IGCSE Maths: Sine Rule and Cosine Rule Explained

The sine rule and the cosine rule extend trigonometry to triangles that have no right angle. The sine rule links every side to the angle opposite it through one constant ratio. The cosine rule is Pythagoras plus a correction term that measures exactly how far your angle sits from a right angle.

The sine rule pairs a side with its opposite angle

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Read this as a statement about pairs, not as a string of symbols to memorise. In any triangle the longest side faces the largest angle, and the sine rule makes that intuition numerical: the ratio of a side to the sine of its opposite angle is identical for all three pairs. The practical consequence is a test you can run on any diagram in two seconds. If you can see one complete pair, a side together with the angle facing it, the sine rule is open to you. If you cannot, it is not.

The cosine rule is Pythagoras with a correction term

$$a^2 = b^2 + c^2 - 2bc\cos A$$

Put $A = 90^\circ$ into it. Since $\cos 90^\circ = 0$, the last term vanishes and you are left with $a^2 = b^2 + c^2$, the theorem you have used since Year 9. Now open the angle past $90^\circ$: $\cos A$ turns negative, so $-2bc\cos A$ turns positive and the opposite side comes out longer than Pythagoras would predict. Close the angle below $90^\circ$ and the opposite side comes out shorter. Students who carry that picture stop losing the minus sign halfway through a calculation, because they can see what the term is for.

The area formula belongs to the same family

$$\text{Area} = \frac{1}{2}ab\sin C$$

This is the familiar half base times height, with the height written as $a\sin C$. On CIE 0580 Extended, the sine rule, the cosine rule and this area formula sit inside the same trigonometry section of the syllabus, and examiners regularly ask for two of the three in different parts of one question.

Decide by counting what the triangle gives you. Two angles and any side, or two sides and an angle opposite one of them, means the sine rule. Two sides with the angle between them, or all three sides, means the cosine rule. If neither pattern appears, you are missing a step, not a formula.

That decision is the whole skill, and it is what IGCSE questions are really testing. Once the rule is chosen, the rest is calculator work.

The pattern that catches people out is the one where the information is spread across two triangles. A typical CIE 0580 Extended structured question gives a quadrilateral $ABCD$ with a diagonal drawn in. Part (a) gives two sides and the included angle in triangle $ABC$, so you apply the cosine rule to find the diagonal. Part (b) then uses that diagonal inside triangle $ACD$, where you now have a side and the angle opposite it, so the sine rule takes over. Part (c) asks for the area of the whole quadrilateral, which is $\frac{1}{2}ab\sin C$ applied twice and added. The diagonal is the bridge, and a candidate who does not see it stalls on part (b) with full marks available.

A second trap is the hidden pair. A question may give you angle $A$ and angle $B$ without giving angle $C$, and the side you need is $c$. There is no complete pair yet, so the sine rule looks unusable. It becomes usable the moment you write $C = 180^\circ - A - B$. That single line is often worth a method mark on its own.

Whether these rules are on your syllabus at all depends on your tier, which is covered in the guide to Extended versus Core.

Here is a standard IGCSE Extended question. In triangle $ABC$,$AB = 8.2$cm,$BC = 11.5$ cm and angle $ABC = 63^\circ$. Find $AC$, then angle $BAC$, then the area of the triangle. It uses all three formulas in sequence and is worth roughly seven marks split across three parts.

Part (a): find $AC$

You have two sides and the angle between them, so this is the cosine rule. The side you want is opposite the known angle, which is exactly the configuration the formula is built for.

$$AC^2 = 8.2^2 + 11.5^2 - 2 \times 8.2 \times 11.5 \times \cos 63^\circ$$

$$AC^2 = 67.24 + 132.25 - 188.6\cos 63^\circ = 199.49 - 85.6226... = 113.8673...$$

$$AC = \sqrt{113.8673...} = 10.6708... = 10.7 \text{ cm}$$

The first method mark is for that substitution line, before any arithmetic. The second is for reaching $113.8...$ and taking the square root, which is where candidates who subtract in the wrong order lose credit. The accuracy mark is for $10.7$ to three significant figures.

Part (b): find angle $BAC$

Now you have a complete pair: side $AC = 10.6708...$ with angle $B = 63^\circ$ opposite it. Use the sine rule with the unknown on top.

$$\frac{\sin BAC}{11.5} = \frac{\sin 63^\circ}{10.6708...}$$

$$\sin BAC = \frac{11.5\sin 63^\circ}{10.6708...} = 0.96020...$$

$$BAC = \sin^{-1}(0.96020...) = 73.8^\circ \text{ to 1 decimal place}$$

This is where premature rounding bites. If you carry the rounded $10.7$ into part (b) instead of the full calculator value, you get $\sin BAC = 0.95763...$ and $BAC = 73.3^\circ$. The method marks survive, but the accuracy mark is at risk over half a degree of avoidable error. Store $AC$ in your calculator memory and recall it.

Part (c): find the area

$$\text{Area} = \frac{1}{2} \times 8.2 \times 11.5 \times \sin 63^\circ = 42.0 \text{ cm}^2$$

Notice that part (c) does not need part (a) at all, so a candidate who failed the cosine rule can still take these marks. Finally, sanity check the whole thing: angle $ACB = 180^\circ - 63^\circ - 73.8^\circ = 43.2^\circ$, the smallest angle, and it faces $AB = 8.2$ cm, the shortest side. The triangle is consistent.

