OxfordAQA International GCSE Mathematics: Sine rule, cosine rule and area of a triangle (9260) – 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 OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), section 3.3.2, statement G20. Every question is Extension only. If you need the method first, read the study guide or the revision notes. Course pages: the OxfordAQA IGCSE Mathematics hub and the checklist. Want a quick check of your level? Try the free 10-minute diagnostics.

Give lengths and areas to 3 significant figures and angles to 1 decimal place unless a question says otherwise. Keep full calculator values until the final line.

Questions

1. (Extension only) In triangle PQR, angle P = 35°, angle Q = 82° and QR = 6.8 m. Calculate the length of PR. [3]

2. (Extension only) In triangle XYZ, XY = 9.6 cm, YZ = 7.1 cm and angle XZY = 70°. Calculate angle YXZ. [3]

3. (Extension only) In triangle DEF, DE = 6.4 cm, DF = 10.2 cm and angle EDF = 118°. Calculate the length of EF. [3]

4. (Extension only) A triangle has sides of length 5.8 cm, 7.3 cm and 9.1 cm. Calculate the size of its smallest angle. [3]

5. (Extension only) Two sides of a triangle are 14.2 cm and 9.5 cm. The angle between them is 73°. Calculate the area of the triangle. [2]

6. (Extension only) A triangle has area 36 cm². Two of its sides are 9 cm and 11 cm, and the angle between them is obtuse. Calculate the size of this angle. [3]

7. (Extension only) A triangle has area 58 cm². One side is 12.5 cm and the angle between this side and a second side is 130°. Calculate the length of the second side. [3]

8. (Extension only) In triangle ABC, AB = 7.6 cm, BC = 12.9 cm and angle ABC = 35°.

(a) Calculate the length of AC. [3] (b) Calculate angle ACB. [3]

9. (Extension only) A boat leaves port P and sails 21 km on a bearing of 062° to buoy Q. It then sails 16 km on a bearing of 143° to buoy R.

(a) Show that angle PQR = 99°. [2] (b) Calculate the distance PR. [3] (c) Calculate the bearing of R from P. [3]

10. (Extension only) In quadrilateral ABCD, AB = 6.5 cm, BC = 8.2 cm, CD = 7.9 cm and AD = 9.4 cm. Angle ABC = 104°.

(a) Calculate the length of AC. [3] (b) Calculate angle ADC. [3] (c) Calculate the area of the quadrilateral. [3]

11. (Extension only) In triangle ABC, angle A = 30°, a = 7 cm and b = 12 cm. There are two possible triangles.

(a) Calculate the two possible sizes of angle B. [3] (b) Calculate the length of side c in each triangle. [3]

Answers

1. PR is opposite angle Q, and QR is opposite angle P. PR / sin 82° = 6.8 / sin 35° [1] PR = 6.8 × sin 82° / sin 35° [1] = 11.74… = 11.7 m [1] [3] Examiner insight: write the matching pair first (QR with angle P). A common slip is linking PR to the wrong angle. No third angle is needed here, so there is no need to find it.

2. YZ is opposite angle X, and XY is opposite angle Z. sin X / 7.1 = sin 70° / 9.6 [1] sin X = 7.1 × sin 70° / 9.6 = 0.695… [1] X = 44.0° [1] [3] Examiner insight: the given angle faces the longer side (9.6), so only the acute value works and a single answer is right. A short reason shows your thinking, not just a calculator result.

3. EF² = 6.4² + 10.2² − 2 × 6.4 × 10.2 × cos 118° [1] EF² = 145 + 61.29… = 206.29… [1] EF = 14.4 cm [1] [3] Examiner insight: the cosine of 118° is negative, so the term being subtracted is negative and EF² ends up larger than 145. A result below 12.0 cm would point to a sign slip.

4. The smallest angle is opposite the shortest side (5.8 cm). cos θ = (7.3² + 9.1² − 5.8²) / (2 × 7.3 × 9.1) [1] cos θ = 102.46 / 132.86 = 0.7712… [1] θ = 39.5° [1] [3] Examiner insight: identify the side opposite the angle before substituting, because it is the one subtracted in the numerator. Because all three sides are given, the cosine rule avoids any acute or obtuse doubt.

5. Area = ½ × 14.2 × 9.5 × sin 73° [1] = 64.50… = 64.5 cm² [1] [2] Examiner insight: the angle is between the two given sides, so ½ab sin C applies directly. Put the units as cm², not cm.

