Pythagoras & Right-Angled Trigonometry

Mathematics· 6.1, 6.2, E6.3· 20 min read

1. Pythagoras' Theorem★★☆☆☆⏱ 4 min

📘 Definition

Pythagoras' Theorem

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

For any right-angled triangle, the square of the hypotenuse (c, the longest side opposite the right angle) equals the sum of the squares of the two shorter sides (a, b).

Example:

3-4-5 right triangle: $3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$

Pythagoras' theorem is used when you have two sides of a right-angled triangle and need to find the third, with no angles involved. You can rearrange the formula to find a shorter side: $a^{2} = c^{2} - b^{2}$ .

📐 Worked Example

A right-angled triangle has two shorter sides of length 7 cm and 24 cm. Find the length of the hypotenuse.

1
Label the hypotenuse $c$ , and the shorter sides $a = 7$ cm, $b = 24$ cm.
2
$c^2 = 7^2 + 24^2 = 49 + 576 = 625$
3
Square root both sides to find $c$ : $c = 625 = 25$ cm.

Exam tip:

Always verify your calculated hypotenuse is longer than the other two sides, if not you have rearranged the formula incorrectly.

2. SOHCAHTOA for Right-Angled Trigonometry★★★☆☆⏱ 6 min

📘 Definition

SOHCAHTOA

Mnemonic for the three trigonometric ratios for acute angles in right-angled triangles: SOH = $s in (θ) = \frac{O pp os i t e}{H y p o t e n u se}$ , CAH = $cos (θ) = \frac{A d j a ce n t}{H y p o t e n u se}$ , TOA = $t an (θ) = \frac{O pp os i t e}{A d j a ce n t}$ .

First identify the acute angle you are working with, then label sides relative to this angle: opposite is the side facing the angle, adjacent is the side next to the angle that is not the hypotenuse, and hypotenuse is the longest side.

📐 Worked Example

A right-angled triangle has a hypotenuse of 10 cm and an acute angle of 35°. Find the length of the side opposite the 35° angle, to 1 decimal place.

1
We know the hypotenuse and need the opposite side, so use the SOH ratio.
2
$sin(35°) = \frac{opposite}{10}$
3
Rearrange to solve for the opposite side: $o pp os i t e = 10 \times s in (35°) \approx 10 \times 0.5736 = 5.7$ cm (1 d.p.).

Exam tip:

Always confirm your calculator is set to degrees mode (look for a 'D' on the display) before attempting trig questions, radians mode will give incorrect results.

3. Bearings★★★☆☆⏱ 5 min

Bearings are 3-digit angles measured clockwise from North, used for navigation problems. Most bearings questions for this syllabus can be solved by drawing a right-angled triangle and applying Pythagoras or SOHCAHTOA.

📐 Worked Example

A ship sails 12 km East from point A to point B, then 5 km North to point C. Find the bearing of C from A, to 1 decimal place.

1
Draw right-angled triangle ABC with a right angle at B, AB = 12 km East, BC = 5 km North.
2
First calculate the angle between AB (East) and AC: $t an (θ) = \frac{5}{12} \approx 0.4167$ , so $θ \approx 22.6°$ .
3
Bearing is measured clockwise from North: $90° - 22.6° = 67.4°$ , written as 067.4°.

Exam tip:

Always write bearings as 3-digit numbers, adding leading zeros for angles less than 100° (e.g. 67° becomes 067°) to get full marks.

4. Extended Only: Angles of Elevation/Depression & Perpendicular Distance★★★★☆Extended only⏱ 4 min

Extended tier students need to solve problems involving angles of elevation and depression (these are equal due to alternate angles rule for parallel lines), plus calculate the perpendicular distance from a point to a line (the shortest possible distance, which forms a right angle with the line).

📐 Worked Example

From the top of a 15 m tall building, the angle of depression to a park bench is 28°. Find the horizontal distance from the base of the building to the bench, to 1 decimal place.

1
The angle of depression of 28° equals the angle of elevation from the bench to the top of the building. We have the opposite side (15 m) and need the adjacent side (horizontal distance), so use TOA.
2
$tan(28°) = \frac{15}{x}$
3
Rearrange to solve for x: $x = \frac{15}{t an ( 28° )} \approx \frac{15}{0.5317} = 28.2$ m (1 d.p.).

