Every right-angled trigonometry question from 10 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 6.2. 10 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/31/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 Right-Angled Trigonometry past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q28
The diagram shows a right-angled triangle with hypotenuse 9 cm and the side adjacent to angle x° equal to 6 cm. Calculate the value of x.
Show solution
- From angle x: the 6 cm side is adjacent and the 9 cm side is the hypotenuse, so use cos.
- cos x° = adjacent⁄hypotenuse = 6⁄9 = 0.6666…
- x = cos⁻¹(0.6666…) = 48.18… = 48.2
May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q26
The diagram shows a right-angled triangle, PQR, with the right angle at R. S lies on side PR. Angle PQS = 31° and angle RQS = 44°. QS = 23 cm. Calculate PR.
Show solution
- Work in the small triangle QRS first. It has a right angle at R, angle RQS = 44° and hypotenuse QS = 23 cm.
- QR is adjacent to the 44° angle, so use cos: QR = 23 × cos 44° = 16.544… cm.
- Now use the big triangle PQR. Angle PQR = 31 + 44 = 75°.
- PR is opposite the 75° angle and QR is adjacent, so use tan: PR = QR × tan 75°.
- PR = 16.544… × 3.7320… = 61.74…
- 61.7 cm
- Tip: Keep the full calculator value of QR. Rounding it too early can change the final answer.
May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q25(b,c)
A circle has diameter AC and B is a point on the circumference, so angle ABC = 90°. AB = 14 cm and BC = 8 cm.
(b) Calculate AC. (c) Calculate angle BAC.
Show solution
- Angle ABC = 90°, so AC is the hypotenuse.
- Pythagoras: AC² = 14² + 8² = 196 + 64 = 260.
- AC = √260 = 16.12… = 16.1 cm
- From angle A: BC = 8 cm is opposite and AB = 14 cm is adjacent, so use tan.
- tan(BAC) = 8⁄14 = 0.5714…
- Angle BAC = tan⁻¹(0.5714…) = 29.74… = 29.7°
Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q20
The diagram shows a rectangle with a diagonal. The longest side of the rectangle is 7.6 m. The perimeter of the rectangle is 25.6 metres. Calculate the value of x, the angle between the diagonal and the shorter side.
Show solution
- Perimeter = 2 × long side + 2 × short side. 25.6 = 2 × 7.6 + 2 × short side.
- 2 × short side = 25.6 − 15.2 = 10.4, so the short side = 5.2 m.
- The diagonal makes a right-angled triangle. From angle x, the 7.6 m side is opposite and the 5.2 m side is adjacent, so use tan.
- tan x = 7.6⁄5.2 = 1.4615…
- x = tan⁻¹(1.4615…) = 55.61… = 55.6
Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q24
ABC is a right-angled triangle with the right angle at A. AC = 32 cm and angle ABC = 27°. Calculate BC.
Show solution
- The right angle is at A, so BC (opposite the right angle) is the hypotenuse.
- From angle B, the side AC is opposite. Opposite and hypotenuse means use sin.
- sin 27° = AC⁄BC = 32⁄BC
- Rearrange: BC = 32⁄sin 27° = 32 ÷ 0.4539… = 70.48…
- BC = 70.5 cm
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q26
A right-angled triangle has an angle of 37°. The side adjacent to 37° is 12 cm and the side opposite it is x cm. Calculate the value of x.
Show solution
- From the 37° angle, x is opposite and 12 cm is adjacent. Opposite and adjacent means use tan.
- tan 37° = x⁄12
- Multiply both sides by 12: x = 12 × tan 37° = 12 × 0.7535… = 9.042…
- 9.04
Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q29(b)
A right-angled triangle has hypotenuse 5.4 cm and an angle of 38°. The side adjacent to 38° is y cm. Calculate the value of y.
Show solution
- From the 38° angle, y is adjacent and 5.4 cm is the hypotenuse. Adjacent and hypotenuse means use cos.
- cos 38° = y⁄5.4
- Multiply both sides by 5.4: y = 5.4 × cos 38° = 5.4 × 0.7880… = 4.255…
- 4.26
0580 Right-Angled Trigonometry past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q11
Triangle WXY is right-angled at X. WX = 12.4 cm and angle XWY = 29°. Calculate XY.
Show solution
- The right angle is at X. From angle W, the side XY is opposite and WX is adjacent, so use tan.
- tan 29° = XY⁄12.4
- Multiply both sides by 12.4: XY = 12.4 × tan 29° = 12.4 × 0.5543… = 6.873…
- 6.87 cm
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q7(b)
The point A is on the top of a vertical building. B is on the ground, a horizontal distance of 22 m from the base of the building, and AB = 25 m. Work out the angle of depression of B from A.
Show solution
- The angle of depression from A is the same as the angle of elevation from B, because they are alternate angles.
- At B, the 22 m ground is adjacent and AB = 25 m is the hypotenuse, so use cos.
- cos B = 22⁄25 = 0.88.
- B = cos⁻¹(0.88) = 28.35… = 28.4°
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q13
Two vertical poles, AE and PT, stand on horizontal ground AP. AE = 1.5 m, PT = 12.6 m and AP = 33 m. Calculate the angle of elevation of the point T from the point E.
Show solution
- Draw a horizontal line from E across to the pole PT. This makes a right-angled triangle.
- The vertical side of this triangle is how much higher T is than E: 12.6 − 1.5 = 11.1 m.
- The horizontal side is the distance between the poles: 33 m.
- The angle of elevation θ is at E. The 11.1 m side is opposite θ and the 33 m side is adjacent, so use tan.
- tan θ = opposite⁄adjacent = 11.1⁄33 = 0.3363…
- θ = tan⁻¹(0.3363…) = 18.59…° = 18.6°
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