Learn8300Papers 1–3 · Higher tierGeometry and measuresTrigonometry in three dimensions

Geometry and measures · Subtopic 13 of 17

Trigonometry in Three Dimensions

Every 3D problem is a chain of 2D right-angled triangles. Find a triangle on the base first (often a diagonal), then stand a vertical triangle on it. The angle between a line and a plane is the angle between the line and its shadow on the plane.

Cambridge 4024 Paper 1Cambridge 4024 Paper 2Edexcel 4MA1 Papers 1H, 2HAQA 8300 Papers 1–3≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find face diagonals and space diagonals of cuboids.
Find heights and edge lengths in pyramids.
Identify the angle between a line and a plane.
Draw the 2D right-angled triangle you need.

Pause and think

An ant walks from one bottom corner of a box to the opposite top corner, but a fly goes straight through the air. Whose path is a single straight line, and how could you find its length?

Section 2 of 14

Core idea

Start hereThe longest pencil in a box

What's the longest pencil that fits in a pencil case? Not along the base diagonal, but corner to opposite corner through the middle. 3D problems become 2D ones once you spot the hidden right-angled triangle.

The big idea

Find a right-angled triangle inside the solid, draw it flat, use Pythagoras or trig. Line–plane angle: use the line's projection on the plane.

1. The cuboid

Cuboid: the space diagonal and its angle with the base

Base diagonal

=\sqrt{6^2+4^2}=\sqrt{52}

. Then

\tan\theta=\dfrac{3}{\sqrt{52}}

, so

\theta=22.6^\circ

. The space diagonal is

\sqrt{52+9}=7.81

Space diagonal

$d_{\text{base}}^2=a^2+b^2$
Pythagoras on the base.
$d^2=d_{\text{base}}^2+c^2=a^2+b^2+c^2$
Vertical edge is perpendicular to the base diagonal.

Worked exampleCuboid $6\times4\times3$

Space diagonal and its angle with the base.

Goal: Length; angle.
Strategy: Two right triangles.

$\sqrt{36+16}=\sqrt{52}=7.21$
Base diagonal (projection).
$\sqrt{52+9}=7.81$
Space diagonal.
$\tan\theta=\frac3{7.21}\Rightarrow22.6^\circ$
Height over projection.

Answer

7.81

;

22.6^\circ

✓ Check

\sqrt{36+16+9}=\sqrt{61}

✓.

Worked exampleSquare-based pyramid

Base 8, apex 10 above the centre.

Goal: Slant edge; angle with the base.
Strategy: Centre to corner is half the diagonal.

$\tfrac12\sqrt{128}=5.66$
Half diagonal.
$\sqrt{100+32}=11.5$
$5.66^2=32$ .
$\tan^{-1}\tfrac{10}{5.66}=60.5^\circ$
Angle with the base.

Answer

11.5

;

60.5^\circ

✓ CheckSteep pyramid: angle above

45^\circ

✓.

Common misconceptionWrong right angle Check the right angle is really there (vertical edge ⟂ anything in the base).

Pause and think

Longest rod in a $2\times2\times1$ box?

Summary

Find and draw the right-angled triangle.
Line–plane angle via projection.

Projection: Shadow of a line on a plane.

HistoryMeasuring the Earth with a shadow. Around

240

BC Eratosthenes noticed that at noon on midsummer's day the Sun was directly overhead in Syene, but cast a shadow of about

7^\circ

in Alexandria. Using that angle and the distance between the cities, he estimated the Earth's circumference remarkably well.

Fun factThe angle of a pyramid. The faces of the Great Pyramid of Giza rise at about

52^\circ

to the ground. Builders kept that slope using a ratio called the seked: how far the face moved in for each unit of height, an ancient version of the tangent ratio.

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