Bearings

Video lesson4 worked examples

Theory

Bearings use trigonometry (often the sine or cosine rules) to calculate a distance or direction.

Bearings are always measured clockwise from North and are expressed as three digits (e.g., $045^\circ$ ).

A common error candidates make is treating non-right-angled bearing scenarios as right-angled triangles, which leads to incorrectly using basic SOH-CAH-TOA or Pythagoras instead of the sine/cosine rules.

The Golden Rule: always sketch the situation with a North line at each point. Use the North lines and the angle facts (angles round a point, alternate angles between parallel Norths) to find the angle inside the triangle, then apply the Sine or Cosine Rule. To go back the other way, a back bearing differs from the original by $180^\circ$ .

⚠️ Common Examiner Traps

Three figures: always write bearings with three digits, e.g. $072^\circ$ , not $72^\circ$ .
Back bearing: the reverse direction is $\pm 180^\circ$ — add 180 if the bearing is under 180, subtract if it is over.
Triangle angle vs bearing: the angle you calculate inside the triangle is usually not the bearing — combine it with the North line to get the actual bearing.
Not right-angled: unless a right angle is given, use the Sine or Cosine Rule, not SOH-CAH-TOA.

Worked examples

Example 1

Ship A leaves port on a bearing of $040^\circ$ for 10 km. Ship B leaves the same port on a bearing of $100^\circ$ for 15 km. Find the distance between the two ships.

Step 1: Find the angle between their paths at the port:

100^\circ - 040^\circ = 60^\circ

Step 2: Use the Cosine Rule to find the distance ( $d$ ) between them:

d^2 = 10^2 + 15^2 - 2(10)(15)\cos(60^\circ)

Step 3: Calculate:

d^2 = 100 + 225 - 300(0.5) = 325 - 150 = 175

Answer: $d = \sqrt{175} \approx 13.2\text{ km}$ .

Example 2

A hiker walks 5 km on a bearing of $060^\circ$ , then 8 km on a bearing of $150^\circ$ . How far is the hiker from the starting point?

Step 1: Find the interior angle of the triangle at the turning point. The back-bearing of the first leg is $060^\circ + 180^\circ = 240^\circ$ , and the second leg is on a bearing of $150^\circ$ . So the interior angle is $240^\circ - 150^\circ = 90^\circ$ .

Step 2: Since the paths meet at $90^\circ$ , this forms a right-angled triangle, so use Pythagoras:

d^2 = 5^2 + 8^2 = 25 + 64 = 89

Answer: $d = \sqrt{89} \approx 9.4\text{ km}$ .

Example 3

Town B is 20 miles from Town A on a bearing of $050^\circ$ . Town C is 30 miles from Town A on a bearing of $130^\circ$ . Find the distance between Town B and Town C.

Step 1: Angle at A is $130^\circ - 050^\circ = 80^\circ$ .

Step 2: Find distance BC using Cosine Rule:

BC^2 = 20^2 + 30^2 - 2(20)(30)\cos(80^\circ)

BC^2 \approx 400 + 900 - 1200(0.1736) \approx 1091.6

Answer: $BC \approx 33.0\text{ miles}$ .

Example 4

Back Bearing

The bearing of a harbour H from a boat B is $118^\circ$ . Calculate the bearing of the boat B from the harbour H.

Step 1: The reverse (back) bearing differs by $180^\circ$ . Since the original bearing $118^\circ$ is less than $180^\circ$ , add:

118^\circ + 180^\circ = 298^\circ

Step 2: Check it is a valid three-figure bearing (between 000° and 360°) — it is.

Answer: the bearing of B from H is $298^\circ$ .

Ready to practise?

9 Bearings questions with answers and video solutions