The sine rule can give two answers because $\sin\theta = \sin(180^\circ - \theta)$. When a question gives two sides and an angle that is not between them, an acute angle and its obtuse partner both satisfy the equation and both can build a real triangle. Your calculator only ever shows the acute one.

Here is the situation in numbers. Suppose angle $A = 35^\circ$, side $a = 7$ cm and side $b = 9$ cm, and you are asked for angle $B$.

$$\sin B = \frac{9\sin 35^\circ}{7} = 0.73745...$$

$$B = 47.5^\circ \quad \text{or} \quad B = 180^\circ - 47.5^\circ = 132.5^\circ$$

Both are genuine. With $B = 132.5^\circ$ the angles still total less than $180^\circ$, since $35^\circ + 132.5^\circ = 167.5^\circ$, leaving $12.5^\circ$ for the third angle. Two different triangles fit the same three pieces of information.

How IGCSE questions resolve it

Three things tell you which answer the examiner wants. The diagram: CIE 0580 figures are printed NOT TO SCALE, but an angle drawn clearly obtuse is still a signal about the shape being described. The wording: a question that needs the obtuse case usually says so, with a phrase such as angle $B$ is obtuse. The arithmetic: if the second angle would push the total past $180^\circ$, it is impossible and the acute answer is the only one.

The cosine rule does not have this problem

Because $\cos\theta$ is negative for obtuse angles and positive for acute ones, $\cos^{-1}$ returns a single value between $0^\circ$ and $180^\circ$. It carries the sign information that the sine loses. So when you know all three sides and need an angle, especially the largest one, reach for $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ rather than the sine rule. The same $\sin\theta = \sin(180^\circ - \theta)$ idea reappears later in the syllabus when you are asked to solve an equation such as $\sin x = 0.4$ for $0^\circ \leqslant x \leqslant 360^\circ$, which is why examiners keep testing it.

IGCSE mark schemes award a method mark for a correct substitution into the correct formula and an accuracy mark for the final value. Writing the cosine rule with your own numbers already in it earns credit even if the arithmetic then goes wrong. A correct answer with no working scores full marks, but a wrong answer with no working scores nothing.

That asymmetry is the practical reason to show the substitution line on every sine and cosine rule question, however confident you feel. It costs eight seconds and insures the whole question.

The conventions worth knowing

Mark schemes mark M for method, A for accuracy, and use short annotations such as oe (or equivalent), cao (correct answer only) and ft (follow through). Follow through matters enormously in these multi-part trigonometry questions: if part (a) is wrong but part (b) is carried out correctly using your own wrong value, the method marks in part (b) are still available. Students who abandon a question after a bad first part throw those marks away. The conventions are unpacked further in the guide to IGCSE Maths mark schemes.

Accuracy rules that decide the A mark

The front cover of CIE 0580 papers instructs candidates to give answers to three significant figures unless the question states otherwise, and to give angles in degrees to one decimal place. So $10.6708...$ cm is written $10.7$ cm and $73.78...^\circ$ is written $73.8^\circ$. Exact answers, and answers that terminate early, are written exactly rather than padded. When a part says show that, the given answer is the target and you must produce more accuracy than the target to prove you got there.

Whether the formulas are printed for you depends on the board

This is one of the sharper differences between the two most common IGCSE maths specifications, and it changes how you revise. The full comparison sits in CIE 0580 versus Edexcel 4MA1.

Sine and cosine rule questions appear only on Extended (CIE) or Higher (Edexcel) papers, and always where a calculator is allowed. They almost never come as a bare triangle. They arrive inside a structured multi part question, wrapped in bearings, in a quadrilateral split by a diagonal, or in a three dimensional solid.

Bearings are the most common dressing

A bearing is measured clockwise from north and written with three figures, so due east is $090^\circ$ and a bearing of $47^\circ$ is written $047^\circ$. A ship sails from $A$ to $B$ on one bearing and from $B$ to $C$ on another, and you are asked for the distance $AC$. The real work is turning the two bearings into the interior angle at $B$ using parallel north lines and angles on a straight line. Once that angle exists, the cosine rule finishes the job in one line. Candidates lose marks on the geometry of the bearings, not on the trigonometry.

Three dimensional problems hide a triangle

In a pyramid or a cuboid question, the instruction is always the same: extract the relevant triangle and redraw it flat at the side of the page with its known lengths marked. Many of these triangles are right angled and need only $\tan$, $\sin$ or $\cos$, but the sloping faces of a pyramid frequently produce a non right triangle where the cosine rule is the only way in.

How to practise this efficiently

Work by topic rather than by whole paper. Take ten sine and cosine rule questions from different past series and do only those, in one sitting, with the mark scheme closed. You will see the same three dressings repeat and the choice of rule will become automatic. The method for mining past papers this way is set out in the guide to using IGCSE Maths past papers. Because these are calculator questions, the habits that protect your marks are calculator habits: degree mode, memory recall, and rounding only at the end.