Geometry · GCSE Maths
The sine rule
GCSE Maths sine rule: use a / sin A = b / sin B when you have a side and its opposite angle, including the ambiguous case on Higher if two triangles are possible.
Pair each side with the angle opposite it. a / sin A = b / sin B = c / sin C. Use this when you do not have a contained pair of sides sandwiching an angle (that would be cosine rule).
The important bits
What you need to know
- 1
The sine rule is a / sin A = b / sin B = c / sin C, where side a is opposite angle A, and so on. Label the triangle with this convention before you substitute.
- 2
Use the sine rule when you know a side and the angle opposite it, plus one more side or angle. ASA, AAS, and some SSA cases are sine-rule territory.
- 3
To find a side: a / sin A = b / sin B, so a = sin A × (b / sin B). To find an angle: sin A = a × (sin B / b), then A = sin⁻¹(…).
- 4
You only need two of the three fractions. Choose the pair that contains the unknown and a complete known pair (a side and its opposite angle).
- 5
Angles in a triangle add to 180°. After finding one angle with the sine rule, the third angle is often 180° minus the two you know — cheaper than a second sine-rule calculation.
- 6
Higher, ambiguous case (SSA): if you know two sides and a non-included acute angle, sin⁻¹ can give an acute angle and 180° minus that angle. Check which of those, if either, fits with the other given angle.
- 7
Do not use the sine rule as a first resort in a right-angled triangle when SOHCAHTOA already has a right angle and two pieces. You can, but it is slower and easier to mis-label.
- 8
Keep the pairing honest: side a with angle A. Swapping so that a sits opposite B is how a numerically tidy answer fails the diagram.
Quotations worth analysing
Short evidence. Real method.
“a / sin A = b / sin B”
Write this with your letters filled in before rearranging. The method mark is on the pairing of each side with the angle opposite it.
“The ambiguous case”
sin⁻¹(0.5) = 30° or 150°. Both can be angles in a triangle. You must test 180° − θ against the given angle before discarding one.
“Side a is opposite angle A”
If you label the side next to A as a, every subsequent line is the sine rule for a different triangle. Annotate the diagram first.
Go deeper
Know a pair, then scale
Suppose angle A = 40°, side a = 7 cm, angle B = 60°. Then 7 / sin 40° = b / sin 60°, so b = 7 × sin 60° / sin 40°. That is one line of rearrangement: multiply both sides by sin 60°. You need a complete pair (here a and A) to scale to the unknown pair (b and B). If instead you know a, A and b, you can find B: sin B = b × sin A / a, then B = sin⁻¹ of that. The calculator will offer an acute angle. On Foundation that is usually the intended angle. On Higher, if B is clearly obtuse, take 180° minus the calculator value. Keep four decimal places on the sines if you write them, then round as asked. The third angle is cleaner from 180° − A − B than from a third sine-rule fraction.
Go deeper
The ambiguous case, Higher
You are given acute angle A, side a, and side b (SSA). Compute sin B = b sin A / a. If that value is between 0 and 1, there is at least one angle B. The calculator’s acute B₁ may work. The obtuse candidate is B₂ = 180° − B₁, possible only if A + B₂ < 180° so a third angle remains. Sketch both. A diagram that looks obtuse wants B₂; a question that says “angle B is acute” wants B₁. If sin B > 1, no triangle. If sin B = 1, a right angle, one triangle. Write “two possible angles, 30° and 150°; 150° + 40° = 190° > 180°, so discard 150°” when that is the case — the discard sentence scores.
Go deeper
Sine rule versus cosine rule at the start
Start by listing what you have: three sides (SSS) or two sides and the included angle (SAS) means cosine rule first. Two angles and a side, or two sides and a non-included angle, means sine rule. Right-angled with two sides or a side and an acute angle: SOHCAHTOA. Using the sine rule on SAS is possible only after you have found another angle, which is the long way round. Using the cosine rule when you already have a side and its opposite angle is possible but heavier. The decision is a ten-second look at the given letters, not a personality preference. If a later part asks for the area, ½ab sin C needs two sides and the included angle — you may need the sine or cosine rule first to produce that included angle.
See the idea in action
In triangle ABC, angle A = 40°, angle B = 73°, side a = 8.0 cm. Find side b to 3 s.f. Step 1: Side a opposite A, side b opposite B. We have a pair (a, A) and we want (b, B). Step 2: a / sin A = b / sin B, so 8.0 / sin 40° = b / sin 73°. Step 3: b = 8.0 × sin 73° / sin 40°. Step 4: sin 73° = 0.9563… , sin 40° = 0.6428… , so b = 8.0 × 1.487… = 11.9 cm to 3 s.f. Step 5: Angle C = 180° − 40° − 73° = 67°, which can check via 8.0 / sin 40° = c / sin 67° if wanted. Size check: angle B > angle A, so side b > side a, and 11.9 > 8.0.
Exam technique
Turn knowledge into marks
Write a / sin A = b / sin B with numbers under the letters before you rearrange. If two angles are possible, write both and discard the one that makes the angle sum exceed 180°.
Common mistakes
Do not give these marks away
- 01
Pairing a side with an angle that is not opposite it, so the sine rule is applied to the wrong letters.
- 02
Using the cosine rule when a side and its opposite angle are already known, or SOHCAHTOA on a triangle that is not right-angled.
- 03
Taking the calculator’s acute sin⁻¹ as the only possible angle in an SSA question, without testing 180° minus that angle.
In triangle ABC, A = 40°, a = 7 cm, B = 60°. Side b is
A7 × sin 60° / sin 40°
B7 × sin 40° / sin 60°
C7 × 60 / 40
D7 / sin 60°
Show the answer
7 × sin 60° / sin 40°. b / sin 60° = 7 / sin 40°, so b = 7 sin 60° / sin 40°. Multiplying by sin 40° / sin 60° swaps the pair. 7 × 60 / 40 treats the sine rule as if it were lengths over angles without sine.
Quick questions
If this is the bit you searched
When do I use the sine rule?
When you know a side and the angle opposite it, plus one more side or angle, in a triangle that need not be right-angled. ASA, AAS and some SSA cases.
What is the ambiguous case?
SSA: two sides and a non-included acute angle. sin⁻¹ can give two angles, θ and 180° − θ. Keep a candidate only if the three angles would still sum to 180°.
Do I need all three fractions of the sine rule?
No. Use the two that contain your unknown and a known opposite pair. The third angle is often easier from 180° minus the other two.
Can I use the sine rule in a right-angled triangle?
Yes, because it is true in every triangle, but SOHCAHTOA is usually shorter if a right angle is given. Do not force SOHCAHTOA onto a non-right-angled triangle.