Overview
What you’ll learn
- Find any side of a right-angled triangle with Pythagoras’ theorem, including recognising common triples.
- Label opposite, adjacent and hypotenuse correctly and apply SOH CAH TOA for sides and angles.
- Recall the exact values of sin, cos and tan at 0°, 30°, 45°, 60° and 90°.
- Choose and apply the sine rule, cosine rule or ½ab sin C in any triangle.
- Build right-angled triangles inside 3D solids to find diagonals and angles.
Specification checklist (AQA 8300 · Edexcel 1MA1 · OCR J560)
Statements are paraphrased from the shared DfE subject content and grouped by tier. Foundation content is studied by every student; Higher statements are examined on Higher-tier papers only. QuizQuip is not affiliated with AQA, Pearson Edexcel or OCR.
Foundation — every student
- 16.1Use Pythagoras’ theorem (a² + b² = c²) in right-angled triangles to find the hypotenuse or a shorter side.
- 16.2Use the trigonometric ratios in right-angled triangles — sin θ = opposite⁄hypotenuse, cos θ = adjacent⁄hypotenuse, tan θ = opposite⁄adjacent — to find unknown sides and, using inverse functions, unknown angles.
- 16.3Solve problems in context by choosing and combining Pythagoras’ theorem and trigonometry in right-angled triangles, in two dimensions.
Higher — Higher tier only
- 16.4Know and use the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90°, and of tan θ for θ = 0°, 30°, 45° and 60°.
- 16.5Apply the sine rule (a⁄sin A = b⁄sin B), the cosine rule (a² = b² + c² − 2bc cos A) and the area formula Area = ½ab sin C in non-right-angled triangles.
- 16.6Apply Pythagoras’ theorem and trigonometric ratios in three-dimensional problems, using right-angled triangles found inside solids.
Key points
- 1Pythagoras: a² + b² = c², where c is the hypotenuse — the side opposite the right angle, always the longest.
- 2Finding a shorter side subtracts: b² = c² − a². Triples worth spotting: 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and their multiples).
- 3SOH CAH TOA: sin = opposite⁄hypotenuse, cos = adjacent⁄hypotenuse, tan = opposite⁄adjacent — label the sides from the angle in the question.
- 4Finding an angle needs the inverse function: tan θ = 5⁄8 gives θ = tan⁻¹(0.625) ≈ 32.0°.
- 5Exact values (Higher): sin 30° = ½, sin 45° = √2⁄2, sin 60° = √3⁄2; cos 60° = ½; tan 45° = 1, tan 60° = √3, tan 30° = 1⁄√3.
- 6Sine rule (Higher): a⁄sin A = b⁄sin B = c⁄sin C — use it with an angle and its opposite side known.
- 7Cosine rule (Higher): a² = b² + c² − 2bc cos A for two sides and the included angle; Area = ½ab sin C.
- 8In 3D (Higher), hunt for the right-angled triangle inside the solid: face diagonals first, then the space diagonal.
Notes · 1 of 4
Pythagoras’ theorem
Squares on the sides: two smaller squares build the biggest.
In any right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides: a² + b² = c². Finding the hypotenuse adds: legs 9 and 12 give c = √(81 + 144) = √225 = 15. Finding a shorter side subtracts: hypotenuse 13 with one leg 5 gives b = √(169 − 25) = √144 = 12. The theorem only works where a right angle actually is — the hypotenuse always sits opposite it and is always the longest side, which makes “my answer is longer than the hypotenuse” an instant error check.
Recognising Pythagorean triples saves time on non-calculator papers: 3-4-5, 5-12-13, 8-15-17, 7-24-25, and any multiple of them (6-8-10, 9-12-15, 10-24-26 …). Applied questions hide right angles in ladders, ramps, rectangles and compass journeys — sketch the triangle, mark the right angle, then decide whether the unknown is the hypotenuse (add) or a leg (subtract).
Notes · 2 of 4
SOH CAH TOA: sides and angles
Label from the angle, pick the ratio, rearrange.
Trigonometry starts by labelling the sides from the angle you care about: the opposite side faces the angle, the adjacent side sits next to it (never the hypotenuse), and the hypotenuse stays opposite the right angle. Then choose the ratio containing your two known sides: SOH (sin = opp⁄hyp), CAH (cos = adj⁄hyp), TOA (tan = opp⁄adj). Finding a side substitutes and multiplies: a 6 m ladder at 70° to the ground reaches 6 × sin 70° = 5.64 m up the wall.
