Sine & Cosine Rule & Area of Triangles (Edexcel IGCSE Maths): Revision notes

Section 1

Why do we need these rules?

Right-angled trig (SOH CAH TOA) only works when there is a right angle. Most triangles in exam questions are non-right-angled (scalene), so you need two extra tools:

The sine rule — for triangles where you know a side and its opposite angle.
The cosine rule — for triangles where you know all three sides, or two sides and the included angle.

Label the triangle consistently: side $a$ is opposite angle $A$ , side $b$ is opposite angle $B$ , side $c$ is opposite angle $C$ . Getting this labelling right is the single biggest source of marks lost.

Key termsNon-right-angled triangleIncluded angle

Exam tip

Before doing anything, sketch the triangle and label the sides/angles using lower-case letters opposite the matching capital-letter angles.

Section 2

When and how do I use the sine rule?

The sine rule states:

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

Use it when you know an opposite pair (a side and the angle directly across from it), plus one more piece of information.

Finding a side: flip to $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ , rearrange to $a = \dfrac{b \sin A}{\sin B}$ .
Finding an angle: use the reciprocal form $\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$ so the unknown sine ends up on top.

The angles in a triangle always sum to $180°$ , so once you have two angles you can find the third by subtraction.

Key termsSine ruleOpposite pair

Example

$a = 8$ cm, $A = 40°$ , $B = 65°$ . Find $b$ : $b = \dfrac{a \sin B}{\sin A} = \dfrac{8 \sin 65°}{\sin 40°} \approx 11.3$ cm.

Common mistake

Using the sine rule when you don't have a full opposite pair — check you have a side AND its matching angle before starting.

Section 3

The ambiguous case — what is it and how do I spot it?

When you're given two sides and a non-included angle (SSA) and use the sine rule to find another angle, calculator's inverse sine can give the wrong answer — there may be two possible triangles.

If $\sin\theta = k$ gives one solution $\theta_1$ (calculator answer), a second valid solution is $\theta_2 = 180° - \theta_1$ , provided $\theta_2$ still allows the angles to sum sensibly to less than $180°$ .

This only happens in SSA (side-side-angle) setups — always check whether the question hints at an obtuse alternative, e.g. "find the two possible values of...".

Key termsAmbiguous case

Common mistake

Forgetting the second solution $180° - \theta_1$ when the question gives SSA information — this is a very common dropped mark.

Exam tip

Always check: does $180° - \theta_1$ plus the known angle still add to less than $180°$ ? If yes, it's a valid second triangle.

Section 4

When and how do I use the cosine rule?

The cosine rule states:

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

Use it when the sine rule won't work, specifically:

SAS (two sides and the included angle) — to find the third side, using the formula directly.
SSS (all three sides) — to find any angle, using the rearranged form:

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

Notice angle $A$ is opposite side $a$ , and sides $b, c$ are the two either side of angle $A$ (the included angle).

Key termsCosine ruleSASSSS

Example

$b = 7$ , $c = 5$ , $A = 60°$ . Find $a$ : $a^2 = 7^2 + 5^2 - 2(7)(5)\cos 60° = 49 + 25 - 35 = 39$ , so $a = \sqrt{39} \approx 6.24$ .

Think of it like this

Think of the cosine rule as Pythagoras' theorem with a correction term $-2bc\cos A$ — when $A = 90°$ , $\cos A = 0$ and it collapses back to $a^2 = b^2 + c^2$ .

Section 5

How do I find the area of a non-right-angled triangle?

When you know two sides and the included angle, use:

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

where $C$ is the angle between sides $a$ and $b$ . This replaces $\text{Area} = \frac12 \times \text{base} \times \text{height}$ when you don't have a perpendicular height.

Exam questions often combine this with the cosine or sine rule in one multi-step problem: e.g. find a missing angle first, then use it to calculate the area.

Key termsIncluded angleArea formula

Example

$a = 6$ , $b = 9$ , $C = 50°$ . Area $= \frac12 (6)(9)\sin 50° \approx 20.7$ units $^2$ .

Common mistake

Using the wrong angle — it must be the one enclosed between the two sides you multiply, not any angle in the triangle.

Must Know

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ — use with an opposite side/angle pair.
Cosine rule (side): $a^2 = b^2+c^2-2bc\cos A$ — use for SAS.
Cosine rule (angle): $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ — use for SSS.
Area: $\text{Area} = \frac12 ab\sin C$ — use with two sides and the included angle.
Watch for the ambiguous case (SSA): a second angle $180° - \theta$ may also be valid.
Always sketch and label the triangle first — correct labelling prevents wrong-formula errors.

That's the notes covered.

Carry on to the next subtopic.