IGCSE Mathematics: Trigonometry — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Trigonometry revision notes

Tier note: Pythagoras’ theorem (C6.1) and sine, cosine and tangent in right-angled triangles (C6.2) are Core. Items marked (Extended) are Extended only: exact trigonometric values (E6.3), the graphs of trigonometric functions (E6.4), the sine and cosine rules (E6.5) and Pythagoras and trigonometry in 3D (E6.6). All other questions use Core content.

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Questions

1. A right-angled triangle has two shorter sides of length 5 cm and 12 cm. Calculate the length of the hypotenuse. [2]

2. In a right-angled triangle, the hypotenuse is 15 cm and the side opposite angle x is 7 cm. Calculate angle x, correct to 1 decimal place. [3]

3. (Extended) Write down the exact value of sin 30°, without using a calculator. [1]

4. (Extended) Sketch the general shape of the graph of y = cos x for 0° ≤ x ≤ 360°, and state the value of cos 0° and cos 180°. [2]

5. (Extended) In triangle ABC, angle A = 40°, angle B = 65° and side a (opposite A) = 8 cm. Use the sine rule to find side b (opposite B), correct to 3 significant figures. [3]

6. (Extended) In triangle PQR, PQ = 7 cm, QR = 9 cm, and the angle between them, angle Q, is 55°. Use the cosine rule to find PR, correct to 3 significant figures. [3]

7. (Extended) A cuboid has length 6 cm, width 8 cm and height 10 cm.

(a) Calculate the length of the space diagonal of the cuboid. [2] (b) Calculate the angle this diagonal makes with the base of the cuboid, correct to 1 decimal place. [3]

8. (Extended) In triangle XYZ, XY = 6 cm, YZ = 8 cm and XZ = 11 cm. Use the cosine rule to find angle Y, correct to 1 decimal place. [3]

9. In a right-angled triangle the hypotenuse is 10 cm and one angle is 35°. Calculate the length of the side opposite the 35° angle. [2]

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Answers

1. hyp² = 5² + 12² = 25 + 144 = 169 [1] → hyp = √169 = 13 cm [1].

2. sin x = 7 ÷ 15 [1] = 0.4667 [1] → x = sin⁻¹(0.4667) = 27.8° [1].

3. sin 30° = ½ [1].

4. The graph is a smooth wave starting at its maximum, falling to its minimum at 180°, and returning to its maximum at 360° [1]. cos 0° = 1; cos 180° = −1 [1].

5. b ÷ sin B = a ÷ sin A → b = (a × sin B) ÷ sin A [1] = (8 × sin 65°) ÷ sin 40° [1] = (8 × 0.9063) ÷ 0.6428 = 11.3 cm [1].

6. PR² = PQ² + QR² − 2 × PQ × QR × cos Q [1] = 7² + 9² − 2(7)(9)cos 55° = 49 + 81 − 126(0.5736) [1] = 130 − 72.27 = 57.73 → PR = 7.60 cm [1].

7. (a) Diagonal = √(6² + 8² + 10²) [1] = √200 = 14.1 cm (3 s.f.) [1]. (b) Base diagonal = √(6² + 8²) = √100 = 10 cm [1]. tan(angle) = 10 ÷ 10 = 1 [1] → angle = 45.0° [1].

8. Rearranging the cosine rule to find an angle: cos Y = (XY² + YZ² − XZ²) ÷ (2 × XY × YZ) [1] = (6² + 8² − 11²) ÷ (2 × 6 × 8) = (36 + 64 − 121) ÷ 96 = −21 ÷ 96 = −0.2188 [1] → Y = cos⁻¹(−0.2188) = 102.6° [1].

9. Opposite = hypotenuse × sin 35° [1] = 10 × 0.5736 = 5.74 cm [1].

Common mistake: using cos instead of sin. Label the sides (opposite, adjacent, hypotenuse) from the given angle before choosing the ratio.

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Where marks are usually lost

• Applying Pythagoras’ theorem to the wrong pair of sides — the hypotenuse is always the longest side and is always opposite the right angle.
• Mixing up which side is opposite, adjacent and hypotenuse relative to the marked angle, especially when the triangle is drawn in an unfamiliar orientation.
• Giving an “exact value” as a rounded decimal approximation instead of a surd (e.g. writing 1.414 instead of √2) when the question specifically asks for it. This does not apply to a terminating decimal that equals the exact value precisely – 0.5 is exactly equal to ½, not a rounding of it, so either form is acceptable unless the question specifically demands fraction notation.
• Using the sine rule when the triangle doesn’t have a matching angle–side pair available, when the cosine rule is needed instead (and vice versa).
• In 3D problems, trying to use a 2D right-angled triangle directly on the solid without first finding the correct 2D “base diagonal” or cross-section to work in.
• Leaving the calculator in the wrong angle mode (degrees vs radians), which silently produces a plausible-looking but wrong answer.

Examiner report insight

• Choosing a longer, valid method (e.g. Pythagoras followed by sine or tangent, or the sine/cosine rule) when the triangle is right-angled and a single direct ratio would answer the question – the longer route is more prone to a rounding-related accuracy loss, not just slower.
• In a “show that” question, showing at least one more decimal place of working than the given answer – matching the given answer with no extra working shown does not prove it, even if the method was genuinely used.
• In 3D problems worked in multiple stages, keeping an intermediate length in exact (surd) form rather than rounding it before the final trigonometric step – early rounding compounds into a final answer that misses the required accuracy.

Source: Cambridge International, 0580 Mathematics Principal Examiner Report, June 2024 series, Papers 12, 21, 22, 23 (verified 2026-09-02).

Approaching trigonometry questions

Before choosing a method, identify what the triangle actually gives you: a right angle points to SOHCAHTOA or Pythagoras; a matching angle-side pair points to the sine rule; two sides with the included angle, or all three sides, points to the cosine rule – rearranged for an angle when the unknown is an angle rather than a side, as in a “find angle Y given three sides” question. A negative value inside cos⁻¹ is not a sign of an error – it correctly signals an obtuse angle, and should be carried through rather than assumed to be a mistake.