IGCSE Mathematics 0580 · Topic 6.4

IGCSE Mathematics: Sine and Cosine Rules Practice Questions

Use the sine rule when you have a matching pair of a side and its opposite angle. Use the cosine rule when you have three sides, or two sides with the angle between them.

Cambridge IGCSE Mathematics (0580) · Topic 6.4: Sine and Cosine Rules

Topic 6.4 of Cambridge IGCSE Mathematics 0580 handles triangles with no right angle. Choosing the wrong rule wastes several minutes, so the selection test is worth learning before anything else. The questions below cover both rules and the area formula.

What you need to know for Sine and Cosine Rules

IGCSE Mathematics Sine and Cosine Rules questions and answers

4 exam-style questions written to the 0580 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

In triangle ABC, angle A is 40 degrees, angle B is 75 degrees and side b is 12 cm. Calculate the length of side a, correct to 3 significant figures.

Show the worked answer
Answer: 7.99 cm
  1. Side b and angle B form a matching pair, and side a is opposite angle A, so use the sine rule.
  2. a divided by sin A = b divided by sin B, so a = b sin A divided by sin B.
  3. a = 12 x sin 40 divided by sin 75 = 12 x 0.6428 divided by 0.9659.
  4. a = 7.7136 divided by 0.9659 = 7.986, which is 7.99 cm to 3 significant figures.
How the marks are awarded. 1 mark for selecting and stating the sine rule. 1 mark for correct substitution. 1 mark for 7.99 cm with the unit.
Where students lose the mark. Pairing a side with the wrong angle. Side a must be paired with angle A, not with whichever angle appears next to it on the diagram.
Question 2[4 marks]

In triangle ABC, b = 8 cm, c = 11 cm and the angle A between them is 52 degrees. Calculate the length of side a, correct to 3 significant figures.

Show the worked answer
Answer: 8.76 cm
  1. Two sides and the angle between them are known, and there is no matching side and opposite angle pair, so use the cosine rule.
  2. a2 = b2 + c2 minus 2bc cos A = 64 + 121 minus 2 x 8 x 11 x cos 52.
  3. 2 x 8 x 11 = 176, and cos 52 = 0.61566, so the last term is 108.36.
  4. a2 = 185 minus 108.36 = 76.64, so a = the square root of 76.64 = 8.755, which is 8.76 cm.
How the marks are awarded. 1 mark for selecting the cosine rule. 1 mark for correct substitution. 1 mark for a squared equal to 76.6. 1 mark for 8.76 cm.
Where students lose the mark. Forgetting to take the square root at the end and giving 76.6 cm. The formula gives a squared, not a.
Question 3[4 marks]

A triangle has sides of 7 cm, 9 cm and 12 cm. Calculate the largest angle, correct to 1 decimal place.

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Answer: 96.4 degrees
  1. The largest angle is opposite the longest side, so it is opposite the 12 cm side. Call that angle C, with a = 7 and b = 9.
  2. Use the cosine rule rearranged for an angle: cos C = (a2 + b2 minus c2) divided by 2ab.
  3. cos C = (49 + 81 minus 144) divided by (2 x 7 x 9) = minus 14 divided by 126 = minus 0.11111.
  4. C = the inverse cosine of minus 0.11111 = 96.38 degrees, which is 96.4 degrees to 1 decimal place. The negative cosine correctly indicates an obtuse angle.
How the marks are awarded. 1 mark for identifying the angle opposite the longest side. 1 mark for correct substitution into the rearranged cosine rule. 1 mark for cos C = minus 0.111. 1 mark for 96.4 degrees.
Where students lose the mark. Discarding the negative sign on the cosine. A negative cosine means the angle is obtuse, which is exactly what is expected for the largest angle here.
Question 4[3 marks]

A triangle has sides of 9 cm and 14 cm with an angle of 35 degrees between them. Calculate its area, correct to 3 significant figures.

Show the worked answer
Answer: 36.1 cm2
  1. Two sides and the included angle are known, so use area = half ab sin C.
  2. Area = half x 9 x 14 x sin 35.
  3. half x 9 x 14 = 63, and sin 35 = 0.57358.
  4. Area = 63 x 0.57358 = 36.13, which is 36.1 cm2 to 3 significant figures.
How the marks are awarded. 1 mark for using area = half ab sin C. 1 mark for correct substitution. 1 mark for 36.1 cm2 with the unit.
Where students lose the mark. Using half base times height with 9 and 14. That formula needs a perpendicular height, and 14 is a slanted side, not the height.

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Sine and Cosine Rules FAQs

When do I use the sine rule and when the cosine rule?

Use the sine rule whenever you have a side together with the angle opposite it, forming a matching pair. Use the cosine rule when you have three sides, or two sides with the angle between them, because in those cases no matching pair exists.

What is the cosine rule?

For a side, a squared equals b squared plus c squared minus 2bc cos A, where A is the angle between the known sides b and c. For an angle, cos A equals b squared plus c squared minus a squared, all divided by 2bc, where a is the side opposite the angle.

What does a negative cosine mean?

It means the angle is obtuse, between 90 and 180 degrees. This happens legitimately when finding the largest angle of a triangle, so the negative value should be kept and the inverse cosine taken as normal rather than the sign being dropped.

How do I find the area of a non-right-angled triangle?

Use half ab sin C, where a and b are two sides and C is the angle between them. The standard half base times height formula needs a perpendicular height, which is usually not given for a triangle with no right angle.

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Written to the published Cambridge IGCSE Mathematics (0580) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.