3D Trigonometry
3D trigonometry finds lengths and angles inside solid objects by picking out a right-angled triangle in a vertical or horizontal plane. The angle between a line and a plane is the angle between the line and its shadow on that plane. In NSW Year 10 Advanced maths this applies to prisms, cubes, pyramids, tents and towers, with bearings (NESA outcome MA5‑TRG‑P‑01).
You will find diagonals and angles in cylinders, prisms and cubes, the angle between a space diagonal and the base, and angles between edges, faces and the base of a square pyramid. Practical problems include a conical tent, a flagpole at the corner of a sports field, an LED strip in a shed, a plane seen from a control tower and a pitched roof panel. Harder questions find the height of a tower or mast from two points on the ground. It is part of Year 10 Advanced Trigonometry.
Every question with a fully worked solution.
- 3D Trigonometry - Video - 3D Trigonometry Watch
Theory
A 3D trigonometry problem is always solved in flat, right-angled triangles. Find a triangle on the ground, then a triangle standing up from it, and carry the shared side across. This Year 10 Advanced maths page covers space diagonals, the angle between a line and a plane, angles in pyramids and cones, and elevation and bearing problems in three dimensions. It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
In a 3D shape, look for right-angled triangles that lie flat in one plane: on the floor, on a face, or standing vertically inside the solid. Each one can be solved with Pythagoras' theorem or sine, cosine and tangent.
The space diagonal of a rectangular prism joins opposite corners through the inside. The base diagonal lies on the floor.
The angle between a line and a plane is found by dropping a perpendicular from the top of the line to the plane. The line, the perpendicular and the segment joining their feet (the projection) make a right-angled triangle. The angle is at the point where the line meets the plane.
In a right pyramid or cone the apex is directly above the centre of the base, so the height meets the base at right angles. Distances from the centre use half a side, half a diagonal or the radius.
For a rectangular prism \(l \times w \times h\):
| Square pyramid, side \(s\) | Distance from centre \(O\) |
|---|---|
| to the midpoint of an edge (face angle) | \(\dfrac{s}{2}\) |
| to a corner (edge angle) | \(\dfrac{s\sqrt{2}}{2}\) |
How to solve a 3D problem
- Sketch the solid and mark every right angle: vertical edges meet the floor at \(90^\circ\), and the height of a right pyramid or cone meets the base at its centre.
- Pick the right-angled triangle that contains the unknown. Redraw it flat, on its own.
- Find any missing side of that triangle in another right-angled triangle first, usually on the floor (a base diagonal or a ground distance).
- Solve and keep exact surds or unrounded values until the last step.
Useful set-ups
- Line and plane: drop a perpendicular from the top end, join the feet, and use tan (or sin with the line as hypotenuse).
- Cone or tent: radius, height and slant form a right-angled triangle; a ridge tent is a triangular prism, so use half the base across and the full length along.
- Sloping plane (a roof): a point at distance \(d\) up a roof of pitch \(\theta\) is \(d\sin\theta\) higher. Points on a horizontal edge of the roof are all at the same height.
- Elevation and bearings: draw a plan of the ground with the bearings, solve it, then draw the vertical triangle with the angle of elevation.
- Height unknown in two triangles: write each ground distance as \(\dfrac{h}{\tan\theta}\) and join them with Pythagoras' theorem or a sum (Example 4).
(i) the length of the space diagonal \(AG\)
(ii) the angle \(\theta\) between \(AG\) and the base, correct to the nearest degree.
Work in two right-angled triangles: first the floor (triangle \(ABC\)), then the upright triangle \(ACG\).
| \(AC^2\) | \(=\) | \(12^2 + 9^2 = 225\) |
| \(AC\) | \(=\) | \(15\text{ cm}\) |
| \(AG^2\) | \(=\) | \(15^2 + 20^2 = 625\) |
| \(AG\) | \(=\) | \(\mathbf{25}\text{ cm}\) |
(ii) In triangle \(ACG\), \(GC = 20\) is opposite \(\theta\) and \(AC = 15\) is adjacent.
| \(\tan\theta\) | \(=\) | \(\dfrac{20}{15}\) |
| \(\theta\) | \(=\) | \(53.130\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{53^\circ}\) |
Shortcut: \(AG = \sqrt{l^2 + w^2 + h^2}\).
(i) the exact height \(TO\)
(ii) the angle between a sloping edge and the base
(iii) the angle between a triangular face and the base. Give angles to one decimal place.
