Understanding bearings
What does a bearing of 125° mean?
North = 0° (or 360°). Bearings increase clockwise. 125° is between east (90°) and south (180°).
Year 11
Bearings, navigation and trigonometric applications
Interpret multi-step geometry contexts and solve real navigation or design problems strategically.
Lesson overview
Geometry — bearings and 3D trigonometry
A bearing is a direction expressed as an angle measured clockwise from north, given as a three-figure bearing (e.g., 145°). Bearings are used in navigation, surveying and engineering to describe direction uniquely. Navigation problems often involve multiple stages: finding a distance using trigonometry, then using that distance and another bearing to locate a second point. Three-dimensional problems extend the trigonometry to include heights, slopes and angles above the horizontal. Key skills: drawing accurate diagrams, identifying which trigonometric ratio (sin, cos, tan) or rule (sine rule, cosine rule) applies, applying angle of elevation and depression, and solving for unknown distances or angles.
Using sine rule in a navigation context
From point A, a ship sails 50 km on a bearing of 060° to point B. From B, it sails 70 km on a bearing of 150° to point C. Find the distance AC.
Identify the angle at B from the bearing change. Use the cosine rule.
Angle of elevation and depression
From a point 100 m away from a building, the angle of elevation to the top is 35°. What is the height of the building?
Angle of elevation from horizontal up to the top. Use tan for height/distance.
Bearing return problem
From A, a walker goes 5 km on a bearing of 045° to reach B. What bearing should they use to return to A?
The return bearing is always 180° opposite (add 180°, or subtract if already > 180°).
Worked example — multi-step navigation
A surveyor at point A measures that a building top B is 80 m away on a bearing of 040°. From the building (now at point B), looking back, the surveyor at A is on a bearing of 220°. The building is 50 m tall. If the surveyor measures from ground level, find the angle of elevation from A to the top of the building.
- Identify that the surveyor is 80 m from the building base (horizontal distance).
- The building height is 50 m.
- Use tan(angle) = opposite / adjacent = height / distance = 50 / 80 = 0.625.
- Find the angle: angle = arctan(0.625) ≈ 32.0°.
- This is the angle of elevation from point A to the building top.
Try it
Use the examples carefully, then choose the answer.
Question 1
A bearing of 270° points in which direction?
💡 0° = North, 90° = East, 180° = South, 270° = West.
Question 2
If a ship travels from A to B on a bearing of 110°, what is the bearing from B back to A?
💡 The return bearing is 110° + 180° = 290°.
Question 3
From a point 150 m from a tree, the angle of elevation to the top is 25°. What is the height of the tree (to nearest m)?
💡 height = 150 × tan(25°) ≈ 150 × 0.4663 ≈ 70 m.
Question 4
Two points P and Q are 20 km apart. From P, Q is on a bearing of 080°. From Q, P is on a bearing of 260°. Are these bearings consistent?
💡 Return bearing = initial bearing ± 180°.
Question 5
A ladder leans against a wall 5 m high. The base of the ladder is 3 m from the wall. At what angle does the ladder meet the ground (angle of elevation, to 1 d.p.)?
💡 tan(θ) = . θ = arctan().
Common mistakes
Watch for these when working through the lesson.
- Forgetting that bearings are always measured clockwise from north, not from the current direction.
- Mixing up angle of elevation (up from horizontal) with angle of depression (down from horizontal).
- Not using the right trigonometric ratio (using sin instead of tan, or vice versa).
- Forgetting the bearing return rule: add 180° (or subtract 180° if > 180°).
Related topics
These ideas fit closely with this lesson.
- Trigonometry in right-angled triangles
- Sine rule and cosine rule
- Coordinate geometry and distance
- 3D geometry and Pythagoras in 3D