5. Extended Only: Exact Trig Values★★★☆☆Extended only⏱ 3 min

For the Extended non-calculator Paper 2, you must memorize exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°. These are often tested in low-difficulty, high-mark questions.

Angle	sin θ	cos θ	tan θ
0°	0	1	0
30°	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{1}{3}$
45°	$\frac{2}{2}$	$\frac{2}{2}$	1
60°	$\frac{3}{2}$	$\frac{1}{2}$	$3$
90°	1	0	Undefined

📐 Worked Example

Without a calculator, calculate the exact value of $s in (30°) + cos (60°)$ .

1
Recall exact values: $s in (30°) = \frac{1}{2}$ , $cos (60°) = \frac{1}{2}$ .
2
$\frac{1}{2} + \frac{1}{2} = 1$

Exam tip:

Memorize the 30-60-90 and 45-45-90 special triangles to derive exact values if you forget them during the non-calculator paper.

6. Common Pitfalls

Wrong move:

Adding the hypotenuse squared to another side in Pythagoras' formula.

Why:

The hypotenuse is the longest side, so it must be the only term on one side of the equation.

Correct move:

Write $c^{2} = a^{2} + b^{2}$ where c is always the hypotenuse, rearrange for shorter sides as $a^{2} = c^{2} - b^{2}$ .

Wrong move:

Labeling opposite/adjacent sides relative to the right angle instead of the given acute angle.

Why:

Trig ratios are defined relative to the acute angle you are calculating for, not the 90° angle.

Correct move:

Identify the acute angle first, then label opposite (facing it) and adjacent (next to it, not hypotenuse) relative to that angle.

Wrong move:

Writing bearings as 1 or 2-digit numbers (e.g. 67° instead of 067°).

Why:

CIE mark schemes explicitly require 3-digit bearings for full marks.

Correct move:

Add leading zeros to bearings less than 100° to make them 3 digits.

Wrong move:

Using radians mode on your calculator instead of degrees mode.

Why:

All angles in CIE IGCSE Maths 0580 are in degrees, so radians mode produces incorrect results.

Correct move:

Check for a 'D' symbol on your calculator display to confirm degrees mode is active before starting trig questions.

Wrong move:

Confusing angles of elevation and depression, or not recognizing they are equal.

Why:

This leads to labeling the wrong angle in your triangle, resulting in incorrect ratio use.

Correct move:

Draw horizontal lines at both the observer and object, mark equal alternate angles to set up your triangle correctly.

7. Quick Reference Cheatsheet

Concept	Rule/Value	Tier
Pythagoras' Theorem	$a^{2} + b^{2} = c^{2}$ , c = hypotenuse	Core
SOH Ratio	$s in (θ) = \frac{O pp os i t e}{H y p o t e n u se}$	Core
CAH Ratio	$cos (θ) = \frac{A d j a ce n t}{H y p o t e n u se}$	Core
TOA Ratio	$t an (θ) = \frac{O pp os i t e}{A d j a ce n t}$	Core
Bearings	3-digit, measured clockwise from North	Core
Angle of Elevation/Depression	Equal via alternate angles rule	Extended
Exact sin 30°	$\frac{1}{2}$	Extended
Exact cos 45°	$\frac{2}{2}$	Extended
Exact tan 60°	$3$	Extended

8. Frequently Asked

How do I know when to use Pythagoras vs SOHCAHTOA?

Use Pythagoras if you only need to work with sides (no angles given or required). Use SOHCAHTOA if you have at least one acute angle given or need to calculate an angle.

Do Core tier students need to memorize exact trig values?

No, exact trig values are only required for the Extended tier of CIE IGCSE Maths 0580 for Papers 2 and 4.

What's Next

Now that you have mastered right-angled trigonometry for CIE IGCSE Maths 0580, the next key topic is non-right-angled trigonometry using the sine and cosine rules, required for both Core and Extended tiers. You should also practice applying these skills to mixed 2D geometry problems, which are common in Section B of both calculator and non-calculator papers. Extended students should prioritize extra practice on exact trig value questions for Paper 2, as these are low-difficulty high-mark questions if memorized correctly.