The sine rule and cosine rule GCSE topic covers two formulas for solving triangles that are not right-angled, when ordinary SOHCAHTOA cannot be used. The sine rule links each side to the sine of its opposite angle, while the cosine rule connects all three sides to one included angle — together they solve any triangle.
What are the sine rule and cosine rule?
Both rules apply to any triangle, not just right-angled ones, but they solve different types of problem. Labelling a triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively, the two formulas are:
$$\text{Sine rule: } \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
$$\text{Cosine rule: } a^2 = b^2 + c^2 - 2bc\cos A$$
Both formulas appear on the GCSE formula sheet, so you do not need to memorise them, but you must know which one to pick and how to rearrange it.
When do you use the sine rule instead of the cosine rule?
Choosing the correct rule depends on which information the question gives you:
- Count what you know. List the sides and angles you have been given, out of the three sides and three angles that describe the triangle.
- Check for an opposite pair. If you know an angle and the side directly opposite it, plus one more side or angle, use the sine rule.
- Check for three sides, or two sides and the included angle. If you know all three sides, or two sides and the angle between them, use the cosine rule.
- Match the rearranged version to what you are finding. Use the rule to find a missing side or a missing angle, rearranging as needed.
How do you use the sine rule to find a missing side?
Worked example: In triangle ABC, angle $A = 40°$, angle $B = 65°$, and side $a = 8$ cm. Find side $b$.
Set up the sine rule using the known opposite pair ($a$ and $A$) and the unknown pair ($b$ and $B$):
$$\frac{a}{\sin A} = \frac{b}{\sin B} \implies \frac{8}{\sin 40°} = \frac{b}{\sin 65°}$$
Rearrange to make $b$ the subject:
$$b = \frac{8 \times \sin 65°}{\sin 40°} = \frac{8 \times 0.9063}{0.6428} \approx 11.28 \text{ cm}$$
So side $b$ is approximately 11.3 cm to 3 significant figures.
How do you use the cosine rule to find a missing side or angle?
The cosine rule has two versions, one for finding a side and a rearranged version for finding an angle:
$$a^2 = b^2 + c^2 - 2bc\cos A \qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
Worked example: In triangle ABC, $b = 7$ cm, $c = 9$ cm, and angle $A = 55°$. Find side $a$.
Substitute the known values into the first version:
$$a^2 = 7^2 + 9^2 - 2(7)(9)\cos 55° = 49 + 81 - 126 \times 0.5736 \approx 57.73$$
$$a = \sqrt{57.73} \approx 7.60 \text{ cm}$$
Worked example: In a triangle with sides $a = 6$ cm, $b = 8$ cm and $c = 10$ cm, find angle $C$ (opposite side $c$).
Use the rearranged cosine rule, remembering that $C$ is opposite $c$:
$$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{36 + 64 - 100}{2(6)(8)} = \frac{0}{96} = 0$$
$$C = \cos^{-1}(0) = 90°$$
This confirms the well-known 6-8-10 triangle is right-angled, since it is a scaled-up 3-4-5 triangle.
How do you find the area of a triangle using the sine rule?
When a right-angled height is not available, use the sine-rule area formula, which needs two sides and the included angle:
$$\text{Area} = \frac{1}{2}ab\sin C$$
Worked example: A triangle has sides of 6 cm and 9 cm with an included angle of 50°. Find its area.
$$\text{Area} = \frac{1}{2}(6)(9)\sin 50° = 27 \times 0.7660 \approx 20.7 \text{ cm}^2$$
The area is approximately 20.7 cm², correct to 3 significant figures.
Frequently asked questions
Can you use the sine rule and cosine rule on right-angled triangles?
Yes — both rules still give correct results on right-angled triangles, but SOHCAHTOA is usually quicker because it needs fewer steps and less calculator work. Save the sine rule and cosine rule for triangles where there is no right angle, which is when SOHCAHTOA cannot be applied at all.
What is the ambiguous case of the sine rule?
The ambiguous case happens when the sine rule is used with two sides and a non-included angle, and there are two possible triangles that fit the given information — one with an acute angle and one with an obtuse angle. GCSE questions usually specify which triangle is required, but higher tier students should check whether $180° - \theta$ also gives a valid answer.
How do you remember which sides and angles go where in the formulas?
Keep the labelling consistent: side $a$ is always opposite angle $A$, side $b$ opposite angle $B$, and side $c$ opposite angle $C$. Writing this labelling onto your triangle sketch before substituting into either formula prevents the most common error, which is mixing up which side pairs with which angle.
Do I need to memorise the sine rule and cosine rule formulas for the exam?
No — both formulas are printed on the GCSE mathematics formula sheet provided in the exam, for both foundation and higher tier where relevant. What you do need to practise is recognising which triangle information points to which rule, and rearranging the formula confidently to solve for the unknown side or angle.
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