(i) \(OC\) is half a diagonal of the base.
| \(OC\) | \(=\) | \(\dfrac{1}{2}\sqrt{6^2 + 6^2} = 3\sqrt{2}\) |
| \(TO^2\) | \(=\) | \(9^2 - (3\sqrt{2})^2 = 81 - 18 = 63\) |
| \(TO\) | \(=\) | \(\mathbf{3\sqrt{7}}\text{ m}\) |
(ii) In triangle \(TOC\), the edge is the hypotenuse:
| \(\cos\beta\) | \(=\) | \(\dfrac{3\sqrt{2}}{9}\) |
| \(\beta\) | \(=\) | \(61.874\ldots^\circ \approx \mathbf{61.9^\circ}\) |
(iii) In triangle \(TOM\), \(OM = 3\) m is half a side:
| \(\tan\alpha\) | \(=\) | \(\dfrac{3\sqrt{7}}{3} = \sqrt{7}\) |
| \(\alpha\) | \(=\) | \(69.295\ldots^\circ \approx \mathbf{69.3^\circ}\) |
The face is steeper than the edge, because \(OM\) is shorter than \(OC\).
(i) the height of the tower, correct to one decimal place
(ii) the angle of elevation of \(T\) from \(B\), correct to the nearest degree.
Split the problem: a triangle on the ground, then a vertical triangle. From \(B\), west is \(270^\circ\), so \(\angle ABF = 300^\circ - 270^\circ = 30^\circ\), and \(\angle FAB = 90^\circ\).
| \(AF\) | \(=\) | \(120\tan 30^\circ = 69.282\ldots\) |
| \(h\) | \(=\) | \(AF\tan 40^\circ = 58.134\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{58.1}\text{ m}\) |
(ii) The ground distance \(BF\) is the hypotenuse of the ground triangle:
| \(BF\) | \(=\) | \(\dfrac{120}{\cos 30^\circ} = 138.564\ldots\) |
| \(\tan\phi\) | \(=\) | \(\dfrac{h}{BF} = \dfrac{58.134\ldots}{138.564\ldots}\) |
| \(\phi\) | \(=\) | \(22.760\ldots^\circ \approx \mathbf{23^\circ}\) |
Write both ground distances in terms of \(h\). East and south are at right angles, so use Pythagoras' theorem in the ground triangle \(FPQ\).
| \(FP\) | \(=\) | \(\dfrac{h}{\tan 34^\circ}\) |
| \(FQ\) | \(=\) | \(\dfrac{h}{\tan 26^\circ}\) |
| \(\dfrac{h^2}{\tan^2 34^\circ} + \dfrac{h^2}{\tan^2 26^\circ}\) | \(=\) | \(300^2\) |
| \(h^2\left(\dfrac{1}{\tan^2 34^\circ} + \dfrac{1}{\tan^2 26^\circ}\right)\) | \(=\) | \(90\,000\) |
| \(h^2 \times 6.4017\ldots\) | \(=\) | \(90\,000\) |
| \(h\) | \(=\) | \(118.569\ldots\) |
| \(\therefore\ h\) | \(\approx\) | \(\mathbf{118.6}\text{ m}\) |
Check: \(FP \approx 175.8\) m and \(FQ \approx 243.1\) m, and \(\sqrt{175.8^2 + 243.1^2} \approx 300\) m.
Common pitfalls
Frequently asked questions
How do you find the space diagonal of a rectangular prism?
Use Pythagoras' theorem twice: first find the base diagonal, then use it with the height. In one step the space diagonal is the square root of length squared plus width squared plus height squared.
How do you find the angle between a line and a plane?
Drop a perpendicular from the top end of the line to the plane and join the foot to where the line meets the plane. This makes a right-angled triangle; the angle at the meeting point is the angle between the line and the plane.
What is the difference between the angle of a face and the angle of an edge in a pyramid?
The angle between a triangular face and the base uses the distance from the centre to the middle of a base edge. The angle between a sloping edge and the base uses the distance from the centre to a corner, which is longer, so the edge angle is smaller.
How do you do 3D trigonometry problems with bearings?
Draw a plan view of the ground and use the bearings to solve the ground triangle. Then draw the vertical triangle that stands on one of its sides and use the angle of elevation.
Is 3D trigonometry in the Year 10 Standard or Advanced course?
Both. It prepares you for Mathematics Advanced in Years 11 and 12.