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The sine rule says a/sin A = b/sin B = c/sin C; the cosine rule finds a side from two sides and the included angle. Free notes and 7 questions answered.
Sine and Cosine Rules is part of Sine and cosine rules in scalene triangles in GCSE AQA Higher Further Maths.
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GCSE AQA Higher
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The sine rule says a/sin A = b/sin B = c/sin C; the cosine rule finds a side from two sides and the included angle. Free notes and 7 questions answered.
Use this lesson to stay inside Sine and cosine rules in scalene triangles while connecting the detail back to the wider Geometry topic.
Read the core explanations from Sine and Cosine Rules before testing yourself from memory.
SOHCAHTOA is powerful, but it only works in right-angled triangles. Most triangles you meet in GCSE problems — and in the real world — have no right angle at all. That is where the sine rule and the cosine rule come in.
Both rules work in any triangle, including scalene triangles where every side and every angle is different. The skill the examiner is testing is knowing which rule to reach for and then applying it without errors.
The standard convention is critical: lowercase letters label sides, uppercase letters label the opposite angles. Side $a$ sits opposite angle $A$, side $b$ opposite $B$, and side $c$ opposite $C$. Get this pairing wrong and every formula falls apart.
Think back to the ratio idea from SOHCAHTOA — a side divided by a trigonometric function of its related angle. The sine rule extends that idea across the whole triangle: every side-to-sine-of-opposite-angle ratio is equal.
When you need to find a missing angle, flip the formula so the sines are on top. This avoids dividing by a sine and keeps the arithmetic cleaner.
Worked example: Sine Rule — Finding a Missing Side
In triangle $ABC$, angle $A = 48^\circ$, angle $B = 73^\circ$, and side $b = 9.4\text{ cm}$. Find side $a$.
Answer: $a \approx 7.31\text{ cm}$ (to 3 significant figures)
There are two situations where the sine rule cannot help: when you know two sides and the angle between them (the included angle), or when you know all three sides and need an angle. For both, you need the cosine rule.
Notice that if $C = 90^\circ$ then $\cos C = 0$, and the cosine rule reduces to $c^2 = a^2 + b^2$ — Pythagoras' theorem. The cosine rule is genuinely a generalisation of a result you already know.
Worked example: Cosine Rule — Finding a Missing Side
In triangle $PQR$, $p = 7\text{ cm}$, $q = 5\text{ cm}$, and angle $R = 110^\circ$. Find side $r$.
Answer: $r \approx 9.90\text{ cm}$ (to 3 significant figures)
Worked example: Cosine Rule — Finding a Missing Angle
A triangle has sides $a = 6\text{ cm}$, $b = 8\text{ cm}$, $c = 10\text{ cm}$. Find angle $C$.
Answer: $C = 90^\circ$
You already know that area $= \frac{1}{2} \times \text{base} \times \text{height}$. The problem is that the perpendicular height is rarely given directly in a scalene triangle. The sine area formula replaces the height using trigonometry, so you only need two sides and the angle between them.
Worked example: Area Using the Sine Formula
Triangle $ABC$ has $a = 5\text{ cm}$, $b = 7\text{ cm}$, and angle $C = 38^\circ$. Calculate the area.
Answer: $\text{Area} \approx 10.8\text{ cm}^2$ (to 3 significant figures)
Every triangle problem gives you a starting set of information. Matching that set to the correct rule is the most important decision in the question. Here is a compact reference for that decision.
| Information given | What to find | Rule to use |
|---|---|---|
| Two angles + any side | Missing side | Sine rule |
| Two sides + non-included angle | Missing angle or side | Sine rule |
| Two sides + included angle | Missing side | Cosine rule |
| Two sides + included angle | Area | Area = $\frac{1}{2}ab\sin C$ |
| Three sides | Missing angle | Cosine rule (angle form) |
The decision table and formula bank are only useful if you can apply them under exam conditions. Work through these problems, selecting the correct rule before you start each one.
The sine and cosine rules are reliable tools, but the exam is designed to catch specific errors. The mistakes below account for a large proportion of lost marks on these questions.
Mastering the sine and cosine rules means you can solve any triangle — not just right-angled ones — which is the core specification requirement for scalene triangle trigonometry at GCSE.
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
The cosine rule finds a side when two sides and the included angle are known: r² = p² + q² − 2pq cos R. For p = 7 cm, q = 5 cm and R = 110°, r² = 49 + 25 − 70 × cos 110° ≈ 97.94, so r ≈ 9.90 cm.
The sine rule states that in any triangle each side divided by the sine of its opposite angle is equal: a/sin A = b/sin B = c/sin C. For angle A = 48°, angle B = 73° and b = 9.4 cm, a = 9.4 × sin 48° ÷ sin 73° ≈ 7.31 cm.
When all three sides are known, use cos C = (a² + b² − c²) ÷ (2ab). For sides 6 cm, 8 cm and 10 cm, cos C = (36 + 64 − 100) ÷ 96 = 0, so C = cos⁻¹(0) = 90°.
The sine rule is used when two angles and any side are known, or two sides and a non-included angle. The cosine rule is used when two sides and the included angle are known and the third side is needed, or when all three sides are known and an angle is needed.
The standard convention uses lowercase letters for sides and uppercase letters for the opposite angles: side a is opposite angle A, side b opposite B, and side c opposite C. A wrong pairing makes every formula fall apart.
The cosine of an obtuse angle is negative, so in the cosine rule, subtracting 2pq cos R adds to p² + q². This makes the side opposite an obtuse angle the longest side, as with cos 110° ≈ −0.3420.
The angle opposite the 10 cm side of a 6-8-10 triangle is 90°. Using cos C = (a² + b² − c²)/(2ab) = (36 + 64 − 100)/96 = 0, so C = 90°, a multiple of the 3-4-5 Pythagorean triple.