Labelling a triangle
Both rules rely on one convention: each side is named with the lower-case letter of the angle it sits opposite. So side \(a\) is across from angle \(A\), side \(b\) from angle \(B\), and side \(c\) from angle \(C\). Get this right first and everything else slots into place.
- Opposite side
- The side that does not touch a given angle. Angle \(A\) and side \(a\) form an opposite pair, and it is these pairs that the Sine Rule links together.
The Sine Rule
The Sine Rule connects each side to the sine of its opposite angle. Every ratio in it is equal:
You only ever need two of the three ratios at a time. Use the side-up version when you are finding a side, and flip it to \(\dfrac{\sin A}{a}=\dfrac{\sin B}{b}\) when you are finding an angle. The rule works whenever you have a complete opposite pair (a side and its opposite angle) plus one more piece.
In triangle \(ABC\), \(A = 35^\circ\), \(B = 65^\circ\) and side \(a = 8\) cm. Find side \(b\).
The Cosine Rule
When you do not have a complete opposite pair, the Cosine Rule takes over. It links all three sides to one angle:
Use it as written to find a side when you know two sides and the angle between them. To find an angle when you know all three sides, rearrange to:
A triangle has \(a = 6\) cm, \(b = 9\) cm and the included angle \(C = 55^\circ\). Find side \(c\).
Order of operations
In the Cosine Rule, do the multiplication \(2ab\cos C\) as a single block before subtracting. Do not square-root until the very end, and never subtract \(2ab\) first and then multiply by cos.
The area formula
You do not need the perpendicular height to find a triangle's area. If you know two sides and the angle between them, use:
The pattern matches the Cosine Rule exactly: two sides and their included angle. For example, with \(a=7\), \(b=10\) and \(C=40^\circ\), the area is \(\tfrac12(7)(10)\sin 40^\circ = 35 \times 0.6428 = 22.5\) cm\(^2\).
Choosing which to use
The whole skill is matching the tool to what you are given. This table is the decision you make at the start of every question:
| You are given | You want | Use |
|---|---|---|
| A side and its opposite angle, plus one more | A side or an angle | Sine Rule |
| Two sides and the included angle | The third side | Cosine Rule |
| All three sides | Any angle | Cosine Rule |
| Two sides and the included angle | The area | \(\tfrac12 ab\sin C\) |
Where this is assessed
Picking the correct rule is Criterion D (applying maths to solve a problem), while a clean, correctly-rounded, well-set-out solution is Criterion C. Show the substitution line before the number, examiners reward it.
Check yourself
1. You know all three sides of a triangle and want the largest angle. Which rule, and why? +
Use the Cosine Rule rearranged to \(\cos C = \dfrac{a^2+b^2-c^2}{2ab}\). The Sine Rule needs a known angle to start, and here you have none, so only the Cosine Rule works. The largest angle sits opposite the longest side.
2. Find the area of a triangle with sides \(a = 5\) cm, \(b = 8\) cm and included angle \(C = 30^\circ\). +
\(\text{Area} = \tfrac12 ab\sin C = \tfrac12(5)(8)\sin 30^\circ = 20 \times 0.5 = \mathbf{10\ \text{cm}^2}\).
3. In a triangle, \(b = 12\) cm, \(B = 80^\circ\) and \(a = 9\) cm. Find angle \(A\). +
Finding an angle, so use \(\dfrac{\sin A}{a}=\dfrac{\sin B}{b}\). Then \(\sin A = \dfrac{a\sin B}{b} = \dfrac{9\sin 80^\circ}{12} = \dfrac{9 \times 0.9848}{12} = 0.7386\). So \(A = \sin^{-1}(0.7386) = \mathbf{47.6^\circ}\).
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