Finding an angle divides first, inverts second: opposite 8 with adjacent 5 gives tan θ = 1.6, so θ = tan⁻¹(1.6) = 58.0° (to 1 d.p.). GCSE answers round angles to 1 d.p. and lengths to 3 significant figures unless told otherwise — and never round a value you still need: keep full accuracy in the calculator through every later step.
Notes · 3 of 4
Exact trigonometric values (Higher)
A half, a root two, a root three — no calculator needed.
Five angles carry exact values the exam expects from memory: 0°, 30°, 45°, 60° and 90°. One table covers them:
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | ½ | √2⁄2 | √3⁄2 | 1 |
| cos θ | 1 | √3⁄2 | √2⁄2 | ½ | 0 |
| tan θ | 0 | 1⁄√3 | 1 | √3 | — |
Two memory hooks: sin rises 0 → ½ → √2⁄2 → √3⁄2 → 1 while cos falls in exactly the opposite order (cos θ = sin(90° − θ)), and tan is the quotient sin⁄cos — which is why tan 30° = (½)⁄(√3⁄2) = 1⁄√3 and tan 60° = √3. Non-calculator questions mix these exact values with surd arithmetic: sin 45° × √2 = (√2⁄2) × √2 = 1.
Notes · 4 of 4
Any triangle, and 3D solids (Higher)
Sine rule, cosine rule, ½ab sin C — and diagonals through space.
Without a right angle, two rules take over. The sine rule, a⁄sin A = b⁄sin B, needs one matched angle-and-side pair: with A = 40°, a = 7 cm and B = 65°, b = 7 sin 65°⁄sin 40° = 9.87 cm. The cosine rule, a² = b² + c² − 2bc cos A, needs two sides and the angle between them: with 8 cm, 11 cm and 52°, the third side is √(64 + 121 − 176 cos 52°) = 8.8 cm (1 d.p.). Areas follow ½ab sin C: ½ × 8 × 11 × sin 52° = 34.7 cm².
Choosing the rule is a two-question checklist. Do I know a side and its opposite angle? → sine rule. Do I know two sides and the angle sandwiched between them? → cosine rule. In 3D problems, no new mathematics appears — just harder triangle-hunting: a cuboid 3 cm × 4 cm × 12 cm has a face diagonal √(3² + 4²) = 5 on the base, and then a space diagonal √(5² + 12²) = √169 = 13 cm rising through the box. Draw each right-angled triangle separately, in 2D, before calculating.
Revise
Key terms
The words examiners expect you to use precisely.
- hypotenuse
- The longest side of a right-angled triangle, opposite the right angle.c in a² + b² = c²; the denominator of sin and cos.
- Pythagoras’ theorem
- In a right-angled triangle the squares of the shorter sides sum to the square of the hypotenuse.a² + b² = c²; subtract to find a shorter side.
- SOH CAH TOA
- The memory hook for the three ratios: sin = opp⁄hyp, cos = adj⁄hyp, tan = opp⁄adj.Label opposite and adjacent from the angle in the question.
- exact trig values
- The surd and fraction values of sin, cos and tan at 0°, 30°, 45°, 60° and 90°.sin 30° = ½, cos 60° = ½, tan 45° = 1, tan 60° = √3.
- sine rule
- In any triangle, a⁄sin A = b⁄sin B = c⁄sin C.Use when an angle and its opposite side are both known.
- cosine rule
- In any triangle, a² = b² + c² − 2bc cos A.Use with two sides and the included angle; also rearranges to find an angle.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Adding the legs instead of their squares: c = a + b.
Pythagoras compares squares, not lengths.
Instead: c = √(a² + b²); 9 and 12 give 15, never 21.
Mistake: Adding when finding a shorter side.
The hypotenuse square is the largest, so the leg is found by subtraction.
Instead: b² = c² − a²; flipping the subtraction goes negative.
Show 10 more mistakesShow fewer mistakes
Mistake: Labelling adjacent from the wrong angle.
Opposite and adjacent swap depending on which acute angle you use.
Instead: Mark the angle in the question first, then label.
Mistake: Using cos for opposite⁄hypotenuse.
