3D Trigonometry (AA SL)
Finding lengths and angles inside 3D solids - cuboids, pyramids, and combinations of the two - always comes down to spotting the right flat, right-angled triangle hidden inside the shape, then applying ordinary Pythagoras and trigonometry to it. It's part of the broader 3D Geometry topic.
11 questions on this sub-topic.
The key idea
Covered under IB syllabus reference SL3.1. There's no separate "3D trig formula" - you find a right-angled triangle inside the solid and use the same tools as always.
Pythagoras in 3D
\(d = \sqrt{x^2+y^2+z^2}\)
Extends straight from 2D Pythagoras - use it to find a space diagonal or any straight-line distance across a solid.
SOHCAHTOA in 3D
\(\sin,\ \cos,\ \tan\) as usual
Once you've identified a right-angled triangle inside the solid (e.g. a vertical edge and a diagonal on the base), the trig ratios work exactly as in 2D.
Need the full syllabus wording and volume/surface-area formulas? See 3D Geometry.
Worked examples
The diagram shows a cuboid \(ABCDEFGH\) where \(AB = 8\) cm, \(BC = 6\) cm and \(AE = 3\) cm.
(a) Find the length of \(AC\).
(b) Find the angle that \(AG\) makes with the plane \(ABCD\), giving your answer to the nearest degree.
Worked solution
(a) Triangle \(ABC\) is right-angled at \(B\) (opposite sides of a rectangle): \(AC=\sqrt{AB^2+BC^2}=\sqrt{64+36}=\sqrt{100}=10\) cm. A1
(b) \(AG\) is the space diagonal. Triangle \(ACG\) is right-angled at \(C\) (since \(CG\) is vertical and \(AC\) is horizontal), with \(AC=10\) and \(CG=AE=3.\) M1
The angle \(\theta\) that \(AG\) makes with plane \(ABCD\) is \(\angle GAC\), so \(\tan\theta=\dfrac{CG}{AC}.\) M1
\(\tan\theta=\dfrac{3}{10}.\) A1
\(\theta=\tan^{-1}\!\left(\tfrac{3}{10}\right)\approx 16.7°\approx 17°.\) A1
\(VABCD\) is a right pyramid with a square base of side 10 cm. The apex \(V\) is directly above the centre \(M\) of the base, and the vertical height \(VM = 12\) cm.
(a) Find the length \(AM\), where \(M\) is the centre of the base.
(b) Find the length of the slant edge \(VA.\)
(c) Find the size of angle \(AVB\), the angle at the apex between two adjacent slant edges, to 1 decimal place.
Worked solution
(a) \(AM\) is half the base diagonal: \(AM=\tfrac12\sqrt{10^2+10^2}.\) M1
\(AM=\tfrac12\sqrt{200}\approx 7.07\) cm. A1
(b) Triangle \(VMA\) is right-angled at \(M\): \(VA=\sqrt{VM^2+AM^2}.\) M1
\(VA=\sqrt{12^2+7.07^2}\approx 13.93\) cm. A1
(c) Triangle \(AVB\) is isosceles with \(VA=VB\approx13.93\) and \(AB=10\): \(\cos(A\hat VB)=\dfrac{VA^2+VB^2-AB^2}{2\cdot VA\cdot VB}.\) M1
\(A\hat VB\approx 42.1°.\) A1
Common mistakes
- Trying to work in 3D directly. There's no shortcut - always isolate a single flat, right-angled triangle inside the solid first, then solve that triangle on its own using ordinary 2D methods.
- Using the wrong length as the "opposite" or "adjacent" side. Redraw the specific triangle you've identified separately if it helps - it's easy to misjudge which 3D edge plays which role once several lines cross in the same diagram.
- Rounding a length too early. An intermediate length like a diagonal or slant edge should be carried through unrounded (or stored on the GDC) into the next step, or small errors compound into a wrong final angle.
Ready to practise properly?
11 3D trigonometry questions, marked instantly like the real exam.
Quick answers
How do I find a length or angle inside a 3D solid?
Find a right-angled triangle inside the solid that contains the length or angle you need, then apply Pythagoras' theorem or basic trigonometry (SOHCAHTOA) to that flat triangle - the same way you would in 2D.
What is the angle between a line and a plane in 3D?
It's the angle between the line and its own projection (shadow) onto the plane, found by dropping a perpendicular from a point on the line to the plane and using the right-angled triangle this creates.