Right-Angled Triangles — Pythagoras & Trigonometry (Edexcel IGCSE Maths): Revision notes

Section 1

How do I find a missing side with Pythagoras' theorem?

Pythagoras' theorem applies only to right-angled triangles. It links the three sides:

$a^2 + b^2 = c^2$

where $c$ is the hypotenuse (the longest side, always opposite the right angle), and $a$ , $b$ are the two shorter sides.

Finding the hypotenuse: add the squares of the two shorter sides, then square root. $c = \sqrt{a^2 + b^2}$
Finding a shorter side: subtract the known shorter side's square from the hypotenuse's square, then square root. $a = \sqrt{c^2 - b^2}$

Always identify the hypotenuse first — it is the side you never subtract.

Key termshypotenusePythagoras' theorem

Common mistake

Do not just add or subtract squares randomly — check whether you are finding the hypotenuse (add) or a shorter side (subtract). Subtracting when you should add gives a negative number under the root.

Example

A ladder leans against a wall. Base = 3 m, height reached = 4 m. Ladder length $= \sqrt{3^2+4^2} = \sqrt{25} = 5$ m.

Section 2

How do I label sides for trigonometry (sin, cos, tan)?

Once an angle (other than the right angle) is chosen as the reference angle $\theta$ , the three sides get specific labels:

Hypotenuse (H): the longest side, opposite the right angle — same as in Pythagoras.
Opposite (O): the side directly across from $\theta$ .
Adjacent (A): the remaining side, next to $\theta$ (touching it, but not the hypotenuse).

The labels change if you swap which angle you are working from, so always re-label O and A when the angle changes.

Key termsoppositeadjacent

Exam tip

Draw the triangle and mark H first (opposite the right angle), then O (across from the angle you're using), then A (whatever's left).

Common mistake

The hypotenuse is fixed by the right angle — it never changes. Only O and A swap depending on which angle you pick.

Section 3

How do I remember and use SOH CAH TOA?

SOH CAH TOA gives the three trigonometric ratios:

$\sin\theta = \frac{O}{H} \qquad \cos\theta = \frac{A}{H} \qquad \tan\theta = \frac{O}{A}$

Finding a missing side: decide which two sides are involved (H with O, H with A, or O with A), pick the matching ratio, substitute, then rearrange.

Finding a missing angle: once you have a ratio value, use the inverse function:

$\theta = \sin^{-1}\left(\frac{O}{H}\right), \quad \theta = \cos^{-1}\left(\frac{A}{H}\right), \quad \theta = \tan^{-1}\left(\frac{O}{A}\right)$

Make sure your calculator is in degree mode unless the question specifies radians.

Key termsSOH CAH TOAinverse trigonometric function

Example

A right-angled triangle has hypotenuse 10 cm and one angle 30°. To find the side opposite 30°: $\sin 30° = \frac{O}{10}$ , so $O = 10 \times \sin 30° = 5$ cm.

Common mistake

When rearranging, if the unknown is on the bottom of the fraction (e.g. finding A in $\tan\theta = O/A$ ), you must multiply then divide — don't forget to flip the equation correctly.

Section 4

What are the exact trig values I need to memorise?

Exam questions often ask for exact values (surds/fractions) rather than decimals, derived from a 45°-45°-90° triangle and a 30°-60°-90° triangle:

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
0°	0	1	0
30°	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
45°	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	1
60°	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
90°	1	0	undefined

Notice sin and cos values mirror each other (sin 30° = cos 60°, etc.) because the two non-right angles in a triangle always sum to 90°.

Key termsexact valuesurd

Think of it like this

Think of the 45° triangle as an isosceles right-angled triangle with legs 1 and hypotenuse $\sqrt{2}$ , and the 30-60-90 triangle as an equilateral triangle of side 2 cut in half.

Exam tip

If asked to 'give your answer as an exact value' or 'in surd form', do not round — leave $\sqrt{}$ and fractions as they are.

Must Know

$a^2 + b^2 = c^2$ , where $c$ is always the hypotenuse (opposite the right angle).
Label sides relative to the chosen angle: Hypotenuse, Opposite, Adjacent.
$\sin\theta = O/H$ , $\cos\theta = A/H$ , $\tan\theta = O/A$ — SOH CAH TOA.
Use inverse functions ( $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ) to find a missing angle.
Memorise exact values for 0°, 30°, 45°, 60°, 90° for sin, cos and tan.
Always check your calculator is in degree mode before solving.

That's the notes covered.

Carry on to the next subtopic.