours.
- Two ships from one port: the angle between their paths is the difference of their bearings.
- Two lookouts sight one point: work out each angle inside the triangle from the bearings; check the angle sum.
- A point due east of one point and due south of another: it makes the right angle you need.
- Turns that are not right angles: split every leg into east and north parts, add them, then use Pythagoras' theorem and \(\tan^{-1}\) (Example 4).
(i) Write it as a compass bearing.
(ii) Find the bearing of \(H\) from \(T\), as a true bearing and as a compass bearing.
(i) \(128^\circ\) is past east (\(90^\circ\)) but not yet south (\(180^\circ\)), so measure from south towards east.
| \(180^\circ - 128^\circ\) | \(=\) | \(52^\circ\) |
| \(128^\circ\) | \(=\) | \(\mathbf{S\,52^\circ\,E}\) |
(ii) The back bearing points the opposite way: add \(180^\circ\).
| \(128^\circ + 180^\circ\) | \(=\) | \(\mathbf{308^\circ}\) |
| \(\text{compass}\) | \(=\) | \(\mathbf{N\,52^\circ\,W}\) |
For a compass back bearing, swap both letters and keep the angle: S\(\,52^\circ\,\)E becomes N\(\,52^\circ\,\)W. Check: \(360^\circ - 52^\circ = 308^\circ\).
\(236^\circ\) is \(236^\circ - 180^\circ = 56^\circ\) past south, towards west. Make a right-angled triangle with the flight as the hypotenuse.
| \(\text{south}\) | \(=\) | \(3.6\cos 56^\circ = 2.0130\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{2.01}\text{ km}\) |
| \(\text{west}\) | \(=\) | \(3.6\sin 56^\circ = 2.9845\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{2.98}\text{ km}\) |
South is adjacent to the \(56^\circ\) angle at the start (cos); west is opposite it (sin).
(i) Explain why the turn is a right angle.
(ii) Find the ship's distance from \(P\), to two decimal places.
(iii) Find its true bearing from \(P\), to the nearest degree.
(i) The ship turns clockwise by \(165^\circ - 75^\circ = 90^\circ\), so the two legs meet at a right angle.
(ii) The two legs are the shorter sides, so use Pythagoras' theorem.
| \(d\) | \(=\) | \(\sqrt{11^2 + 6^2} = \sqrt{157}\) |
| \(\) | \(=\) | \(12.5299\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{12.53}\text{ km}\) |
(iii) At \(P\), \(6\) km is opposite \(\theta\) and \(11\) km is adjacent.
| \(\theta\) | \(=\) | \(\tan^{-1}\left(\dfrac{6}{11}\right) = 28.6104\ldots^\circ\) |
| \(\text{bearing}\) | \(=\) | \(75^\circ + 28.6104\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{104^\circ}\) |
\(\theta\) is measured from the first leg, which is already \(75^\circ\) clockwise from north, so add it on.
(i) How far is she from the start, to one decimal place?
(ii) What true bearing takes her straight back to the start, to the nearest degree?
The turn is \(80^\circ\), not a right angle, so split each leg into an east part and a north part, then add.
| \(\text{east}\) | \(=\) | \(4\sin 30^\circ + 6\sin 110^\circ = 7.6381\ldots\) |
| \(\text{north}\) | \(=\) | \(4\cos 30^\circ + 6\cos 110^\circ = 1.4119\ldots\) |
\(\cos 110^\circ\) is negative: the second leg heads south. Equivalently, \(110^\circ\) is \(20^\circ\) south of east, giving \(6\cos 20^\circ\) east and \(6\sin 20^\circ\) south.
(i)
| \(d\) | \(=\) | \(\sqrt{7.6382^2 + 1.4120^2}\) |
| \(\) | \(=\) | \(7.7675\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.8}\text{ km}\) |
(ii) The finish is east and a little north of the start. Its bearing from the start is
| \(90^\circ - \tan^{-1}\left(\dfrac{\text{north}}{\text{east}}\right)\) | \(=\) | \(79.5266\ldots^\circ\) |
| \(\text{back bearing}\) | \(=\) | \(79.5266\ldots^\circ + 180^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{260^\circ}\) |
Keep the unrounded angle until the last step, then round.
Common pitfalls
Frequently asked questions
What is a true bearing?
A true bearing is an angle measured clockwise from north, written with three figures. East is \(090^\circ\), south is \(180^\circ\) and west is \(270^\circ\).
What is a compass bearing?
A compass bearing starts at north or south and turns an angle towards east or west, such as N\(\,40^\circ\,\)E or S\(\,15^\circ\,\)W. The angle is never more than \(90^\circ\).
How do you find a back bearing?
Add \(180^\circ\) if the bearing is less than \(180^\circ\), or subtract \(180^\circ\) if it is more. For a compass bearing, swap N with S and E with W and keep the angle.
How do you work out how far east or north someone has travelled?
Use the journey as the hypotenuse of a right-angled triangle. For a distance \(d\) on true bearing \(\beta\), east \(= d\sin\beta\) and north \(= d\cos\beta\).
How do you find the bearing back to the start after two legs?
Find the total distance east and north, use \(\tan^{-1}\) to find the angle of the finish from the start, write that as a true bearing, then add or subtract \(180^\circ\).