OCR GCSE Mathematics: Pythagoras, right-angled trigonometry and exact trigonometric values (J560) – 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 from your board or school.

These questions follow OCR GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), statements 10.05a, 10.05b and 10.05c. Questions 1 to 8 are in the Foundation tier and Higher tier content. Questions 9 to 11 are Higher tier only. Read the study guide first and keep the revision notes to hand. See also the OCR GCSE Mathematics hub, the printable checklist and the free 10-minute diagnostics. Unless told otherwise, round lengths to 3 significant figures and angles to 1 decimal place. Diagrams are not drawn: sketch each one yourself.

Questions

1. A rectangular field measures 45 m by 28 m.

(a) Find the length of the diagonal path across the field. [2] (b) How much shorter is the diagonal path than walking along two sides? [1]

2. A right-angled triangle has hypotenuse 14.2 cm and one short side 8.7 cm. Find the length of the other short side. [3]

3. A triangle has sides 9.6 cm, 7.2 cm and 12 cm.

(a) Show that the triangle is right-angled. [2] (b) Find its area, and the perpendicular distance from the right angle to the longest side. [2]

4. In triangle PQR, angle Q = 90°, angle P = 41° and PR = 15.6 cm.

(a) Find QR. [2] (b) Find PQ. [2]

5. A ramp rises 0.9 m over a horizontal distance of 4.8 m.

(a) Find the angle the ramp makes with the horizontal. [2] (b) Find the length of the ramp. [2]

6. A kite string is pulled straight at 36° to the horizontal. The kite is 28 m above the hand holding the string. Find the length of the string. [3]

7.

(a) Write down the exact values of sin 30°, cos 45° and tan 60°. [3] (b) Show that (tan 60°)² + (cos 30°)² = 15/4. [3]

8. Triangle ABC has a right angle at B, angle A = 60° and AB = 2√3 cm. Give answers in exact form.

(a) Find BC. [2] (b) Find AC. [1] (c) Write the perimeter in the form a + b√3. [1]

9. (Higher tier only) In triangle ABC, D is a point on AC and BD is perpendicular to AC. AB = 9.4 cm, angle BAD = 62° and DC = 6.5 cm.

(a) Find BD. [2] (b) Find angle BCD. [2] (c) Find BC. [2]

10. (Higher tier only) A cuboid ABCDEFGH has a rectangular base ABCD with AB = 9 cm and BC = 4 cm. The height AE is 6 cm, and G is the vertex opposite A.

(a) Find the length of AC. [2] (b) Find the length of AG. [2] (c) Find the angle between AG and the base ABCD. [2]

11. (Higher tier only) A pyramid has a square base of side 8 cm. Its apex is 10 cm vertically above the centre of the base.

(a) Find the length of a sloping edge. [3] (b) Find the angle between a sloping edge and the base. [2] (c) Find the angle between a sloping face and the base. [2]

Answers

1. (a) d² = 45² + 28² = 2025 + 784 = 2809 [1], so d = 53 m [1] (b) 45 + 28 = 73, and 73 − 53 = 20 m [1] Examiner insight: A perfect square under the root is a sign you have added correctly. Write the units with the final answer.

2. a² = 14.2² − 8.7² [1] = 201.64 − 75.69 = 125.95 [1], so a = 11.2 cm [1] Examiner insight: The hypotenuse is the largest square. Subtracting in the other order gives a negative number, a clear signal to swap.

3. (a) 9.6² + 7.2² = 92.16 + 51.84 = 144 [1]; 12² = 144, so the squares match and the triangle is right-angled [1] (b) Area = ½ × 9.6 × 7.2 = 34.56 cm² [1]. Using 12 as the base: ½ × 12 × h = 34.56, so h = 5.76 cm [1] Examiner insight: A show-that answer reads best when both sides are evaluated and a conclusion is written in words.

