Rukasha Maths

GCSE maths Beta

Pythagoras and right-angled trigonometry · Foundation notes

Foundation · Geometry · AQA references G20, G21

Pythagoras and trigonometry connect measurements in right-angled triangles. First mark the right angle and identify the hypotenuse, the side opposite it and the longest side. Pythagoras uses the squares of all three lengths, so it is useful when two lengths are known. Trigonometry uses an angle and a ratio of two sides. Opposite and adjacent are named relative to the chosen angle, so their labels can change when you choose a different angle. Draw a labelled sketch, choose the equation using the known measurements, and make sure your calculator is in degrees.

Learning objectives

  • Calculate a missing side in a right-angled triangle.
  • Choose a trigonometric ratio for a side or angle.

Key ideas

The hypotenuse is opposite the right angle. Pythagoras applies only to right-angled triangles.

a2+b2=c2(hypotenuse c)a^2+b^2=c^2\quad\text{(hypotenuse }c\text{)}

Add squares to find the hypotenuse; subtract squares to find a shorter side. Then take a square root.

Label opposite and adjacent relative to the chosen acute angle.

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}

Use the tangent ratio when the hypotenuse is not involved.

tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

For a missing angle, use the appropriate inverse ratio in degree mode.

Worked examples

Example 1

Find the unknown shorter side of a right-angled triangle.

c=13 cm,a=5 cm:x2+52=132c=13\text{ cm},\quad a=5\text{ cm}:\qquad x^2+5^2=13^2
x2=169−25=144x^2=169-25=144

A length takes the positive root.

x=144=12 cmx=\sqrt{144}=12\text{ cm}

Answer

12 cm12\text{ cm}

Example 2

Find the side opposite the acute angle in a right-angled triangle.

θ=30∘,c=10 cm:sin⁡30∘=o10\theta=30^\circ,\quad c=10\text{ cm}:\qquad\sin30^\circ=\frac{o}{10}
o=10sin⁡30∘=10×12=5 cmo=10\sin30^\circ=10\times\frac12=5\text{ cm}

Answer

5 cm5\text{ cm}

Question styles to prepare for

Original practice guidance aligned to the specification, not a count of past-paper appearances or a prediction. The AO labels below describe the intended practice focus, not official AQA mark allocations.

AO1 · Methods

Find a missing length or angle in a right-angled triangle.

AO2 · Reasoning

Choose and justify a method using the sides given and the unknown.

AO3 · Problem solving

Extract a right-angled triangle from a larger diagram and interpret the result.

When a task asks you to explain, include a mathematical reason. When solving a problem, show how the information becomes a calculation, equation or diagram, then interpret your result.

Avoid this mistake

A squared length is not the final length. Take the square root and check that the hypotenuse is the longest side.

Try it yourself

Attempt each question before opening its answer.

Check 1: A right-angled triangle has shorter sides 8 cm and 15 cm. Find the hypotenuse.

Answer: 17 cm

c² = 8² + 15² = 64 + 225 = 289, so c = √289 = 17.

Check 2: In a right-angled triangle, an acute angle is 30° and the hypotenuse is 10 cm. Find the opposite side.

Answer: 5 cm

sin 30° = opposite/10. Since sin 30° = 1/2, the opposite side is 10 × 1/2 = 5 cm.

Explain it in your own words

  • Explain why the hypotenuse square must be the largest one.
  • Draw one triangle and label opposite and adjacent for each acute angle.
  • Write which ratio you would use when given opposite and adjacent.

Keep practising

Download the compact topic guide (PDF) · Practise this topic in Blitz · All foundation notes

Specification source

AQA Geometry specification · Assessment objectives. Checked 16 October 2026. Topic references identify related content; these notes do not cover every specification statement.