Bearings
A true bearing gives a direction as an angle measured clockwise from north, written with three digits such as \(065^\circ\). In NSW Year 10 Advanced maths bearings are combined with right-angled trigonometry to find distances north, south, east and west, and the bearing back to a starting point (NESA outcome MA5‑TRG‑C‑02).
You will turn bearings clockwise and anticlockwise, find the angle between two bearings, and use reverse bearings and parallel lines to find angles in a triangle. You will find how far north and east a drone has flown, a bearing on a map with a scale, the bearing to fly back to an airport and the time a ship takes to reach a point. Harder problems involve a kayaker's two legs, two ships leaving port and a runner's three legs. It is part of Year 10 Advanced Trigonometry.
Theory
A true bearing gives a direction as an angle measured clockwise from north, written with three digits such as \(065^\circ\). This Year 10 Advanced maths page shows how to draw bearing problems, use back bearings, find distances north, south, east and west with sine and cosine, and solve multi-leg trips, speed problems and problems with two bearings to one point.
A true bearing is measured clockwise from north, from \(000^\circ\) to \(360^\circ\), and is written with three digits: east is \(090^\circ\), south \(180^\circ\), west \(270^\circ\).
A compass bearing starts at N or S and turns towards E or W, by less than \(90^\circ\). S\(50^\circ\)E means \(50^\circ\) east of south, which is the true bearing \(180^\circ - 50^\circ = 130^\circ\). N\(20^\circ\)W is \(360^\circ - 20^\circ = 340^\circ\).
"The bearing of \(B\) from \(A\)" means stand at \(A\) and measure from the north line at \(A\). The back bearing, of \(A\) from \(B\), differs by \(180^\circ\): add \(180^\circ\) if the bearing is under \(180^\circ\), subtract if it is over.
Turning clockwise adds to a bearing; turning anticlockwise subtracts. Add or subtract \(360^\circ\) to land between \(000^\circ\) and \(360^\circ\).
For a trip of length \(d\) at an angle \(\theta\) from the north–south line:
| True bearing \(B\) | Quadrant | Angle in the triangle |
|---|---|---|
| \(0^\circ\)–\(90^\circ\) | NE | \(B\) east of north |
| \(90^\circ\)–\(180^\circ\) | SE | \(180^\circ - B\) east of south |
| \(180^\circ\)–\(270^\circ\) | SW | \(B - 180^\circ\) west of south |
| \(270^\circ\)–\(360^\circ\) | NW | \(360^\circ - B\) west of north |
How to solve a bearings problem
- Draw a north line at every point where a bearing is measured. Mark each bearing clockwise from north.
- Find the angle inside the right-angled triangle, using the quadrant table or a back bearing.
- Solve with sine, cosine, tangent or Pythagoras' theorem.
- Convert any angle back to a three-figure true bearing, then check the direction is in the right quadrant.
Special cases
- Angle between two bearings from one point: subtract them, or split the angle at the north line and add the two parts.
- Angle at a turning point: back bearing of the first leg minus the bearing of the second leg. If two legs differ by \(90^\circ\), the triangle is right-angled.
- Several legs: find the total east and total north (west and south count as negative), then use Pythagoras' theorem and tan. A leg that is opposite another (bearings \(180^\circ\) apart) cancels part of it.
- Maps and speed: convert a map length with the scale, and use distance \(=\) speed \(\times\) time. Change minutes to hours by dividing by \(60\).
- Two bearings to one point, or an unknown leg such as \(x\) and \(2x\): write each length in terms of the unknown and form an equation (Example 4).
(i) How far south and how far east of \(H\) is it, correct to one decimal place?
(ii) On what bearing must it fly to return to \(H\)?
(i) \(128^\circ\) is \(180^\circ - 128^\circ = 52^\circ\) east of south. The \(36\) km path is the hypotenuse, south is adjacent to \(52^\circ\) and east is opposite.
| \(\text{south}\) | \(=\) | \(36\cos 52^\circ = 22.163\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{22.2}\text{ km}\) |
| \(\text{east}\) | \(=\) | \(36\sin 52^\circ = 28.368\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{28.4}\text{ km}\) |
(ii) The return trip is the back bearing:
| \(128^\circ + 180^\circ\) | \(=\) | \(\mathbf{308^\circ}\) |
(i) the true bearing of \(L\) from \(B\)
(ii) the true bearing of \(B\) from \(L\).