4. (a) QR is opposite P, and PR is the hypotenuse, so sin 41° = QR/15.6 [1]; QR = 15.6 × sin 41° = 10.2 cm [1] (b) PQ is adjacent to P, so cos 41° = PQ/15.6 [1]; PQ = 15.6 × cos 41° = 11.8 cm [1] Examiner insight: Naming the sides first (opposite, adjacent, hypotenuse) is the quickest way to avoid picking the wrong ratio.

5. (a) tan θ = 0.9/4.8 [1]; θ = tan⁻¹(0.1875) = 10.6° [1] (b) Length² = 0.9² + 4.8² = 0.81 + 23.04 = 23.85 [1]; length = 4.88 m [1] Examiner insight: Part (b) can also be done with sine, but using the given lengths avoids relying on a rounded angle.

6. The height is opposite the 36° angle and the string is the hypotenuse: sin 36° = 28/L [1]; L = 28 / sin 36° [1] = 47.6 m [1] Examiner insight: When the unknown is the denominator, dividing is needed. Check that the answer is longer than 28 m, because a hypotenuse must exceed a short side.

7. (a) sin 30° = 1/2 [1], cos 45° = √2/2 [1] (also written 1/√2), tan 60° = √3 [1] (b) tan² 60° = (√3)² = 3 [1]; cos² 30° = (√3/2)² = 3/4 [1]; 3 + 3/4 = 15/4 [1] Examiner insight: Exact-value questions ask for surds or fractions, so a rounded decimal answers a different question.

8. (a) tan 60° = BC/AB, so BC = 2√3 × √3 [1] = 6 cm [1] (b) cos 60° = AB/AC, so AC = 2√3 ÷ 1/2 = 4√3 cm [1] (c) Perimeter = 2√3 + 6 + 4√3 = 6 + 6√3 cm [1] Examiner insight: Combine surds only when they are like terms. Here 2√3 + 4√3 = 6√3, but the 6 stays separate.

9. (a) In triangle ABD, sin 62° = BD/9.4 [1]; BD = 9.4 × sin 62° = 8.30 cm [1] (b) In triangle BDC, tan C = BD/DC = 8.2997/6.5 [1]; C = 51.9° [1] (c) BC² = 8.2997² + 6.5² [1]; BC = 10.5 cm [1] Examiner insight: Part (c) is the check on part (b): sin 51.9° × 10.5 should return about 8.3. Keep the unrounded BD when you use it again.

10. (a) AC² = 9² + 4² = 97 [1]; AC = √97 = 9.85 cm [1] (b) AG² = AC² + CG² = 97 + 36 = 133 [1]; AG = √133 = 11.5 cm [1] (c) The shadow of AG on the base is AC, so tan θ = 6/√97 [1]; θ = 31.4° [1] Examiner insight: Using the exact √97 rather than 9.85 gives a more reliable angle. Mark the right angle at C in triangle ACG before you start.

11. (a) Half the base diagonal = ½ × √(8² + 8²) = 4√2 [1]; edge² = (4√2)² + 10² = 32 + 100 = 132 [1]; edge = √132 = 11.5 cm [1] (b) tan θ = 10/(4√2) [1]; θ = 60.5° [1] (c) The base of the right-angled triangle runs from the centre to the middle of a base edge, which is 4 cm: tan φ = 10/4 [1]; φ = 68.2° [1] Examiner insight: Parts (b) and (c) use different horizontal distances. Say which point on the base you are measuring to, since a clear diagram makes that visible.

Where marks are usually lost

• Adding the squares of the sides when a short side is wanted.
• Leaving the answer as a square without taking the square root.
• Labelling opposite and adjacent from the right angle instead of from the angle in the question.
• Multiplying when the unknown is the denominator of the ratio.
• Writing a decimal where an exact answer was requested.
• Rounding part-way through, then using the rounded value in the next step.
• Missing units on lengths and areas.
• In 3D, picking the half-diagonal of the base when the distance to the middle of an edge is needed, or the other way round.

Next steps

• Re-read the revision notes and the study guide.
• Find your weak topics with the free 10-minute diagnostics.
• Tick off the unit on the printable checklist and return to the course hub.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), first assessment in 2017, published by OCR. Statements 10.05a, 10.05b and 10.05c.