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

Section 1

What is Pythagoras' theorem?

In any right-angled triangle, Pythagoras' theorem relates the three sides: $a^2 + b^2 = c^2$ where $c$ is the hypotenuse (the longest side, opposite the right angle) and $a, b$ are the other two sides.

To find the hypotenuse: $c = \sqrt{a^2 + b^2}$
To find a shorter side: $a = \sqrt{c^2 - b^2}$

Key termshypotenusePythagoras' theorem

Example

A triangle has legs 6 cm and 8 cm. The hypotenuse is $\sqrt{6^2+8^2} = \sqrt{100} = 10$ cm.

Section 2

What are the sine, cosine and tangent ratios?

For an acute angle $\theta$ in a right-angled triangle, label the sides relative to $\theta$ : opposite (opp), adjacent (adj), and hypotenuse (hyp). The three trigonometric ratios are:

Ratio	Formula
sine	$\sin\theta = \frac{\text{opp}}{\text{hyp}}$
cosine	$\cos\theta = \frac{\text{adj}}{\text{hyp}}$
tangent	$\tan\theta = \frac{\text{opp}}{\text{adj}}$

Use whichever ratio connects the side you know with the side or angle you want to find.

Key termsoppositeadjacenthypotenusesinecosinetangent

Think of it like this

Remember SOH CAH TOA: Sine=Opp/Hyp, Cosine=Adj/Hyp, Tangent=Opp/Adj.

Section 3

How do I solve two-dimensional problems, including bearings?

To find a missing side: identify which two sides are involved, choose the matching ratio (or Pythagoras if both are legs/hypotenuse), and rearrange to solve.

To find a missing angle: rearrange using the inverse function, e.g. $\theta = \sin^{-1}\left(\frac{\text{opp}}{\text{hyp}}\right)$ .

Bearings problems often combine trigonometry with right-angled triangles formed between compass directions — sketch a right-angled triangle from the given information before applying the ratios.

Always give angles correct to one decimal place unless told otherwise.

Key termsinverse trigonometric function

Exam tip

State the ratio you are using (e.g. 'using SOH') before substituting values — this signals your method clearly for M1 method marks.

Common mistake

Rounding intermediate answers before the final step — this causes accuracy errors; keep full calculator values until the last line.

Section 4

What is the shortest distance from a point to a line? (Extended)

The perpendicular distance from a point to a line is always the shortest possible distance between them. This fact is often used to justify why a particular right-angled triangle gives the minimum distance in a problem.

Key termsperpendicular distance

Section 5

What are angles of elevation and depression? (Extended)

The angle of elevation is measured upwards from the horizontal to a point above
The angle of depression is measured downwards from the horizontal to a point below

Both angles are always measured from a horizontal line, and by alternate angles, the angle of elevation from one point equals the angle of depression from the other when the two horizontals are parallel.

Key termsangle of elevationangle of depression

Example

From the top of a 20 m cliff, the angle of depression to a boat is 15°. The horizontal distance to the boat is $\frac{20}{\tan 15°} \approx 74.6$ m.

Must Know

Pythagoras: $a^2+b^2=c^2$ , with $c$ the hypotenuse
SOH CAH TOA: $\sin\theta=\frac{opp}{hyp}$ , $\cos\theta=\frac{adj}{hyp}$ , $\tan\theta=\frac{opp}{adj}$
Use inverse trig functions ( $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ) to find angles
Give angles to 1 decimal place; keep full calculator accuracy until the final answer
Perpendicular distance from a point to a line is always the shortest distance (Extended)
Angle of elevation (looking up) equals angle of depression (looking down) between two parallel horizontals (Extended)

That's the notes covered.

Carry on to the next subtopic.