Pythagoras and Trigonometry in 2D Figures Revision Notes | GCSE AQA Higher Maths
Find the hidden right-angled triangle, then use Pythagoras if no angle is involved, or a trigonometric ratio if one is. Free GCSE Maths notes and 9 FAQs.
Updated 24 Mar 2026 · Revise Right Team
Pythagoras and Trigonometry in 2D Figures is part of Mensuration and calculation in GCSE AQA Higher Maths.
3 min
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GCSE AQA Higher
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What this lesson covers
Find the hidden right-angled triangle, then use Pythagoras if no angle is involved, or a trigonometric ratio if one is. Free GCSE Maths notes and 9 FAQs.
Apply Pythagoras' theorem to find missing sides in right-angled triangles within 2D figures
Apply trigonometric ratios (sine, cosine, tangent) to find unknown angles and lengths in right-angled triangles
Identify which theorem or ratio to use given the information provided in a 2D problem
Calculate angles and lengths in composite 2D shapes that contain embedded right-angled triangles
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Use this lesson to stay inside Mensuration and calculation while connecting the detail back to the wider Geometry and measures topic.
Geometry and measures
Mensuration and calculation
GCSE AQA Higher Maths
Revision notes
Read the core explanations from Pythagoras and Trigonometry in 2D Figures before testing yourself from memory.
1. Why This Matters
Architects calculating roof pitches, engineers designing ramps, and surveyors measuring land all rely on the same two tools you are about to master: Pythagoras' theorem and trigonometry. In 2D problems, right-angled triangles are hidden inside rectangles, isosceles triangles, and composite shapes — your job is to spot them and extract the information you need.
This lesson covers both tools together because exam questions frequently require you to decide which one to reach for. By the end, you will have a reliable decision process and the worked practice to back it up.
2. The Two Tools and When to Use Each
Before reaching for a formula, ask yourself one question: do I have an angle involved? If the answer is no — you have two sides and need the third — Pythagoras' theorem is your tool. If the answer is yes — an angle is either given or required — you need a trigonometric ratio.
A reliable memory aid for the three ratios is SOH CAH TOA: Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent. Before substituting any values, label the triangle sides relative to the angle you are working with — this prevents the most common labelling errors.
3. Finding a Missing Side
There are two distinct scenarios for finding a missing side: you either use Pythagoras' theorem (no angle given) or a trigonometric ratio (an angle is given). The worked examples below cover both, starting with Pythagoras.
Worked example: Using Pythagoras' Theorem to Find a Missing Side
A right-angled triangle has legs of length 5 cm and 12 cm. Calculate the length of the hypotenuse. Give your answer to 1 decimal place.
$a^2 + b^2 = c^2$ — State the formula first — this earns the method mark [M1] even if arithmetic goes wrong later.
$5^2 + 12^2 = c^2$ — The two legs are $a = 5$ and $b = 12$. The hypotenuse $c$ is unknown. [M1 for correct substitution]
$5 \times 5 = 25$ and $12 \times 12 = 144$, so $25 + 144 = c^2$ — Expand each square longhand to avoid errors. $25 + 144 = 169$.
$c = \sqrt{169} = 13$ — This is an exact answer — no rounding needed here. [A1 for correct answer with units]
$c = 13 \text{ cm}$ — Always include units. Missing units can cost a mark on GCSE Higher exam papers.
Answer: The hypotenuse is 13 cm.
Now consider the case where an angle is given and you need a missing side. This is where SOH CAH TOA takes over from Pythagoras.
Worked example: Using Trigonometry to Find a Missing Side
In a right-angled triangle, the hypotenuse is 10 cm and one angle is $38^\circ$. Calculate the length of the side opposite the $38^\circ$ angle. Give your answer to 1 decimal place. (calculator-allowed exam question)
Hypotenuse $= 10$ cm (given). Side required $=$ opposite. Adjacent is not involved. — Labelling first prevents choosing the wrong ratio.
Opposite and hypotenuse $\Rightarrow$ use $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ — SOH: Sine links Opposite and Hypotenuse. [M1 for selecting sine]
$\sin 38^\circ = \dfrac{\text{opposite}}{10}$ — $\theta = 38^\circ$ and hypotenuse $= 10$ cm.
$\text{opposite} = 10 \times \sin 38^\circ$ — Multiply both sides by 10. [M1 for correct rearrangement]
$\text{opposite} = 10 \times 0.6157\ldots = 6.157\ldots$ — Keep the full calculator display until the final step — never round mid-calculation.
$\text{opposite} \approx 6.2 \text{ cm}$ — Round to 1 decimal place as instructed. [A1 for 6.2 cm]
Answer: The opposite side is 6.2 cm (to 1 d.p.).
4. Finding a Missing Angle
When two sides are known and you need an angle, the method is almost identical to finding a side — except the final step uses an inverse trigonometric function. On your calculator, this is the $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ button (sometimes labelled as SHIFT or 2nd followed by the ratio button).
Worked example: Using Trigonometry to Find a Missing Angle
A right-angled triangle has an opposite side of 7 cm and an adjacent side of 9 cm. Calculate the angle between the adjacent side and the hypotenuse. Give your answer to 1 decimal place. (calculator-allowed exam question)
Opposite $= 7$ cm, Adjacent $= 9$ cm. Hypotenuse is not given. — The angle sits between the adjacent side and the hypotenuse, so label carefully.
Opposite and adjacent $\Rightarrow$ use $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$ — TOA: Tangent links Opposite and Adjacent. [M1 for selecting tangent]
$\tan\theta = \dfrac{7}{9}$ — Write the ratio with the numbers substituted in.
