Every exact trigonometric values question from 7 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 6.3. 7 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 Exact Trigonometric Values past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q18
y = sin 60° × tan 30°. Find the value of y.
Show solution
- Use the exact values: sin 60° = √3⁄2 and tan 30° = 1⁄√3.
- y = √3⁄2 × 1⁄√3.
- The √3 on the top cancels the √3 on the bottom: y = 1⁄2
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q8
Using values from the list 1⁄√3, 1⁄2, 1⁄√2, √3, √3⁄2, complete: cos 45° = … and tan 60° = …
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- For 45°, use a right-angled isosceles triangle with shorter sides 1 and hypotenuse √2.
- cos 45° = adjacent⁄hypotenuse = 1⁄√2: cos 45° = 1⁄√2
- For 60°, use half an equilateral triangle with sides 1, √3 and 2. The side opposite 60° is √3 and the side next to it is 1.
- tan 60° = opposite⁄adjacent = √3⁄1: tan 60° = √3
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q19
A right-angled triangle has an angle of 60°. The side opposite the 60° angle is 9 cm and the side adjacent to it is x cm. Find the exact value of x. Give your answer in the form a√b.
Show solution
- From the 60° angle, the 9 cm side is opposite and the x cm side is adjacent, so use tan.
- tan 60° = 9⁄x. The exact value of tan 60° is √3, so √3 = 9⁄x.
- Rearrange: x = 9⁄√3.
- Rationalise the denominator: 9⁄√3 × √3⁄√3 = 9√3⁄3 = 3√3
Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q17
Triangle ABC is right-angled at B. BC = 12 cm and angle ACB = 30°.
(a) Find the length of AC. Give your answer in the form a√b.
(b) Find the shortest distance from B to AC.
Show solution
- From the angle C: BC is adjacent and AC is the hypotenuse, so use cos.
- cos 30° = BC⁄AC = 12⁄AC, so AC = 12⁄cos 30°.
- The exact value of cos 30° is √3⁄2, so AC = 12 ÷ √3⁄2 = 12 × 2⁄√3 = 24⁄√3.
- Rationalise the denominator: 24⁄√3 × √3⁄√3 = 24√3⁄3 = 8√3 cm
- The shortest distance from B to AC is the perpendicular line from B to AC.
- That perpendicular makes a small right-angled triangle with hypotenuse BC = 12 cm and the 30° angle at C.
- The perpendicular is opposite the 30° angle: distance = 12 × sin 30° = 12 × 1⁄2 = 6 cm
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q26
Solve tan x = −1⁄√3 for 0° ≤ x ≤ 360°.
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- Ignore the minus sign first: tan 30° = 1⁄√3, so the reference angle is 30°.
- tan is negative in the second quadrant (90° to 180°) and the fourth quadrant (270° to 360°).
- Second quadrant: 180 − 30 = 150°. Fourth quadrant: 360 − 30 = 330°.
- x = 150°, x = 330°
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q20
A right-angled triangle has an angle of 30°. The side opposite the 30° angle is 6 cm and the adjacent side is x cm. Find the exact value of x.
Show solution
- From the 30° angle, the 6 cm side is opposite and the x cm side is adjacent, so use tan.
- tan 30° = opposite⁄adjacent = 6⁄x
- The exact value of tan 30° is 1⁄√3, so 6⁄x = 1⁄√3.
- Cross-multiply: x = 6 × √3.
- x = 6√3
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q22
(a) Write down the exact value of tan 60°.
(b) Solve 2 sin x − 1 = 0 for 0° ≤ x ≤ 360°.
Show solution
- Learn this exact value: tan 60° = √3
- Rearrange: 2 sin x = 1, so sin x = 1⁄2.
- The first solution is sin⁻¹(1⁄2) = 30°.
- The sine graph is symmetrical about 90°, so the second solution is 180 − 30 = 150°.
- x = 30° or x = 150°
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