OCR GCSE Mathematics: Sine rule and cosine rule (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 cover sections 10.05d (sine rule) and 10.05e (cosine rule) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), plus the area formula ½ab sin C from 10.03a. All of this content is Higher tier only, so every question below is Higher tier only. Give lengths and areas to 3 significant figures and angles to 1 decimal place unless a question says otherwise. Diagrams are not drawn to scale: sketch your own.

Learn the methods first in the study guide and the revision notes. The OCR GCSE Mathematics hub and the printable checklist show where this unit sits.

Questions

1. (Higher tier only) In triangle PQR, angle P = 52°, angle Q = 67° and the side QR = p = 14.6 cm. Calculate the length of PR = q. [3]

2. (Higher tier only) In triangle ABC, a = 18.4 cm, b = 15.1 cm and angle A = 79°. Calculate angle B. [3]

3. (Higher tier only) Two sides of a triangle have lengths 9.8 cm and 13.5 cm. The angle between them is 57°. Calculate the length of the third side. [3]

4. (Higher tier only) A triangle has sides of length 7.2 cm, 10.4 cm and 15.1 cm. Calculate the size of its largest angle. [4]

5. (Higher tier only) For each situation, write down whether you would start with the sine rule or the cosine rule.

(a) You know two angles and the side opposite one of them, and you want the side opposite the other. [1] (b) You know all three sides and you want an angle. [1] (c) You know two sides and the angle between them, and you want the third side. [1] (d) You know two sides and an angle opposite one of those sides, and you want the angle opposite the other side. [1]

6. (Higher tier only) Two sides of a triangle are 8.6 cm and 11.3 cm long. The angle between them is 124°. Calculate the area of the triangle. [3]

7. (Higher tier only) In triangle ABC, angle A = 35°, a = 7.5 cm and b = 11 cm. Find both possible values of angle B, and show that each one is possible in a triangle. [5]

8. (Higher tier only) ABCD is a quadrilateral. The diagonal AC splits it into triangles ABC and ACD. In triangle ABC, AB = 6.4 cm, BC = 8.1 cm and angle ABC = 105°. In triangle ACD, angle CAD = 58° and angle ADC = 71°. Calculate

(a) the length of AC, [3] (b) the length of CD. [3]

9. (Higher tier only) A boat leaves harbour H and sails 15 km on a bearing of 038° to a buoy P. It then sails 11 km on a bearing of 125° to a lighthouse Q.

(a) Show that angle HPQ = 93°. [2] (b) Calculate the distance HQ. [3] (c) Calculate the bearing of Q from H. [3]

10. (Higher tier only) A triangular plot of land has two sides of length 14 m and 9 m. Its area is 50 m². The angle between the two given sides is acute.

(a) Calculate the size of the angle between the two given sides. [3] (b) Calculate the length of the third side of the plot. [3]

11. (Higher tier only)

(a) Show that the cosine rule a² = b² + c² - 2bc cos A becomes Pythagoras’ theorem when A = 90°. [2] (b) In a triangle, two sides have lengths 5 cm and 8 cm and the angle between them is 60°. Show that the third side is exactly 7 cm. [3]

Answers

1. Use the sine rule with the side you want on top: q / sin 67° = 14.6 / sin 52° [1]. So q = 14.6 × sin 67° / sin 52° [1]. This gives 17.05…, so q = 17.1 cm [1]. [3] Examiner insight: Pair each side with the angle opposite it before you write anything. The side QR faces angle P, and the side PR faces angle Q, so the pairing here is p with 52° and q with 67°.

2. sin B / 15.1 = sin 79° / 18.4 [1]. sin B = 15.1 × sin 79° / 18.4 = 0.8055… [1]. B = sin⁻¹(0.8055…) = 53.66…, so B = 53.7° [1]. [3] Examiner insight: The given angle faces the longer side, so the obtuse alternative cannot fit and one answer is enough. Saying why you ignore 126.3° shows reasoning that a bare answer does not.

3. x² = 9.8² + 13.5² - 2 × 9.8 × 13.5 × cos 57° [1]. x² = 278.29 - 144.12… = 134.17… [1]. x = 11.58…, so x = 11.6 cm [1]. [3] Examiner insight: Write the substitution before the evaluation. If a slip happens in the arithmetic, a clear line of working shows your method is correct.

4. The largest angle faces the longest side, 15.1 cm, so call it C. cos C = (7.2² + 10.4² - 15.1²) / (2 × 7.2 × 10.4) [1]. This is (51.84 + 108.16 - 228.01) / 149.76 [1] = -68.01 / 149.76 = -0.4541… [1]. C = cos⁻¹(-0.4541…) = 117.0…, so C = 117.0° [1]. [4] Examiner insight: A negative cosine is the signal that the angle is obtuse. Checking that the angle you quote is over 90° catches sign errors in the numerator.

