Bearings
Bearings in NSW Year 10 Core maths describes direction in two ways (NESA outcome MA5‑TRG‑C‑02). A true bearing is a three-figure angle measured clockwise from north, such as \(070^\circ\), and a compass bearing starts at north or south and turns towards east or west, such as S\(\,40^\circ\,\)W.
You will read bearings from diagrams, convert between true and compass bearings, and find the reverse bearing of one point from another. You will also use trigonometry and Pythagoras' theorem to find how far north, south, east or west a boat, plane or hiker has travelled, the bearing back to a starting point, and the distance between two ships sailing on different bearings. It is part of Year 10 Core Trigonometry.
Theory
A true bearing is an angle measured clockwise from north and written with three figures, such as \(075^\circ\). This Year 10 maths page covers compass bearings, converting between the two, back bearings, and using trigonometry to find how far north, south, east or west something is, and the bearing of one place from another.
A true bearing gives a direction as the angle turned clockwise from north, from \(000^\circ\) to \(360^\circ\). It is written with three figures: east is \(090^\circ\), south is \(180^\circ\) and west is \(270^\circ\).
A compass bearing starts from north or south, whichever is closer, then turns towards east or west. N\(\,20^\circ\,\)E means face north and turn \(20^\circ\) towards east. A compass bearing angle is never more than \(90^\circ\).
“The bearing of \(B\) from \(A\)” is measured at \(A\). Draw a north line at \(A\) and turn clockwise until you reach \(B\).
The back bearing (the bearing of \(A\) from \(B\)) points the opposite way, so it differs by \(180^\circ\).
Converting a true bearing \(T\) to a compass bearing:
| Direction between | True bearing \(T\) | Compass bearing |
|---|---|---|
| N and E | \(0^\circ\) to \(90^\circ\) | N\(\,T\,\)E |
| S and E | \(90^\circ\) to \(180^\circ\) | S\(\,(180^\circ - T)\,\)E |
| S and W | \(180^\circ\) to \(270^\circ\) | S\(\,(T - 180^\circ)\,\)W |
| N and W | \(270^\circ\) to \(360^\circ\) | N\(\,(360^\circ - T)\,\)W |
Back bearing:
How to solve a bearings problem
- Sketch with a north arrow at every point where a bearing is measured. North, south, east and west meet at right angles.
- Find the right-angled triangle and the angle inside it: \(180^\circ - T\), \(T - 180^\circ\), \(360^\circ - T\) or \(90^\circ - T\), depending on the quadrant.
- Solve with sine, cosine, tangent or Pythagoras' theorem.
- Turn an angle back into a bearing from north, and check it lands in the right quadrant.
Useful facts
- Reading a diagram: add the quarter turns. A line \(45^\circ\) past south is \(180^\circ + 45^\circ = 225^\circ\).
- Walk east then south (or any two compass directions): \(\tan\theta = \dfrac{\text{second leg}}{\text{first leg}}\), then add to or subtract from \(090^\circ\), \(180^\circ\), \(270^\circ\) or \(360^\circ\).
- To fly or sail back to the start, use the back bearing.
- Walk on a bearing until you are a set distance south (or west): that distance is a side of the triangle and the path is the hypotenuse, so divide by cos or sin.
- North lines at two points are parallel, so alternate and co-interior angles move angles between them.
- Two bearings \(90^\circ\) apart from the same point (or a bearing and a back bearing \(90^\circ\) apart) give a right angle, so Pythagoras' theorem finds the distance.
(i) Write the true bearing \(238^\circ\) as a compass bearing.
(ii) Write N\(\,35^\circ\,\)W as a true bearing.
(iii) The bearing of \(Q\) from \(P\) is \(238^\circ\). Find the bearing of \(P\) from \(Q\).
