Trigonometry in 3D — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers finding lengths and angles inside three-dimensional solids such as cuboids and pyramids. By the end of this guide you should be able to find the space diagonal of a cuboid, find the angle between an edge or diagonal and a face, and work through a problem that needs more than one stage.

You should also be able to find the angle between a line and a plane using the projection of that line, apply Pythagoras' theorem and the trigonometric ratios in the right order, and keep intermediate values accurate.

The organising idea is that there is no 3D trigonometry — only 2D trigonometry done inside a 3D shape. Every formula you need is one you already have: Pythagoras' theorem and SOH CAH TOA. The whole difficulty is finding the right-angled triangle hiding inside the solid, and once you redraw that triangle on its own as a flat picture, the question becomes an ordinary one. Students who struggle here are almost always trying to work within the 3D sketch instead of extracting the triangle.

Key terms and definitions

Plane — a flat surface within a solid, such as the base of a cuboid.

Space diagonal — the line from one corner of a solid to the opposite corner, passing through the interior.

Base diagonal — the diagonal across the base face, used as a stepping stone to the space diagonal.

Projection — the "shadow" a line casts on a plane, found by dropping perpendicular from its end onto that plane.

Angle between a line and a plane — the angle between the line and its projection on that plane.

Foot of the perpendicular — the point where a vertical line meets the plane below it.

Core concepts

Extract the triangle

The single technique in this topic is to find a right-angled triangle containing what you want, then draw it separately, flat on the page, and label its sides with the lengths you know.

Working inside the 3D sketch tempts you into treating lines as perpendicular when they are not. Redrawing removes the problem entirely, and the working then looks like an ordinary right-angled triangle question because that is what it now is.

Label the separate triangle before calculating anything. The marks follow from the labelled triangle.

The space diagonal of a cuboid

This needs Pythagoras' theorem applied twice.

First find the diagonal across the base, using the length and width as the two shorter sides. Then use that base diagonal together with the height as the two shorter sides of a second, upright right-angled triangle, whose hypotenuse is the space diagonal.

For a cuboid measuring 3 by 4 by 12: the base diagonal squared is 9 + 16 = 25, and the space diagonal squared is 25 + 144 = 169, so the space diagonal is 13.

Keep the intermediate value squared rather than rooting it and squaring it again. It avoids rounding and usually makes the arithmetic exact.

There is also a direct formula: d² = l² + w² + h², giving d = √(3² + 4² + 12²) = √169 = 13. It is the same calculation with the middle step hidden, and either route earns the marks.

The angle between a line and a plane

This is the part of the topic that needs care, because the angle is not between the line and just any line in the plane.

The angle between a line and a plane is the angle between the line and its projection — the shadow it would cast on that plane if light came from directly above.

For the space diagonal of a cuboid, the projection onto the base is the base diagonal. So the angle between the space diagonal and the base is the angle inside the right-angled triangle formed by the base diagonal, the vertical height, and the space diagonal itself.

That triangle is the same one used in the second stage of the space diagonal calculation, which is why the two questions so often appear together.

For a cuboid with base diagonal 5 and height 12, the space diagonal is 13, and the angle θ with the base satisfies tan θ = 12 ÷ 5, or equally sin θ = 12 ÷ 13.

Choosing the ratio

Once the triangle is drawn, the choice of ratio is made exactly as in two dimensions: identify which two sides you know and pick the ratio containing both.

Often more than one route works. With all three sides known you may use whichever ratio you prefer — but using the two sides given in the question, rather than one you calculated, avoids carrying forward any rounding error.

Pyramids

In a square-based pyramid, the important line is the one from the apex straight down to the centre of the base. That line is perpendicular to the base, so it forms a right angle with anything drawn in the base plane.

To reach a corner of the base, use half the base diagonal — not half the base edge. This is the commonest error with pyramids: the horizontal distance from the centre to a corner is half a diagonal, and it must be found with Pythagoras' theorem first.

The slant height to the midpoint of a base edge is different again, using half the base edge. Read the question carefully to see which is wanted.

Angles of elevation in 3D

Problems about a mast, a tower or a pole viewed from a point on the ground are 3D in setting but reduce to a single right-angled triangle: the vertical height, the horizontal distance along the ground, and the line of sight.

