IGCSE · Practice questions · Pythagoras and trigonometry in 3D

Flatten it, then solve.

Extended. Six original Cambridge-style questions on Pythagoras and trigonometry in three dimensions.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

One flat right-angled triangle at a time.

Label the solid and mark every right angle (vertical meets horizontal at \(90^\circ\)).
Base first: find the diagonal or distance in the base with Pythagoras.
Then upright: use the vertical triangle for the slant length or the angle.
Line and plane: drop a perpendicular; the angle is between the line and its projection.
Accuracy: keep surds or full calculator values until the end.

Calculation

[2 marks]

A cuboid measures 7 cm by 5 cm by 4 cm. Calculate the length of a space diagonal.

\(d = \sqrt{7^2 + 5^2 + 4^2} = \sqrt{49 + 25 + 16} = \sqrt{90}\) ✓
\(d = 9.49\) cm (3 s.f.) ✓

Calculation

[3 marks]

For the cuboid in the previous question, with base 7 cm by 5 cm and height 4 cm, calculate the angle between a space diagonal and the base.

Base diagonal \(= \sqrt{7^2 + 5^2} = \sqrt{74}\) cm ✓
\(\tan\theta = \dfrac{4}{\sqrt{74}}\) ✓
\(\theta = 24.9^\circ\) (1 d.p.) ✓

Proof

[2 marks]

Without a calculator, show that the space diagonal of a cube of edge 6 cm is \(6\sqrt{3}\) cm.

\(d^2 = 6^2 + 6^2 + 6^2 = 108\) ✓
\(d = \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) cm ✓

Calculation

[4 marks]

A pyramid has a rectangular base \(ABCD\) with \(AB = 10\) cm and \(BC = 6\) cm. Its vertex \(V\) is 8 cm vertically above the centre \(M\) of the base.

(a) Calculate the length of the edge \(VA\).

(b) Calculate the angle between \(VA\) and the base.

\(AC = \sqrt{10^2 + 6^2} = \sqrt{136}\), so \(AM = \tfrac{1}{2}\sqrt{136} = \sqrt{34} = 5.830\ldots\) cm ✓
(a) \(VA = \sqrt{8^2 + 34} = \sqrt{98} = 9.90\) cm (3 s.f.) ✓
(b) \(\tan(VAM) = \dfrac{8}{\sqrt{34}}\), so angle \(VAM = 53.9^\circ\) (1 d.p.) ✓

Calculation

[3 marks]

A rectangular field \(ABCD\) has \(AB = 40\) m and \(BC = 30\) m. A vertical mast \(AP\) of height 15 m stands at \(A\).

Calculate the angle of elevation of \(P\) from \(C\).

\(AC = \sqrt{40^2 + 30^2} = 50\) m ✓
\(\tan(PCA) = \dfrac{15}{50}\) ✓
Angle of elevation \(= 16.7^\circ\) (1 d.p.) ✓

Analysis

[5 marks]

A room is a cuboid 5 m long, 4 m wide and 3 m high. A string is stretched from a corner of the floor to the opposite corner of the ceiling.

(a) Calculate the length of the string.

(b) Calculate the angle between the string and the floor.

(a) Length \(= \sqrt{5^2 + 4^2 + 3^2} = \sqrt{50} = 7.07\) m (3 s.f.) ✓
(b) Floor diagonal \(= \sqrt{5^2 + 4^2} = \sqrt{41}\); \(\tan\theta = \dfrac{3}{\sqrt{41}}\), so \(\theta = 25.1^\circ\) (1 d.p.) ✓
(c) The distance from the string's far end straight back to the end wall is the 5 m length, so \(\sin\phi = \dfrac{5}{\sqrt{50}}\). ✓
The projection on the wall is the wall's diagonal \(\sqrt{4^2 + 3^2} = 5\) m, so the right-angled triangle has two sides of 5 m: it is isosceles, and \(\phi = 45^\circ\) exactly. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.