3D Trigonometry
3D trigonometry looks harder than it is. The whole skill comes down to one move: find a right-angled triangle hiding inside the solid, then use ordinary SOHCAHTOA on it. Once you can spot that triangle in a cuboid, a pyramid or a cone, every IGCSE question becomes a familiar 2D problem. This page shows you how to find the triangle, work out a missing length with Pythagoras in 3D, and then calculate the angle a line makes with a flat surface. Worked examples and a practice room are below.The one idea behind every 3D trigonometry question
There is no new trigonometry to learn here. Every 3D question is solved by finding a right-angled triangle inside the solid and then using the same sine, cosine and tangent ratios you already know. The challenge is purely about seeing that triangle in a drawing of a box, a pyramid or a cone.
The angle between a line and a plane
This is the most common 3D trigonometry question, and the wording trips students up. To find the angle a line (such as a long diagonal) makes with a flat surface (the base):
- Drop a vertical line from the top of the diagonal straight down to the base. This gives a right angle where it meets the base.
- Join that foot point to the start of the diagonal. This line lies flat in the base.
- You now have a right-angled triangle. The required angle sits between the diagonal and its shadow on the base.
The ratios you will use
To find a side in the right triangle you may first need Pythagoras. To find the angle once you know two sides, use the inverse: \(\theta=\tan^{-1}\!\left(\frac{\text{opp}}{\text{adj}}\right)\), and similarly for \(\sin^{-1}\) and \(\cos^{-1}\).
Worked Examples
💡 Example 1: angle in a cuboid
A cuboid has a base measuring 7 cm by 4 cm and a height of 6 cm. Find the angle between the space diagonal and the base, to 1 decimal place.
Base diagonal \(d\) (using Pythagoras on the base):
\[ \begin{array}{rcl} d &=& \sqrt{7^2+4^2} \\ &=& \sqrt{65}\ \text{cm} \end{array} \]The right triangle has this base diagonal as the adjacent side and the height as the opposite side:
\[ \begin{array}{rcl} \tan\theta &=& \dfrac{6}{\sqrt{65}} \\ \theta &=& \tan^{-1}\!\left(\dfrac{6}{\sqrt{65}}\right) \\ &=& 36.7^\circ \end{array} \]What is happening?
The space diagonal, the height and the base diagonal form a right-angled triangle standing upright inside the box.
Keep \(\sqrt{65}\) exact rather than rounding early. Rounding the base diagonal first would shift the final angle.
The opposite side is the vertical height; the adjacent side is the flat base diagonal, so tangent is the natural choice.
💡 Example 2: edge angle in a pyramid
A right pyramid has a square base of side 10 cm. Its apex is 12 cm vertically above the centre of the base. Find the angle a slant edge makes with the base, to 1 decimal place.
Half the base diagonal (centre to a corner):
\[ \begin{array}{rcl} \tfrac{1}{2}\sqrt{10^2+10^2} &=& \tfrac{1}{2}\sqrt{200} \\ &=& 5\sqrt{2}\ \text{cm} \end{array} \]Right triangle: vertical height 12 (opposite), half-diagonal \(5\sqrt{2}\) (adjacent):
\[ \begin{array}{rcl} \tan\theta &=& \dfrac{12}{5\sqrt{2}} \\ \theta &=& \tan^{-1}\!\left(\dfrac{12}{5\sqrt{2}}\right) \\ &=& 59.5^\circ \end{array} \]What is happening?
The slant edge runs from a base corner up to the apex. Its shadow on the base runs from that corner to the centre, which is half the base diagonal.
The height drops from the apex straight to the centre, meeting the base at a right angle.
That gives a right triangle with the slant edge as the hypotenuse, so tangent links the known height and half-diagonal.
🔑 Key Points
- Every 3D question reduces to one right-angled triangle. Find it first, then the trigonometry is routine.
- The angle between a line and a plane is measured between the line and its shadow on that plane.
- Keep surds exact (such as \(\sqrt{65}\)) until the final step to avoid rounding error.
- For a square-based pyramid, the shadow of a slant edge is half the base diagonal.
⚠️ Common Pitfalls
- Using a base edge instead of the base diagonal. The shadow of a space diagonal is the diagonal of the base, not a side.
- Forgetting to halve the base diagonal in a pyramid: the apex sits above the centre, not a corner.
- Rounding the first length, then using it again. Carry the exact value through.
- Picking the wrong ratio. Decide which side is opposite and which is adjacent after you have drawn the flat triangle.