$\theta = \tan^{-1}\!\left(\dfrac{7}{9}\right)$ — Use the $\tan^{-1}$ function on your calculator. [M1 for correct inverse operation]
$\theta = \tan^{-1}(0.7\overline{7}) = 37.874\ldots^\circ \approx 37.9^\circ$ — Keep the full display until the final rounding step. [A1 for 37.9°]
Answer: $\theta \approx 37.9^\circ$ (to 1 d.p.)
5. Applying Both Tools in 2D Figures
Exam questions rarely present a standalone right-angled triangle. More often, the triangle is embedded inside a rectangle, isosceles triangle, or other 2D shape. The strategy is always the same: identify the right-angled triangle, label its sides, then apply Pythagoras or trigonometry.
A common example is finding the diagonal of a rectangle, or the perpendicular height of an isosceles triangle. In both cases, you draw in the right angle yourself — this is a key problem-solving step that many students skip.
Worked example: Finding the Diagonal of a Rectangle
A rectangle has length 8 cm and width 6 cm. Calculate the length of the diagonal. Give your answer to 1 decimal place.
The diagonal of a rectangle splits it into two right-angled triangles. The legs are the length and width: $a = 8$ cm, $b = 6$ cm. — Drawing a sketch and marking the right angle is worth doing even in an exam — it clarifies the structure.
$c^2 = 8^2 + 6^2$ — No angle is involved, so Pythagoras is the correct tool. [M1]
$c = \sqrt{100} = 10 \text{ cm}$ — This is exact — no rounding needed. [A1 for 10 cm]
Answer: The diagonal is 10 cm.
Worked example: Finding an Angle in a 2D Figure
An isosceles triangle has a base of 10 cm and equal sides of 13 cm. Calculate the base angle (the angle between a base side and an equal side). Give your answer to 1 decimal place. (calculator-allowed exam question)
Drop a perpendicular from the apex to the midpoint of the base. This creates two congruent right-angled triangles. — The perpendicular bisects the base, giving a half-base of $10 \div 2 = 5$ cm. This is the key construction step.
Hypotenuse $= 13$ cm (the equal side). Adjacent $= 5$ cm (half the base). The base angle $\theta$ is at the base corner. — The adjacent side is next to the angle; the hypotenuse is opposite the right angle.
Adjacent and hypotenuse $\Rightarrow$ use $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{5}{13}$ — CAH: Cosine links Adjacent and Hypotenuse. [M1 for selecting cosine]
$\theta = \cos^{-1}\!\left(\dfrac{5}{13}\right)$ — [M1 for correct inverse operation]
$\theta = \cos^{-1}(0.38461\ldots) = 67.380\ldots^\circ \approx 67.4^\circ$ — Keep the full calculator display until the final rounding. [A1 for 67.4°]
Answer: The base angle is approximately $67.4^\circ$ (to 1 d.p.).
6. Deliberate Practice
Now apply both tools yourself. For each problem, decide first whether Pythagoras or trigonometry is needed, then work through the steps. Check your answers below each question.
7. Exam Technique and Common Mistakes
examiners consistently report the same errors in this topic. Knowing where marks are lost — and how to protect them — is as important as knowing the methods themselves.
Applying Pythagoras' theorem and the trigonometric ratios to right-angled triangles in 2D figures covers the full higher-tier specification requirement for this topic: with a reliable decision process (angles involved? — use trig; no angles? — use Pythagoras), correct side labelling, and careful rounding at the final step, you are well placed to secure all available marks.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-24.
Use Pythagoras' theorem, because the diagonal splits the rectangle into two right-angled triangles with the length and width as the legs. For a rectangle 8 cm by 6 cm, the diagonal is √(64 + 36) = √100 = 10 cm.
How do you find the base angle of an isosceles triangle using trigonometry?
Drop a perpendicular from the apex to the midpoint of the base to make two right-angled triangles, then use cosine. For a base of 10 cm and equal sides of 13 cm, cos θ = 5/13, so θ = cos⁻¹(5/13) ≈ 67.4°.
How do you find an angle using tan?
Use tan θ = opposite ÷ adjacent when the hypotenuse is not involved, then apply tan⁻¹. For an opposite side of 7 cm and an adjacent side of 9 cm, θ = tan⁻¹(7/9) ≈ 37.9°.
How do you find right-angled triangles inside 2D shapes?
Right-angled triangles inside 2D shapes are found by first finding the hidden triangle, drawing in the right angle, labelling its sides, then deciding: if no angle is involved, use Pythagoras; if an angle is given or needed, use a trigonometric ratio. Drawing in the right angle is a key step many students skip.
What is the hypotenuse of a triangle with sides 5 cm and 12 cm?
The hypotenuse of a right-angled triangle with legs 5 cm and 12 cm is 13 cm. Using Pythagoras, 5² + 12² = 25 + 144 = 169, and √169 = 13.
Why do you drop a perpendicular in an isosceles triangle?
A perpendicular is drawn from the apex of an isosceles triangle to split it into two congruent right-angled triangles, because it bisects the base. For a base of 10 cm, each half is 5 cm, so trigonometry or Pythagoras can then be used.
How do you use inverse trig on a calculator?
Inverse trig is used on a calculator with the sin⁻¹, cos⁻¹ or tan⁻¹ button, often reached with SHIFT or 2nd followed by the ratio button. For tan θ = 7/9, tan⁻¹(7/9) = 37.9° to 1 decimal place.
When should you round in trigonometry?
In trigonometry, the full calculator display should be kept until the final step and only then rounded. For example, 10 × sin 38° = 6.157..., which is rounded to 6.2 cm only at the end.
How do you decide between Pythagoras and trigonometry in a 2D figure?
The decision depends on angles: if no angle other than the right angle is involved and two sides are known, use Pythagoras; if an angle is given or needed, use a trig ratio from SOH CAH TOA. Labelling the sides first shows which ratio fits.