E6.6· 12 questions · 54 marks · 65 min · 2013–2025· Structured questions
Every Cambridge IGCSE Mathematics Paper 2 question on pythagoras’ theorem and trigonometry, laid out as 12 A4 pages with the mark scheme below. Nothing is left out. Free to read, no account.
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12 / 12Answers below. Sit the paper first if you are practising.
Pastlit
Mathematics 0580 · Pythagoras’ theorem and trigonometry — Paper 2
IGCSE · topical answer key — answer key (teacher use)
Question
Answer
Marks
4
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9| Question | Answer | Marks | From |
|---|---|---|---|
| 1 | see sheet | 4 | 0580/21 May/June 2013 |
| 2 | see sheet | 3 | 0580/21 May/June 2017 |
| 3 | see sheet | 4 | 0580/22 Oct/Nov 2017 |
| 4 | see sheet | 5 | 0580/21 May/June 2020 |
| 5 | see sheet | 5 | 0580/21 May/June 2020 |
| 6 | see sheet | 6 | 0580/22 May/June 2020 |
| 7 | see sheet | 4 | 0580/22 Feb/March 2023 |
| 8 | see sheet | 4 | 0580/21 Oct/Nov 2023 |
| 9 | see sheet | 2 | 0580/21 May/June 2024 |
| 10 | see sheet | 4 | 0580/21 May/June 2024 |
| 11 | see sheet | 4 | 0580/23 May/June 2024 |
| 12 | see sheet | 9 | 0580/22 Feb/March 2025 |
23 For Examiner′s F C Use 6 cm NOT TO SCALE D A 5 cm E 12 cm B The diagram shows a triangular prism of length 12 cm. Triangle ABC is a cross section of the prism. Angle BAC = 90°, AC = 6 cm and AB = 5 cm. Calculate the angle between the line CE and the base ABED. Answer … [4] _____________________________________________________________________________________
4 marks
Mark scheme: 23 4 M1 for recognition of angle CEA 24.8 or 24.77 to 24.78 M1 for 122 + 52 6 M1 for tan = oe their AE
13 A B NOT TO SCALE 5 cm D C 13 cm F E 4 cm H G The diagram shows a cuboid ABCDEFGH. AE = 5 cm, EH = 4 cm and AG = 13 cm. Calculate the angle between the line AG and the base EFGH of the cuboid. … [3]
3 marks
Mark scheme: 13 22.6 or 22.61 to 22.62 3 5 M2 forr sin [=] ooe 13 or M1 for identifyinng angle AGE
26 P Q NOT TO 2 cm SCALE D C 3 cm A 4 cm B The diagram shows a prism of length 4 cm. The cross section is a right-angled triangle. BC = 3 cm and CQ = 2 cm. Calculate the angle between the line AQ and the base, ABCD, of the prism. … [4]
4 marks
Mark scheme: 26 21.8 or 21.80… 4 2 M3 for tan = oe 32 + 4 2 or M1 for 3 2 + 4 2 or 32 + 4 2 + 2 2 and M1 for recognising angle QAC
13 C B 18 cm NOT TO SCALE 13.5 cm 21 cm D E A C lies on a circle with diameter AD. B lies on AC and E lies on AD such that BE is parallel to CD. AB = 21 cm, CD = 18 cm and BE = 13.5 cm. Work out the radius of the circle. … cm [5]
5 marks
Mark scheme: 13 16.6 or 16.64… 5 18 M2 for 21× = [ AC ] oe 13.5 13.5 18 or M1 for scale factor or oe soi 18 13.5 Then Pythagoras method: and M2 for 28 2 + 18 2 [÷ 2] or ( theirAC ) 2 + 18 2 [÷ 2] or M1 for AD 2 = 28 2 + 18 2 2 2 2 or AD = ( theirAC ) + 18 OR alternative trigonometry method e.g. 21 M1 for tan E = 13.5 18 and M1 for AD = cos their 57.3
19 F G NOT TO SCALE E H B C 5.5 cm A 20 cm D The diagram shows cuboid ABCDEFGH of length 20 cm and width 5.5 cm. The volume of the cuboid is 495 cm 3. Find the angle between the line AG and the base of the cuboid ABCD. … [5]
5 marks
Mark scheme: 19 12.2 or 12.24… 5 4.5 M4 for tan = oe 20 2 + 5.5 2 or M1 for recognising angle GAC 495 M1 for 20 × 5.5 M1 for 20 2 + 5.5 2 or 20 2 + 5.5 2 + (their 4. 5) 2 their 4.5 M1 for tan = oe 20 2 + 5.5 2
27 H G NOT TO SCALE E F 6 cm C D 6 cm A B 8 cm The diagram shows a cuboid. AB = 8cm , AD = 6cm and DH = 6cm . Calculate angle HAF. Angle HAF = … [6]
6 marks
Mark scheme: 27 64.9 or 64.89 to 64.90 6 + 72 − 100 B5 for [cos =]100 2 × 10 × 72 OR M1 for 82 + 62 M1 for 62 + 62 (theirAF ) 2 + ( theirAH ) 2 − ( theirHF ) 2 M2 for 2 × (theirAF ) × (theirAH ) or M1 for (theirHF)2 = (theirAF)2 + (their AH)2 – 2 × (theirAF) × (their AH) cos(HAF) AF, AH etc from correct method
22 Q NOT TO SCALE P D C 4 cm 7 cm A B 5 cm The diagram shows a triangular prism ABCDQP of length 7 cm. The cross-section is triangle PAB with PA = 4 cm , AB = 5 cm and angle PAB = 90° . Calculate the angle between the line PC and the base ABCD. … [4]
4 marks
Mark scheme: 22 24.9 or 24.93 to 24.94 4 4 M3 for tan = oe 5 2 + 7 2 or M2 for 52 + 72 oe or 52 + 72 + 42 oe or M1 for recognition of angle PCA.