(i) \(238^\circ\) is between \(180^\circ\) (south) and \(270^\circ\) (west), so start from S and turn towards W.
| \(238^\circ - 180^\circ\) | \(=\) | \(58^\circ\) |
| \(\text{compass bearing}\) | \(:\) | \(\text{S}\,\mathbf{58^\circ}\,\text{W}\) |
(ii) N\(\,35^\circ\,\)W is \(35^\circ\) back from north towards west, so it is \(35^\circ\) less than a full turn.
| \(360^\circ - 35^\circ\) | \(=\) | \(\mathbf{325^\circ}\) |
(iii) The opposite direction differs by \(180^\circ\). \(238^\circ\) is more than \(180^\circ\), so subtract.
| \(238^\circ - 180^\circ\) | \(=\) | \(058^\circ\) |
The bearing of \(P\) from \(Q\) is \(\mathbf{058^\circ}\). Always write true bearings with three figures.
(i) south
(ii) east of its starting point is it? Answer correct to one decimal place.
Draw a line due south from the start. The angle between south and the path is \(180^\circ - 128^\circ = 52^\circ\). The path is the hypotenuse.
(i) South is adjacent to the \(52^\circ\) angle.
| \(s\) | \(=\) | \(15 \cos 52^\circ\) |
| \(\) | \(=\) | \(9.2349\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{9.2}\text{ km}\) |
(ii) East is opposite the \(52^\circ\) angle.
| \(e\) | \(=\) | \(15 \sin 52^\circ\) |
| \(\) | \(=\) | \(11.8201\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{11.8}\text{ km}\) |
(i) the bearing of the finishing point from the start
(ii) the bearing she must take to ride straight back.
Sketch it. The right angle is at the turn. At the start, the angle \(\theta\) is between due west and the line to the finish: \(4\) km is opposite and \(9\) km is adjacent.
| \(\tan\theta\) | \(=\) | \(\dfrac{4}{9}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\left(\dfrac{4}{9}\right)\) |
| \(\) | \(=\) | \(23.96\ldots^\circ\) |
(i) West is \(270^\circ\). The finish is \(\theta\) back from west towards south, so subtract.
| \(\text{bearing}\) | \(=\) | \(270^\circ - 23.96\ldots^\circ\) |
| \(\) | \(=\) | \(246.03\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{246^\circ}\) |
(ii) The way back is the back bearing: \(246^\circ - 180^\circ = \mathbf{066^\circ}\).
Check: south-west of the start means a bearing between \(180^\circ\) and \(270^\circ\). \(246^\circ\) fits.
(i) Show that \(\angle STF = 90^\circ\) and find \(SF\).
(ii) Find the bearing of \(F\) from \(S\), to the nearest degree.
(i) From \(T\), the direction back to \(S\) is the back bearing \(035^\circ + 180^\circ = 215^\circ\). The direction on to \(F\) is \(125^\circ\).
| \(\angle STF\) | \(=\) | \(215^\circ - 125^\circ = 90^\circ\) |
| \(SF^2\) | \(=\) | \(7^2 + 24^2 = 625\) |
| \(SF\) | \(=\) | \(\mathbf{25}\text{ km}\) |
(ii) In the right-angled triangle, the angle at \(S\) has \(24\) opposite and \(7\) adjacent. Add it to the first bearing, since \(F\) is further round clockwise than \(T\).
| \(\angle TSF\) | \(=\) | \(\tan^{-1}\left(\dfrac{24}{7}\right) = 73.73\ldots^\circ\) |
| \(\text{bearing}\) | \(=\) | \(35^\circ + 73.73\ldots^\circ\) |
| \(\) | \(=\) | \(108.73\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{109^\circ}\) |
Common pitfalls
Frequently asked questions
What is a true bearing?
A true bearing is the angle measured clockwise from north, written with three figures 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 turns clockwise from north and can be any angle up to \(360^\circ\). A compass bearing starts from north or south and turns towards east or west, such as S\(\,40^\circ\,\)W, so its angle is at most \(90^\circ\).
How do you find a back bearing?
Add \(180^\circ\) if the bearing is less than \(180^\circ\). Subtract \(180^\circ\) if it is \(180^\circ\) or more. The answer is the bearing for the return direction.
Why are bearings written with three figures?
Writing every bearing with three figures, such as \(075^\circ\) instead of \(75^\circ\), avoids confusion when bearings are read out or written quickly, for example in navigation.
How do you use trigonometry with bearings?
Draw a north line at each point and find a right-angled triangle made by north–south and east–west lines. Work out the angle inside the triangle from the bearing, then use sine, cosine or tangent.