(iii) How far apart are they?
(i) Let \(\alpha\) be the angle west of north at \(B\). West is opposite and north is adjacent.
| \(\tan\alpha\) | \(=\) | \(\dfrac{7.5}{4}\) |
| \(\alpha\) | \(=\) | \(61.927\ldots^\circ\) |
| \(\text{bearing}\) | \(=\) | \(360^\circ - 61.927\ldots^\circ\) |
| \(\) | \(=\) | \(298.072\ldots^\circ \approx \mathbf{298^\circ}\) |
(ii) Subtract \(180^\circ\) for the back bearing, using the unrounded value:
| \(\) | \(\) | \(298.072\ldots^\circ - 180^\circ = 118.072\ldots^\circ \approx \mathbf{118^\circ}\) |
(iii) Pythagoras' theorem:
| \(BL\) | \(=\) | \(\sqrt{7.5^2 + 4^2} = \mathbf{8.5}\text{ km}\) |
(i) Show that \(\angle PQR = 90^\circ\).
(ii) Find the distance \(PR\) and the bearing of \(R\) from \(P\), correct to one decimal place and the nearest degree.
(i) The bearing of \(P\) from \(Q\) is the back bearing \(115^\circ + 180^\circ = 295^\circ\). The angle at \(Q\) between the two directions is
| \(\angle PQR\) | \(=\) | \(295^\circ - 205^\circ = 90^\circ\) |
(ii) Right-angled at \(Q\), so use Pythagoras' theorem and tan.
| \(PR\) | \(=\) | \(\sqrt{10^2 + 7^2} = \sqrt{149} = 12.206\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{12.2}\text{ km}\) |
| \(\tan\alpha\) | \(=\) | \(\dfrac{7}{10}\) |
| \(\alpha\) | \(=\) | \(34.992\ldots^\circ\) |
The ship turned clockwise at \(Q\), so \(R\) is \(\alpha\) clockwise from the first leg:
| \(\text{bearing}\) | \(=\) | \(115^\circ + 34.992\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{150^\circ}\) |
Let \(M\) be the point on the road closest to \(T\), and \(TM = d\). \(TM\) runs north–south, so it is parallel to the north lines. At \(T\): \(\angle ATM = 50^\circ\) and \(\angle BTM = 360^\circ - 320^\circ = 40^\circ\) (alternate angles).
| \(AM\) | \(=\) | \(d\tan 50^\circ\) |
| \(MB\) | \(=\) | \(d\tan 40^\circ\) |
| \(AM + MB\) | \(=\) | \(3\) |
| \(d(\tan 50^\circ + \tan 40^\circ)\) | \(=\) | \(3\) |
| \(d\) | \(=\) | \(\dfrac{3}{\tan 50^\circ + \tan 40^\circ}\) |
| \(\) | \(=\) | \(1.4772\ldots\) |
| \(\therefore\ d\) | \(\approx\) | \(\mathbf{1.48}\text{ km}\) |
Check: \(50^\circ + 40^\circ = 90^\circ\), so the angle at \(T\) is a right angle.
Common pitfalls
Frequently asked questions
How do you write a true bearing?
Measure the angle clockwise from north and write it with three digits, from \(000^\circ\) to \(360^\circ\). East is \(090^\circ\), south is \(180^\circ\) and west is \(270^\circ\).
What is the difference between a true bearing and a compass bearing?
A true bearing is one angle measured clockwise from north, such as \(130^\circ\). A compass bearing starts at N or S and turns towards E or W, such as S\(50^\circ\)E, which is the same direction.
How do you find a back bearing?
Add \(180^\circ\) to the bearing if it is less than \(180^\circ\), or subtract \(180^\circ\) if it is more. The back bearing of \(075^\circ\) is \(255^\circ\).
How do you find how far north and east a bearing takes you?
Find the angle the path makes with the north-south line. The north or south distance is the length times cos of that angle, and the east or west distance is the length times sin of that angle.
How do you find a bearing from distances east and north?
Draw the right-angled triangle, use inverse tan of east over north to find the angle from the north-south line, then turn that angle into a three-figure bearing for the right quadrant.