A complete guide to three-figure bearings, back bearings, and three-dimensional trigonometry — space diagonals of a cuboid and the angle between a line and a plane.
Two final applications extend trigonometry beyond flat, two-dimensional triangles: bearings, used in navigation to describe direction precisely, and three-dimensional trigonometry, used to find lengths and angles inside solid shapes such as cuboids and pyramids. Both build directly on the sine rule, cosine rule and SOH-CAH-TOA covered earlier — the challenge here is setting up the right triangle within a real-world or 3D context before applying the tools you already know.
A bearing is a way of describing a direction as an angle measured clockwise from North, always written using exactly three figures (e.g. 045°, not 45°). Bearings range from 000° (due North) to 359°.
Common cardinal directions correspond to standard bearings: North = 000°, East = 090°, South = 180°, and West = 270°. Any other direction is given as a three-figure number between these.
A ship travels in a direction 30° East of North. Written as a bearing, this is 030° (note the leading zero, to keep exactly three figures).
To find a bearing on a diagram: draw a North line (usually vertical) at the starting point, then measure the angle clockwise from that North line to the line joining the two points.
The back bearing is the bearing of the return journey — from the destination back to the start. It is found by adding 180° if the original bearing is less than 180°, or subtracting 180° if it is 180° or more.
The bearing of B from A is 075°. Since 075° < 180°, the back bearing (A from B) is 075 + 180 = 255°.
The bearing of D from C is 210°. Since 210° ≥ 180°, the back bearing (C from D) is 210 − 180 = 030°.
Bearing problems are ultimately triangle problems in disguise: once the bearings are converted into an angle inside a triangle, the sine rule, cosine rule or SOH-CAH-TOA can be applied exactly as before.
A ship sails from port P on a bearing of 060° for 40 km to reach point Q. It then sails from Q on a bearing of 150° for 25 km to reach point R. Find the distance PR.
The angle PQR (the angle inside the triangle at Q) is found from the two bearings: turning from a bearing of 060° to continue would point back along 240° (the back bearing), and the new course is 150°, so angle PQR = 240 − 150 = 90°.
Since angle PQR = 90°, this becomes a right-angled triangle problem: PR² = 40² + 25² = 1600 + 625 = 2225, so PR = 47.2 km (3 s.f.), using Pythagoras' theorem directly since the angle is 90°.
Three-dimensional trigonometry problems (typically involving cuboids or pyramids) are solved by identifying a suitable right-angled triangle hidden within the solid, usually by first finding a diagonal across a flat face, then using that diagonal as one side of a second, three-dimensional triangle.
The space diagonal of a cuboid (the longest diagonal, running from one corner through the interior to the opposite corner) is found using a two-step Pythagoras method: first find the diagonal of the base, then use that as one side of a second right-angled triangle with the cuboid's height.
A cuboid has length 6 cm, width 8 cm, and height 5 cm. Find the length of its space diagonal.
Base diagonal = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.
Space diagonal = √(10² + 5²) = √(100 + 25) = √125 = 11.2 cm (3 s.f.).
The angle between a sloping line (such as a space diagonal) and a flat plane (such as the base) is found by dropping a perpendicular from the top of the line down to the plane, then using the right-angled triangle this creates with the line's projection onto the plane.
Using the cuboid from Example 4 (base diagonal 10 cm, height 5 cm), find the angle between the space diagonal and the base.
The right-angled triangle here has the base diagonal (10 cm) as the adjacent side and the height (5 cm) as the opposite side, relative to the required angle.
tan θ = 5 ÷ 10 = 0.5, so θ = tan⁻¹(0.5) = 26.6° (3 s.f.).
Common errors include: measuring a bearing anticlockwise instead of clockwise; forgetting to write a bearing with three figures (writing "60°" instead of "060°"); using the wrong sign when calculating a back bearing; and in 3D problems, using the wrong diagonal or forgetting the perpendicular height when identifying the triangle for a line-to-plane angle.
| Situation | Method |
|---|---|
| Bearing less than 180° | Back bearing = bearing + 180° |
| Bearing 180° or more | Back bearing = bearing − 180° |
| Space diagonal of a cuboid | Two-step Pythagoras: base diagonal, then combine with height |
| Angle between a diagonal and a plane | tan θ = height ÷ base diagonal (or the relevant right-angled triangle) |
Both bearings and 3D trigonometry come down to the same core skill built throughout this Trigonometry subcategory: correctly identifying a triangle within a larger context, then applying SOH-CAH-TOA, the sine rule, or the cosine rule as appropriate.
This article is original EDUSAMBAM educational writing. It is designed as a broad bearings and 3D trigonometry resource covering the navigation and solid-geometry applications of the trigonometric tools introduced earlier in this subcategory. Exact examination requirements can vary between examination boards and syllabuses, so students should also compare their work with the specification and past-paper requirements of their own board.
Recommended study approach: practise reading and writing three-figure bearings and back bearings fluently, then work through cuboid diagonal problems by always finding the base diagonal first before attempting the full space diagonal or a line-to-plane angle.
20 questions covering bearings, back bearings, and 3D trigonometry with cuboid space diagonals. Answer every question, then submit to see your score instantly.