Sine Rule (AA HL)
The sine rule links each side of a triangle to the sine of the angle directly opposite it, letting you find a missing side or angle in any triangle that isn't right-angled. This page covers when to reach for it, a worked example at each difficulty, and the mix-up that costs marks most often. It's part of the broader Triangles & Circular Functions topic.
21 questions on this sub-topic.
The sine rule
Covered under IB syllabus reference SL3.2, alongside the cosine rule and the area formula, excluding the ambiguous case. The sine rule itself is in the formula booklet, so the skill being tested is spotting when it applies.
Sine rule
\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\]
Set up the ratio using the angle-side pair you already know, then cross-multiply to solve for the unknown.
✓ In the formula bookletWhen it applies
You need a complete angle-side pair (an angle and the side opposite it) plus one more angle or side.
Two sides with the angle between them, or all three sides, have no opposite pair to start from - that calls for the cosine rule instead.
Need the cosine rule and area formula too? See Triangles & Circular Functions.
Worked examples
In triangle \(ABC\), \(A=40^\circ,\ B=75^\circ,\ a=8.\) Find \(b.\)
Worked solution
\(\dfrac{b}{\sin75^\circ} = \dfrac{8}{\sin40^\circ}.\) M1 A1
\(b = \dfrac{8\sin75^\circ}{\sin40^\circ} \approx 12.0.\) A1
In triangle \(ABC\), \(A=40^\circ,\ B=65^\circ,\ a=8.\) Find \(b.\)
Worked solution
\(\dfrac{b}{\sin65^\circ} = \dfrac{8}{\sin40^\circ}.\) M1 A1
\(b = \dfrac{8\sin65^\circ}{\sin40^\circ} \approx 11.3.\) A1
Common mistakes
- Reaching for the sine rule when it can't be set up. If you know two sides and the angle between them, or all three sides, there's no complete angle-side pair to build the ratio from - switch to the cosine rule instead.
- Pairing the wrong side with the wrong angle. Each side must sit opposite its matching angle in the ratio - side \(a\) opposite \(A\), side \(b\) opposite \(B\). Writing \(\dfrac{a}{\sin B}\) by mistake is a common slip when the triangle is drawn at an angle.
- Rounding too early. Carry full calculator accuracy through the ratio and only round the final answer - rounding an intermediate sine value can shift the last significant figure of the side length.
- Missing the ambiguous case. Given two sides and a non-included acute angle, \(\sin^{-1}\) can return two valid triangles - check whether the obtuse supplementary angle also fits the given information before ruling it out.
Ready to practise properly?
21 sine-rule questions, marked instantly like the real exam.
Quick answers
What is the sine rule?
\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\). It links each side of a triangle to the sine of the angle opposite it, so you can find a missing side or angle once you know one full angle-side pair plus one more piece of information.
When do I use the sine rule instead of the cosine rule?
Use the sine rule when you know an angle and its opposite side, plus one other angle or side. Use the cosine rule when you know two sides and the angle between them, or all three sides, since the sine rule can't be set up directly in those cases.