21 H G E F 9.2 cm NOT TO SCALE D C 4.5 cm A B 15.1 cm The diagram shows a cuboid ABCDEFGH. AB = 15.1 cm, BC = 4.5 cm and CG = 9.2 cm. Calculate the angle that the diagonal BH makes with the face ADHE. … [4] Question 22 is printed on the next page.
4 marks
Mark scheme: 21 55.9 or 55.85… 4 15.1 M3 for tan[…] = oe 4.5 2 + 9.2 2 or M2 for [AH2 =] 4.52 + 9.22 or [BH2 =] 4.52 + 9.22 + 15.12 or M1 for recognising angle BHA if 0 scored SC1 for [angle BHD =] 59.7[1…] or 59.72
17 B 6.7 cm 81° A NOT TO 5.9 cm SCALE C Calculate the area of triangle ABC. … cm2 [2]
2 marks
Mark scheme: 17 19.5 or 19.52… 2 1 M1 for × 6.7 × 5.9 × sin 81 oe 2
21 9.1 cm E F 6.5 cm A B NOT TO SCALE H G 4 cm D C The diagram shows a cuboid. HD = 4 cm, EH = 6.5 cm and EF = 9.1 cm. Calculate the angle between CE and the base CDHG. … [4]
4 marks
Mark scheme: 21 33.2 or 33.18… 4 6.5 M3 for tan oe 4 2 9.12 or M2 for 4 2 9.12 oe or 4 2 9.12 6.5 2 oe or M1 for recognising the angle ECH
24 W V 4 cm NOT TO SCALE D C 5 cm A 15 cm B The diagram shows a triangular prism with cross-section triangle BCV. Angle BCV = 90° , BC = 5cm , CV = 4cm and AB = 15cm . Calculate the angle between AV and the base ABCD. … [4]
4 marks
Mark scheme: 24 14.2 or 14.19 to 14.20 4 4 M3 for tan = oe 15 2 5 2 or M2 for 15 2 5 2 or15 2 52 4 2 or M1 for recognition of angle VAC
18 NOT TO E SCALE D 18 cm 6 cm 60° A B C 17 cm The quadrilateral ACDE is formed by two right-angled triangles ABE and BCD. AC = 17 cm, AE = 18 cm and BD = 6 cm. (a) Show that CD = 10 cm. [5] (b) Find the perimeter of the quadrilateral ACDE. Give your answer in the form p + k q . … cm [4]
9 marks
Mark scheme: 18(a) AB M1 = cos60 or [AB = ] 18cos 60 18 1 A1 cos 60 = and [AB =] 9 2 Correct use of Pythagoras’ theorem M1 i.e. [CD2 =] 6 2 + (17 −theirAB ) 2 oe Correct evaluation for their AB M1 [CD2 =] 36 + 64 or CD = 36 + 64 100 = 10 A1 Dep on M1A1M1M1 18(b) 39 + 9 3 4 B3 for [BE =] 9 3 or for answer k + 9 3 or for answer equivalent to 39 + 9 3 but not in required form OR BE 3 M2 for = oe or better 18 2 BE or M1 for = sin60 oe or better 18 M1 for 18 + 17 + 10 + their BE – 6 oe OR M2 for 182 −their 92 oe or M1 for BE2 + (their 9)2 = 182 oe M1 for 18 + 17 + 10 + their BE – 6 oe