6. 36 = ½ × 9 × 11 × sin θ [1] sin θ = 72 / 99 = 0.7272… [1] The acute value is 46.7°, so the obtuse angle is 180° − 46.7° = 133.3° [1] [3] Examiner insight: the word “obtuse” tells you to use 180° minus the calculator value. Giving 46.7° ignores the condition in the question.

7. 58 = ½ × 12.5 × x × sin 130° [1] x = 116 / (12.5 × sin 130°) [1] = 12.11… = 12.1 cm [1] [3] Examiner insight: collect the known numbers into one product before dividing. Make sure you divide by sin 130° and by 12.5, not just by one of them.

8. (a) AC² = 7.6² + 12.9² − 2 × 7.6 × 12.9 × cos 35° [1] AC² = 224.17 − 160.61… = 63.55… [1] AC = 7.971… = 7.97 cm [1] (b) sin C / 7.6 = sin 35° / AC [1] sin C = 7.6 × sin 35° / 7.971… = 0.5468… [1] C = 33.1° [1] [6] Examiner insight: use the unrounded AC in part (b). Angle C is opposite AB, the shortest side, so it must be acute, which is why the sine rule is safe here. A check with the cosine rule gives the same angle.

9. (a) The bearing of Q from P is 062°, so the bearing of P from Q is 062° + 180° = 242° [1]. Angle PQR = 242° − 143° = 99° [1] (b) PR² = 21² + 16² − 2 × 21 × 16 × cos 99° [1] PR² = 697 + 105.12… = 802.12… [1] PR = 28.32… = 28.3 km [1] (c) sin P / 16 = sin 99° / 28.32… [1] sin P = 0.5580…, so angle P = 33.9° [1] Bearing of R from P = 062° + 33.9° = 095.9° [1] [8] Examiner insight: a sketch with north lines at P and Q makes the angle in part (a) visible. Bearings are written with three figures. The sine rule is safe in part (c) because angle Q is already obtuse, so angle P must be acute.

10. (a) AC² = 6.5² + 8.2² − 2 × 6.5 × 8.2 × cos 104° [1] AC² = 109.49 + 25.78… = 135.27… [1] AC = 11.63… = 11.6 cm [1] (b) cos D = (7.9² + 9.4² − AC²) / (2 × 7.9 × 9.4) [1] cos D = (150.77 − 135.27…) / 148.52 = 0.1043… [1] D = 84.0° [1] (c) Area ABC = ½ × 6.5 × 8.2 × sin 104° = 25.86… [1] Area ACD = ½ × 7.9 × 9.4 × sin 84.01… = 36.92… [1] Total area = 62.8 cm² [1] [9] Examiner insight: the diagonal is shared by both triangles, so keep its unrounded value. Name each triangle you use in part (c). The total is the sum of the two areas, not their product or their mean.

11. (a) sin B / 12 = sin 30° / 7 [1] sin B = 12 × 0.5 / 7 = 0.8571… [1] B = 59.0° or B = 180° − 59.0° = 121.0° [1] (b) First triangle: C = 180° − 30° − 59.0° = 91.0° [1] c = 7 × sin 91.0° / sin 30° = 14.0 cm [1] Second triangle: C = 180° − 30° − 121.0° = 29.0°, c = 7 × sin 29.0° / sin 30° = 6.79 cm [1] [6] Examiner insight: the question says two triangles exist, so look for the obtuse option. Both are valid because 30° + 121.0° is less than 180°. Work out c separately for each triangle.

Where marks are usually lost

• Linking a side to the wrong angle in the sine rule, so the answer is far too big or too small.
• Using the cosine rule for a side with an angle that is not between the two known sides.
• Entering the cosine rule with the wrong order of operations, such as (b² + c² − 2bc) × cos A.
• Forgetting the square root after finding a² and quoting a squared length.
• Ignoring “obtuse” in a question and quoting the acute value from the calculator.
• Missing the second angle from a sine-rule problem where both fit.
• Rounding AC or another shared length early, so later answers drift.
• Writing a bearing with fewer than three figures, such as 95° for 095°, or measuring it from the wrong north line.
• Forgetting units, or giving an area in cm.

Next steps

• Re-read the revision notes for the formulas and method boxes.
• Go back to the study guide for any step that felt unfamiliar.
• Visit the OxfordAQA IGCSE Mathematics hub and tick off statements on the checklist.
• Take the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class if you would like a teacher to go through these questions with you.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA. Statement G20, section 3.3.2 Mensuration and calculation.