GCSE Maths revision · Higher: geometry

The sine rule

Higher only

The idea

In any triangle, asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, with each side opposite its angle. Use it when you know a side and the angle opposite it.

  1. Label sides with small letters opposite their capital angles.
  2. For a side, use asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}; for an angle, flip it: sin⁡Aa=sin⁡Bb\frac{\sin A}{a} = \frac{\sin B}{b}.
  3. When finding an angle, 180∘180^\circ minus your answer may also work.

Watch out An obtuse angle can share a sine with an acute one: check both.

See it worked

Worked exampleCalculator allowed2 marks

In triangle ABCABC, angle A=40∘A = 40^\circ, angle B=65∘B = 65^\circ and BC=8BC = 8 cm.

Work out the length of ACAC. Give your answer correct to 3 significant figures.

  1. BCBC is opposite angle AA and ACAC is opposite angle BB.
  2. Sine rule. ACsin⁡65∘=8sin⁡40∘\frac{AC}{\sin 65^\circ} = \frac{8}{\sin 40^\circ}
  3.  AC=8sin⁡65∘sin⁡40∘=11.279…AC = \frac{8 \sin 65^\circ}{\sin 40^\circ} = 11.279\ldots

Answer 11.3 cm

Your turn

Question 1Calculator allowed2 marks

In triangle ABC, angle A is 34∘34^\circ, angle B is 79∘79^\circ and BC = 8 cm. Work out AC. Give your answer to 3 significant figures.

Question 2Calculator allowed4 marks

In triangle ABC, angle A is 34∘34^\circ, angle B is 71∘71^\circ and side BC is 9 cm.

(a) Find AC to 3 significant figures.

(b) Find angle C.

Question 3Calculator allowed3 marks

In triangle ABC, AB = 12 cm, AC = 9 cm and angle ABC = 35∘.35^\circ. There are two possible triangles. Work out the larger possible area. Give your answer to 3 significant figures. You must show your working.

In past papers

35 official questions in 35 papers since 2017. The latest:

All GCSE Maths past papers

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