Bearings Problems with Trigonometry
Combine bearings with right-angled triangle trigonometry for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-04). A trip made of legs due north or south and due east or west forms a right-angled triangle, so you can work out how far away the finish is and in which direction.
You will use Pythagoras' theorem for the straight-line distance, sine and cosine for how far north or east the trip has gone, and tangent for the angle, written as a three-figure true bearing — practical for navigating by boat, by plane or on foot.
Theory
A journey made of north–south and east–west legs forms a right-angled triangle. This Year 12 Standard 1 (NSW) guide shows how to use Pythagoras for the straight-line distance and SOH CAH TOA for distances north or east and for the bearing.
Many navigation questions describe a trip as legs due north/south and due east/west. These legs meet at a right angle, so the straight-line (direct) distance is the hypotenuse and can be found with Pythagoras' theorem.
For a trip of \(d\) km on a bearing that makes an angle \(\theta\) with north, the distance north is adjacent to \(\theta\) and the distance east is opposite \(\theta\). So north \(=d\cos\theta\) and east \(=d\sin\theta\).
To find a bearing, use \(\tan\theta=\dfrac{\text{east–west leg}}{\text{north–south leg}}\) to get the acute angle from the north–south line, then use the quadrant to turn it into a three-figure bearing. The back bearing for the return trip differs by \(180^\circ\).
Straight-line distance from two perpendicular legs:
Distances north and east for a trip of \(d\) at angle \(\theta\) from north:
If the bearing points south (between \(090^\circ\text{T}\) and \(270^\circ\text{T}\)), use the acute angle from south. For example, \(80\text{ km}\) on \(120^\circ\text{T}\): \(\theta=180^\circ-120^\circ=60^\circ\), so south \(=80\cos 60^\circ=40\text{ km}\) and east \(=80\sin 60^\circ\approx 69.3\text{ km}\).
The acute angle from the north–south line:
How to solve a bearings problem
- Draw a diagram with a north line at the start, each leg of the trip and the right angle.
- Distance: use Pythagoras for the direct distance, or \(d\cos\theta\) and \(d\sin\theta\) for distances north and east.
- Angle: use \(\tan^{-1}\) with the east–west leg on top and the north–south leg underneath.
- Bearing: use the quadrant to write a three-figure bearing; add or subtract \(180^\circ\) for the return trip.
Sketch the trip. The legs meet at a right angle, so the direct distance is the hypotenuse.
| \(d^2\) | \(=\) | \(9^2+7^2\) |
| \(d^2\) | \(=\) | \(81+49=130\) |
| \(d\) | \(=\) | \(\sqrt{130}=11.40\ldots\) |
| \(d\) | \(\approx\) | \(11.4\text{ km}\) |
The kayaker is about \(11.4\text{ km}\) from the start.
Sketch the trip. The angle from north is \(65^\circ\); north is adjacent (cos) and east is opposite (sin).
| \(\text{North}\) | \(=\) | \(18\cos 65^\circ=7.60\ldots\) |
| \(\text{East}\) | \(=\) | \(18\sin 65^\circ=16.31\ldots\) |
The boat is \(7.6\text{ km}\) north and \(16.3\text{ km}\) east of its start.
Sketch the trip. The finish is north-east of the start, so find the angle east of north.
| \(\tan\theta\) | \(=\) | \(\dfrac{\text{east}}{\text{north}}=\dfrac{10}{7}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\!\left(\dfrac{10}{7}\right)=55.00\ldots^\circ\) |
| \(\text{bearing}\) | \(=\) | \(055^\circ\text{T}\) |
In the NE quadrant the bearing is \(\theta\) itself, so the bearing is \(055^\circ\text{T}\).
Sketch the trip. The finish is south-west, so measure the angle west of south.
| \(d\) | \(=\) | \(\sqrt{16^2+10^2}=\sqrt{356}=18.86\ldots\) |
| \(\tan\theta\) | \(=\) | \(\dfrac{10}{16}\) |
| \(\theta\) | \(=\) | \(32.00\ldots^\circ\) |
| \(\text{bearing}\) | \(=\) | \(180^\circ+32^\circ=212^\circ\) |
| \(\text{back}\) | \(=\) | \(212^\circ-180^\circ=032^\circ\) |
(i) \(18.9\text{ km}\) (ii) \(212^\circ\text{T}\) (iii) \(032^\circ\text{T}\)
Common pitfalls
Frequently asked questions
How do you find the distance between two points using bearings?
If the trip is made of north-south and east-west legs, the legs meet at a right angle. Use Pythagoras: the direct distance squared equals the sum of the squares of the two legs.
How do you work out how far north and east a boat has travelled?
Find the angle the path makes with north. The distance north is the distance travelled times cos of the angle, and the distance east is the distance travelled times sin of the angle.
How do you find a bearing using trigonometry?
Use tan inverse of the east-west distance divided by the north-south distance to get the acute angle from the north-south line. Then use the quadrant to write the three-figure bearing.
Why is my bearing wrong when I use tan inverse?
Tan inverse only gives the acute angle from the north-south line. You must adjust it for the quadrant: subtract from 180 for south-east, add to 180 for south-west, and subtract from 360 for north-west.
How do I find the bearing for the return trip?
The return trip is the back bearing. Add 180 degrees if the original bearing is less than 180 degrees, or subtract 180 degrees if it is more than 180 degrees.