The angle of elevation is measured up from the horizontal, so the height is opposite and the ground distance is adjacent, making tan the usual ratio.

From 20 m away, the elevation to the top of a 15 m tower satisfies tan θ = 15 ÷ 20, giving θ = 36.9°.

Worked examples

Example 1: A space diagonal in two stages

A cuboid measures 9 cm by 12 cm by 8 cm. Find the length of its space diagonal, to three significant figures.

The base diagonal squared is 9² + 12² = 81 + 144 = 225, so the base diagonal is 15 cm.

Now use that with the height in the upright triangle: 15² + 8² = 225 + 64 = 289.

The space diagonal is √289 = 17 cm.

The direct formula gives the same result: √(81 + 144 + 64) = √289 = 17.

Example 2: The angle between a diagonal and the base

A cuboid has a base diagonal of 5 cm and a height of 12 cm. Find the angle between its space diagonal and the base, to one decimal place.

The projection of the space diagonal onto the base is the base diagonal, so the angle sits in the right-angled triangle with the base diagonal of 5 as the adjacent side and the height of 12 as the opposite side.

Those two sides call for tan: tan θ = 12 ÷ 5 = 2.4.

Applying the inverse gives θ = 67.4°.

Sense check: the solid is much taller than it is wide, so a steep diagonal is expected.

Example 3: An angle of elevation

A vertical pole of height 10 m is viewed from a point on the ground 10 m from its base. Find the angle of elevation of the top of the pole.

The height is opposite the angle and the ground distance is adjacent, so use tan.

tan θ = 10 ÷ 10 = 1.

Applying the inverse gives θ = 45°.

This is the exact value worth recognising: whenever the opposite and adjacent are equal, the triangle is isosceles and the angle is 45°.

Common mistakes and how to avoid them

Working inside the 3D sketch. Redraw the right-angled triangle flat on the page before calculating anything.

Using a line in the plane that is not the projection. The angle between a line and a plane uses the line's shadow, which for a space diagonal is the base diagonal.

Rounding the base diagonal before using it. Keep it squared, or carry full calculator accuracy into the second stage.

Using half the base edge instead of half the base diagonal in a pyramid. The distance from the centre of the base to a corner is half a diagonal.

Assuming two lines meet at a right angle because they look as though they do. Only use a right angle you can justify, such as a vertical meeting a horizontal plane.

Stopping at the squared value. Pythagoras gives the square of the length; the length is its root.

Forgetting the inverse function when finding an angle. A ratio of 2.4 is not an angle until the inverse tangent is applied.

Exam technique for "Trigonometry in 3D"

Sketch the right-angled triangle separately and label every side you know, before touching the calculator. That drawing is usually worth a mark and makes the rest of the question routine.

Say which triangle you are working in. Naming its vertices makes a multi-stage solution easy to follow and easy to check.

Show the substitution into Pythagoras' theorem or into the ratio as its own line, since that is where the method marks sit.

Keep the intermediate value squared when finding a space diagonal in two stages, or carry full accuracy forward.

Sense-check each answer: the space diagonal is the longest line in a cuboid, and an angle with the base must be less than 90°.

Follow the rounding instruction, and give angles in degrees to the accuracy asked for.

Quick revision summary

There is no separate 3D trigonometry: find the right-angled triangle, redraw it flat, and use Pythagoras' theorem or SOH CAH TOA as usual.

The space diagonal of a cuboid takes Pythagoras twice — base diagonal first, then that with the height — or in one step as d² = l² + w² + h². For 3, 4 and 12 the answer is 13.

The angle between a line and a plane is the angle between the line and its projection on that plane. For a space diagonal, the projection is the base diagonal, so the angle sits in the triangle made by the base diagonal, the height and the space diagonal.

Keep the intermediate base diagonal squared to avoid rounding it.

In a square-based pyramid, the line from the apex to the centre of the base is vertical, and the horizontal distance to a corner is half the base diagonal, not half the base edge.

For an angle of elevation, the height is opposite and the ground distance is adjacent, so use tan and remember the inverse function.

Sense-check: the space diagonal is the longest line in the solid, and equal opposite and adjacent sides give exactly 45°.