IGCSE Maths 0580 Pythagoras and Trigonometry in 3D Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Trigonometry · 6.6

Every pythagoras and trigonometry in 3d question from 6 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 6.6. 6 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/21/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Trigonometry past paper questions.

0580 Pythagoras and Trigonometry in 3D past paper questions: Extended (Papers 2 and 4)

15 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q21(c,d)

A solid cuboid ABCDEFGH has BF = 2 cm, BC = 6 cm and CD = 9 cm.

0580/21 May/June 2026 Q21 – cuboid ABCDEFGH with diagonal CE9 cm6 cm2 cmDCBAHGFENOT TOSCALE

(c) Calculate CE. (d) The angle between CE and the base CDHG is x°. Find the value of sin x°.

Show solution
Part (c)
  1. CE is a space diagonal, so use 3D Pythagoras with all three edges.
  2. CE² = 2² + 6² + 9² = 4 + 36 + 81 = 121.
  3. CE = √121 = CE = 11 cm
Answer CE = 11 cm
Part (d)
  1. E is directly above H on the base CDHG, so CE, CH and the vertical edge EH make a right-angled triangle.
  2. EH = 6 cm is opposite the angle x° and CE = 11 cm is the hypotenuse, so use sin.
  3. sin x° = 6⁄11
Answer sin x° = 6⁄11
26 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q16(c,d)

ABCDE is a square-based pyramid. EC and BD intersect at F. AF is perpendicular to the base BCDE. BC = 16.5 cm and AF = 28 cm.

0580/42 May/June 2026 Q16 – square-based pyramid ABCDE with apex A above F, the centre of the base

(c) Calculate AB. (d) Calculate the angle between AB and the base BCDE.

Show solution
Part (c)
  1. F is the centre of the square base, so BF is half of the diagonal BD.
  2. Pythagoras on the base: BD² = 16.5² + 16.5² = 544.5, so BD = 23.33… cm and BF = 11.667… cm.
  3. Triangle ABF is right-angled at F, with AB as the hypotenuse.
  4. AB² = AF² + BF² = 28² + 11.667…² = 784 + 136.125 = 920.125.
  5. AB = √920.125 = 30.33… = 30.3 cm
Answer 30.3 cm
Part (d)
  1. The angle between AB and the base is angle ABF, because F is directly below A.
  2. From angle B: AF is opposite and AB is the hypotenuse, so use sin.
  3. sin(ABF) = 28⁄30.33… = 0.9230…
  4. Angle ABF = sin⁻¹(0.9230…) = 67.37… = 67.4°
  5. Tip: You could also use tan(ABF) = 28⁄11.667…, which avoids using the rounded AB.
Answer 67.4°
33 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q17

The diagram shows a cuboid, ABCDEFGH. AB = 18 cm, BC = 14 cm and AE = 16 cm. Calculate the length of the diagonal AG.

0580/43 May/June 2026 Q17 – cuboid ABCDEFGH with diagonal AG18 cm14 cm16 cmNOT TOSCALE
Show solution
  1. First find the diagonal of the base, AC: AC² = 18² + 14² = 324 + 196 = 520.
  2. AG is the hypotenuse of a right-angled triangle with sides AC and CG = 16 cm.
  3. AG² = AC² + 16² = 520 + 256 = 776.
  4. AG = √776 = 27.85… = 27.9 cm
  5. Tip: In one step, AG² = 18² + 14² + 16².
Answer 27.9 cm
43 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q19

A box is a cuboid measuring 5 cm by 5 cm by 7 cm. Mala has a straight rod of length 10 cm. Show that this rod does not fit completely inside the box.

0580/22 October/November 2025 Q19 – cuboid box 5 cm by 5 cm by 7 cm5 cm5 cm7 cmNOT TOSCALE
Show solution
  1. The longest straight rod that fits in a cuboid lies along a space diagonal (corner to opposite corner).
  2. Use 3D Pythagoras: diagonal² = 5² + 5² + 7² = 25 + 25 + 49 = 99.
  3. Diagonal = √99 = 9.949… cm.
  4. Compare with 10 cm: √99 is less than √100 = 10.
  5. the longest diagonal is about 9.95 cm, which is less than 10 cm, so the rod does not fit
Answer the longest diagonal is about 9.95 cm, which is less than 10 cm, so the rod does not fit
54 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q27

A prism of length 15 cm has a trapezium cross-section ABCD with angle DAB = 90° and angle ADC = 90°. AB = 10 cm, AD = 5 cm and DC = 7 cm. Calculate the angle the diagonal AG makes with the base ABFE.

0580/41 October/November 2025 Q27 – trapezium prism ABCDEFGH of length 15 cm
Show solution
  1. G is the back corner above C, so it is 5 cm above the base ABFE.
  2. The point directly below G on the base is 7 cm along from the edge AE (the length DC) and 15 cm back along the prism.
  3. Horizontal distance from A to that point (Pythagoras on the base): √(7² + 15²) = √(49 + 225) = √274 = 16.55… cm.
  4. The angle θ between AG and the base is in the right-angled triangle with opposite 5 cm and adjacent √274 cm.
  5. tan θ = 5⁄√274 = 0.3020…
  6. θ = tan⁻¹(0.3020…) = 16.80…° = 16.8°
Answer 16.8°
66 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q24

ABCDEFGH is a cuboid with AE = 6.4 cm, EH = 5.1 cm and the diagonal AG = 13 cm.

0580/43 October/November 2025 Q24 – cuboid ABCDEFGH with diagonal AG5.1 cm6.4 cmNOT TOSCALE

(a) Calculate EF. (b) Calculate the angle between the line AG and the base EFGH.

Show solution
Part (a)
  1. For a cuboid, (space diagonal)² = length² + width² + height².
  2. AG² = EF² + EH² + AE², so 13² = EF² + 5.1² + 6.4².
  3. 169 = EF² + 26.01 + 40.96, so EF² = 169 − 66.97 = 102.03.
  4. EF = √102.03 = 10.10… = 10.1 cm
Answer 10.1 cm
Part (b)
  1. A is directly above E, so the angle between AG and the base is angle AGE.
  2. Triangle AEG is right-angled at E. From G, AE = 6.4 is opposite and AG = 13 is the hypotenuse, so use sin.
  3. sin(AGE) = 6.4⁄13 = 0.4923…
  4. Angle AGE = sin⁻¹(0.4923…) = 29.49… = 29.5°
Answer 29.5°

Question papers used on this page

All pythagoras and trigonometry in 3d questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q21(c,d)
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q16(c,d)
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q17
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q19
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q27
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q24

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