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3D Trigonometry

Nothing new is needed for three dimensions: the trick is to find a flat right-angled triangle hiding inside the solid, then use the same Pythagoras and trigonometry you already know on that single triangle.

MYP 5Extended MathsAdvanced Geometry & TrigonometryCriteria A · C · D~11 min read

Written by Fayek · MYP student

The core idea

A 3D problem is never really three-dimensional once you get going. The whole method is to pull out a single right-angled triangle that lies flat in one plane, redraw it on its own, and solve that. The skill is spotting the triangle, not learning new formulas.

The two everyday tools carry straight over:

\[ a^2 + b^2 = c^2 \qquad\qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \]

Usually you use Pythagoras first to find a length you are missing (often a diagonal across a face), then use trigonometry on a second triangle that contains the angle you want.

The space diagonal

The space diagonal of a cuboid runs from one corner to the opposite corner, cutting right through the middle of the solid. To find it, do Pythagoras twice: once across the base, then once up to the top corner.

For a cuboid with length \(l\), width \(w\) and height \(h\), the base diagonal is \(\sqrt{l^2 + w^2}\). Standing that up against the height gives the space diagonal:

\[ d = \sqrt{l^2 + w^2 + h^2} \]

Keep the exact value

Work with the base diagonal in surd form, such as \(\sqrt{100}\) or \(\sqrt{52}\), and only round at the very end. Rounding the base diagonal early throws off the final angle.

Line and plane angles

The angle between a line and a plane is the angle between the line and its "shadow", its projection straight down onto the plane.

Projection onto a plane
The line you would get by dropping every point of the sloping line vertically onto the plane. For a space diagonal, its projection onto the base is the base diagonal.

So the angle between a space diagonal and the base sits inside a right-angled triangle whose horizontal side is the base diagonal, whose vertical side is the height, and whose slope is the diagonal itself. Then \(\tan\theta = \dfrac{\text{height}}{\text{base diagonal}}\).

A full worked example

Worked example

A cuboid has a base \(8\) cm by \(6\) cm and a height of \(5\) cm. Find (a) the length of the space diagonal, and (b) the angle it makes with the base.

1
Base diagonal first, by Pythagoras on the base: \(\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\) cm.
2
Now the space diagonal, using the base diagonal and the height: \(d = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \approx 11.2\) cm.
3
For the angle, the right-angled triangle has opposite \(=\) height \(=5\) and adjacent \(=\) base diagonal \(=10\): \(\tan\theta = \dfrac{5}{10} = 0.5\).
4
\(\theta = \tan^{-1}(0.5) = 26.57^\circ\).
(a) \(5\sqrt{5}\approx 11.2\) cm    (b) \(\theta \approx 26.6^\circ\)

Where this is assessed

Identifying the right triangle inside the solid is Criterion D (applying maths to an unfamiliar situation), while clean surd or rounded answers with a labelled sub-diagram are Criterion C. Sketch the flat triangle separately, it prevents most mistakes.

Check yourself

1. Find the length of the space diagonal of a cube with side \(4\) cm. +

All three edges are \(4\), so \(d = \sqrt{4^2+4^2+4^2} = \sqrt{48} = \mathbf{4\sqrt{3}\approx 6.93}\) cm.

2. A cuboid has a rectangular base \(12\) cm by \(5\) cm. What is the length of the base diagonal? +

By Pythagoras across the base: \(\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = \mathbf{13}\) cm.

3. A space diagonal of length \(15\) cm sits above a base diagonal of \(12\) cm. Find the angle between the diagonal and the base. +

The base diagonal is the adjacent side and the space diagonal is the hypotenuse, so \(\cos\theta = \dfrac{12}{15} = 0.8\). Then \(\theta = \cos^{-1}(0.8) = \mathbf{36.9^\circ}\).

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