IGCSE Mathematics: Trigonometry (Extended) — Practice Questions (Cambridge 0580)

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 — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Each question practises a skill from the Extended (Supplement) content of the syllabus. After each answer there is a common mistake to watch out for.

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Questions

1. (Extended) In triangle PQR, PQ = 8.5 cm, PR = 11.2 cm and angle QPR = 47°. Calculate the length of QR. [3]

2. (Extended) In triangle ABC, AB = 9 cm, BC = 14 cm and angle BAC = 72°. Calculate angle ACB. [3]

3. (Extended) A cuboid is 12 cm long, 5 cm wide and 4 cm high. A straight line joins a corner of the base to the corner of the top face that is furthest away from it, passing through the inside of the cuboid.

(a) Calculate the length of this line.

(b) Calculate the angle between this line and the base of the cuboid. [4]

4. (Extended) Solve the equation 4 cos x + 3 = 0 for 0° ⩽ x ⩽ 360°. [3]

5. (Extended) Without using a calculator, find the exact value of sin 60° × tan 30° + (cos 45°)². Show your working. [2]

6. (Extended) A circle has centre O and radius 10 cm. AB is a chord of the circle of length 16 cm.

(a) Calculate the perpendicular distance from O to the chord AB.

(b) Calculate angle AOB. [4]

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Answers

1. (Extended) Cosine rule: QR² = 8.5² + 11.2² − 2 × 8.5 × 11.2 × cos 47° [1]. QR² = 72.25 + 125.44 − 129.85… = 67.84… [1]. QR = 8.24 cm (3 s.f.) [1].

Common mistake: Working out 72.25 + 125.44 − 190.4 first and then multiplying by cos 47°. The whole term 2 × 8.5 × 11.2 × cos 47° must be calculated before subtracting.

2. (Extended) Sine rule: sin C / 9 = sin 72° / 14 [1]. sin C = 9 × sin 72° ÷ 14 = 0.6114… [1]. Angle ACB = 37.7° (1 d.p.) [1]. (The obtuse value 142.3° is impossible, because 142.3° + 72° is more than 180°.)

Common mistake: Pairing each side with the wrong angle. AB is opposite angle C and BC is opposite angle A, so match each side with the angle across from it.

3. (Extended) (a) Diagonal of base = √(12² + 5²) = 13 cm [1]. Line = √(13² + 4²) = √185 = 13.6 cm (3 s.f.) [1].

(b) The angle is between the line and the base diagonal: tan θ = 4 ÷ 13 [1]. θ = 17.1° (1 d.p.) [1].

Common mistake: Using an edge of the base (12 cm or 5 cm) instead of the base diagonal. The angle with a plane is measured to the line’s projection on that plane.

4. (Extended) cos x = −3/4 = −0.75 [1]. Principal value x = 138.6° [1]. By the symmetry of the cosine graph, the other solution is 360° − 138.6° = 221.4°, so x = 138.6° or 221.4° [1].

Common mistake: Giving only the calculator value. Sketch y = cos x for 0° to 360° to find every solution in the range.

5. (Extended) sin 60° = √3/2, tan 30° = 1/√3 and cos 45° = 1/√2 [1]. (√3/2) × (1/√3) + (1/√2)² = 1/2 + 1/2 = 1 [1].

Common mistake: Mixing up sin 60° and sin 30°. Sketching the half-equilateral triangle (sides 1, √3, 2) and the isosceles right-angled triangle (sides 1, 1, √2) helps you recall the values.

6. (Extended) (a) The perpendicular from O bisects AB, so each half is 8 cm [1]. Distance = √(10² − 8²) = 6 cm [1].

(b) Half of angle AOB: sin θ = 8 ÷ 10, so θ = 53.13° [1]. Angle AOB = 2 × 53.13° = 106.3° (1 d.p.) [1].

Common mistake: Using the full chord length of 16 cm in the right-angled triangle. The perpendicular from the centre splits the chord into two equal halves of 8 cm.

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Where marks are usually lost

• Rounding intermediate values too early in cosine-rule and sine-rule calculations.
• Using the wrong side–angle pairs in the sine rule.
• Choosing the wrong right-angled triangle in three-dimensional problems.
• Giving only one solution to a trigonometric equation.
• Forgetting that the perpendicular from the centre halves a chord.