The ratio has been mis-copied from SOH CAH TOA.
Instead: sin = opp⁄hyp, cos = adj⁄hyp, tan = opp⁄adj.
Mistake: Leaving the calculator in radians.
Trig functions then answer a different question entirely.
Instead: Set degrees mode for GCSE trigonometry.
Mistake: Rounding halfway through a multi-step calculation.
Early rounding drifts the final answer outside the tolerance.
Instead: Carry full calculator accuracy; round only the final answer.
Mistake: Applying Pythagoras to a triangle with no right angle.
The theorem is a right-angle statement, not a general triangle fact.
Instead: No right angle → sine or cosine rule instead.
Mistake: Quoting sin 30° = √3⁄2 (mixing up with sin 60°).
The exact values are mirrored pairs.
Instead: sin climbs 0, ½, √2⁄2, √3⁄2, 1; cos descends the same list.
Mistake: Writing tan 60° = 1⁄√3.
The reciprocal belongs to tan 30°.
Instead: tan 30° = 1⁄√3 ≈ 0.577; tan 60° = √3 ≈ 1.73.
Mistake: Pairing the sine rule with the wrong opposite side.
a must sit under sin A — its own angle.
Instead: Match each side with the angle facing it.
Mistake: Adding 2bc cos A in the cosine rule.
The term is subtracted; the minus sign encodes the angle’s growth.
Instead: a² = b² + c² − 2bc cos A.
Mistake: Using a face diagonal as the space diagonal in 3D.
The face diagonal lies flat; the space diagonal cuts through the solid.
Instead: Pythagoras twice: face first, then base-to-space.
Check yourself
Quick quiz
30 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Find a hypotenuse with Pythagoras
The two shorter sides of a right-angled triangle measure 6 cm and 8 cm. The longest side is
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Practise
Topical questions
18 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this chapter, in three tiers, each with a model answer, mark scheme and examiner tip. Higher-tier questions are tagged.
Practise topical questionsExam practice
Practice paper
A 30-mark paper in the style of a GCSE paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 16 Practice Paper
30 marks · 40 min · 5 pages
10 multiple-choice questions and structured questions with answer lines. Chapter 16: Pythagoras’ theorem for hypotenuses and shorter sides, SOH CAH TOA for sides and angles, exact trigonometric values, the sine rule, cosine rule and ½ab sin C, the ambiguous case, and 3D diagonal problems.
Mark scheme
PROChapter 16 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
How is this chapter examined?
Paper 1
non-calculator · 1 h 30
Exact fractions, surds and multiples of π expected
Papers 2 and 3
calculator · 1 h 30 each
Non-exact answers to 3 s.f., angles to 1 d.p.
Tiers
Foundation (grades 1–5) · Higher (grades 4–9)
Higher assumes all Foundation content
Want a full paper? Try the Paper 1-style (Foundation), Papers 2 & 3 style (Foundation), Paper 1-style (Higher), Papers 2 & 3 style (Higher) sets.
Questions
Frequently asked questions
How do I know whether to add or subtract in Pythagoras?
Ask which side is unknown. The hypotenuse is the sum: √(a² + b²). A shorter side is the difference: √(c² − a²). A quick sanity check settles it either way — the hypotenuse must come out longer than either leg, a leg shorter than the hypotenuse.
Which of SOH CAH TOA do I use?
Label the two sides you know (or the side you want) with respect to the angle in the question. If the hypotenuse is involved and the opposite side matters, use sin; hypotenuse with adjacent, cos; opposite with adjacent, tan. Writing O, A and H on your sketch first removes almost every wrong choice.
Which exact values do I have to memorise?
Sin and cos at 0°, 30°, 45°, 60° and 90°, and tan at 0°, 30°, 45° and 60°. Learn sin climbing 0, ½, √2⁄2, √3⁄2, 1 and remember cos runs the same list backwards; tan is sin ÷ cos, so tan 30° = 1⁄√3, tan 45° = 1 and tan 60° = √3.
How do I pick between the sine rule and the cosine rule?
Look for a matched pair: if you know a side and the angle directly opposite it, the sine rule works. If instead you know two sides and the angle trapped between them, the cosine rule works. For areas without a right angle, ½ab sin C needs exactly the cosine rule’s data — two sides and the included angle.
Up next
Chapter 17: Perimeter, area and volume