5. (a) Sine rule, because you have a matching side and angle pair. [1] (b) Cosine rule, rearranged for cos of the angle. [1] (c) Cosine rule, because the known angle is between the known sides and there is no pair. [1] (d) Sine rule, because the angle is opposite one of the known sides, which gives a pair. [1] [4] Examiner insight: Look for a side and the angle opposite it. When no such pair exists, only the cosine rule can start the problem.

6. Area = ½ab sin C [1]. Area = ½ × 8.6 × 11.3 × sin 124° [1]. This is 48.59 × 0.8290… = 40.28…, so 40.3 cm² [1]. [3] Examiner insight: The angle must lie between the two sides you multiply. Here that is given directly. Note that sin 124° is positive even though the angle is obtuse.

7. sin B / 11 = sin 35° / 7.5 [1]. sin B = 11 × sin 35° / 7.5 = 0.8412… [1]. B = 57.27…°, so B = 57.3° [1]. The other angle with this sine is 180° - 57.27…° = 122.7…, so B = 122.7° [1]. Check: 35° + 122.7° = 157.7°, which is less than 180°, so this triangle exists. 35° + 57.3° = 92.3°, which is also less than 180° [1]. [5] Examiner insight: A “show that each is possible” request needs a stated check, such as the angle sum for each case. Giving two values without any check only does half the task.

8. (a) AC² = 6.4² + 8.1² - 2 × 6.4 × 8.1 × cos 105° [1]. AC² = 106.57 + 26.83… = 133.40… [1]. AC = 11.55…, so AC = 11.6 cm [1]. [3] (b) Triangle ACD has the pair CD facing 58° and AC facing 71°. CD / sin 58° = AC / sin 71° [1]. CD = 11.55… × sin 58° / sin 71° [1]. This gives 10.35…, so CD = 10.4 cm [1]. [3] Examiner insight: Part (b) depends on part (a). Use your unrounded AC from part (a), not 11.6, to avoid drifting in the last significant figure. State which triangle each rule is applied to.

9. (a) The bearing of H from P is 038° + 180° = 218° [1]. Angle HPQ = 218° - 125° = 93° [1]. [2] (b) HQ² = 15² + 11² - 2 × 15 × 11 × cos 93° [1]. HQ² = 346 + 17.27… = 363.27… [1]. HQ = 19.05…, so HQ = 19.1 km [1]. [3] (c) Use the sine rule for angle PHQ: sin(PHQ) / 11 = sin 93° / 19.05… [1]. sin(PHQ) = 0.5763…, so angle PHQ = 35.19…° (acute, because PQ is shorter than HQ) [1]. The bearing of Q from H is 038° + 35.19…° = 73.19…°, so 073.2° [1]. [3] Examiner insight: A “show that” answer must display each step: the back bearing, then the subtraction. Give the final bearing as three figures, and say why the angle found by the sine rule is acute.

10. (a) ½ × 14 × 9 × sin C = 50 [1]. sin C = 50 / 63 = 0.7936… [1]. C = 52.5° (the acute angle, as stated) [1]. [3] (b) c² = 14² + 9² - 2 × 14 × 9 × cos 52.5…° [1]. c² = 277 - 153.31… = 123.69… [1]. c = 11.12…, so c = 11.1 m [1]. [3] Examiner insight: The word “acute” in the question is there to select one of two possible angles. Underline such words, and quote the angle you used in part (b) unrounded.

11. (a) When A = 90°, cos A = 0 [1]. The term -2bc cos A is then 0, so a² = b² + c², which is Pythagoras’ theorem [1]. [2] (b) Let the third side be a. a² = 5² + 8² - 2 × 5 × 8 × cos 60° [1]. Since cos 60° = ½, a² = 25 + 64 - 40 [1]. a² = 49, so a = 7 cm [1]. [3] Examiner insight: For “show that” with an exact result, write each line of the substitution and state the exact value of the cosine you use. Reaching 7 from a decimal approximation would not demonstrate the exact result.

Where marks are usually lost

• Using the wrong side with the sine of an angle, so the pair is not opposite.
• Starting with the sine rule when the known angle is between the two known sides.
• Multiplying (b² + c² - 2bc) by cos A instead of subtracting 2bc cos A as one term.
• Losing the negative sign of the cosine of an obtuse angle, so a subtraction should be an addition.
• Rounding the answer from one part before using it in the next.
• Finding only the acute angle from sin⁻¹ when the question asks for all possible angles.
• Using an angle in ½ab sin C that is not between the two sides.
• Mixing up the interior angle and the bearing in a bearings problem. Draw north lines and mark the angle.
• Leaving off units, or giving too many decimal places for the accuracy asked.

Next steps

• Re-read the study guide for any method you missed.
• Use the revision notes for a quick recap.
• Check the OCR GCSE Mathematics hub and the printable checklist.
• Try all free 10-minute diagnostics to find the next topic to work on.
• Book a free trial class if you want help from a teacher.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Sections 10.05d, 10.05e and 10.03a. All questions are Higher tier only.