What should I learn from this Further Maths lesson?

By the end of this lesson, you should be able to Apply SOH-CAH-TOA to find unknown sides and angles in right-angled triangles; Identify the hypotenuse, opposite, and adjacent sides relative to a given angle; Calculate missing sides using sine, cosine, and tangent ratios; Find unknown angles using inverse trigonometric functions.

How should I revise Trigonometry Review?

Read each section, then close the notes and explain the idea from memory. Check your explanation against the page and finish with an exam question from Geometry.

Trigonometry Review | GCSE AQA Higher Further Maths

Trigonometry Review is part of Sine and cosine rules in scalene triangles in GCSE AQA Higher Further Maths.

What this lesson covers

Trigonometry Review: Revise trigonometry principles for solving right-angled triangles.

Apply SOH-CAH-TOA to find unknown sides and angles in right-angled triangles
Identify the hypotenuse, opposite, and adjacent sides relative to a given angle
Calculate missing sides using sine, cosine, and tangent ratios
Find unknown angles using inverse trigonometric functions

Key ideas to keep in view

Use this lesson to stay inside Sine and cosine rules in scalene triangles while connecting the detail back to the wider Geometry topic.

Geometry
Sine and cosine rules in scalene triangles
GCSE AQA Higher Further Maths

Revision notes

Read the core explanations from Trigonometry Review before testing yourself from memory.

1. Why Trigonometry Matters

Architects calculate roof angles, surveyors measure land, and engineers design bridges — all using the same three ratios you are about to review. Trigonometry turns a triangle with one known angle and one known side into a fully solved shape.

The key insight is this: in any right-angled triangle, once you fix one of the acute angles, the ratios of the sides are always the same, no matter how big or small the triangle is. That fixed relationship is what sine, cosine, and tangent capture.

2. Labelling the Triangle

Before choosing a ratio, you must label the three sides correctly relative to the angle you are working with — call it $\theta$. The side opposite the right angle is always the hypotenuse (H). The side directly across from $\theta$ is the opposite (O). The side touching $\theta$ that is not the hypotenuse is the adjacent (A).

A common trap: students label the adjacent and opposite correctly for one angle, then forget to re-label when the question switches to the other acute angle. Always re-label from scratch for each new $\theta$.

3. The Three Ratios: SOH-CAH-TOA

The mnemonic SOH-CAH-TOA encodes all three ratios. Each chunk tells you which two sides form the ratio for that function.

Choosing the right ratio is a two-step decision: first identify which two sides are involved (known and unknown), then pick the ratio that connects exactly those two sides. If the hypotenuse is not involved at all, tangent is your ratio.

4. Finding a Missing Side

When a side is unknown, write the chosen ratio as a fraction, substitute the known values, then rearrange to isolate the unknown. The rearrangement follows one of two patterns: multiply across if the unknown is on top, or divide if the unknown is on the bottom.

Worked example: Finding the Opposite Side Using Sine

Triangle $ABC$ has a right angle at $C$. Angle $A = 35^\circ$ and the hypotenuse $AB = 12$ cm. Calculate the length of $BC$ (the side opposite angle $A$). Give your answer to 1 decimal place.

$H = AB = 12$ cm, $\quad O = BC =?$, $\quad$ (opposite is what we need) — BC is directly across from angle A, so it is the opposite. AB is the hypotenuse because it is opposite the right angle at C. [Labelling — no mark yet, but essential for method]
$\sin\theta = \dfrac{O}{H}$ — we have H and need O, so sine fits. — SOH: Sine = Opposite over Hypotenuse. [M1 — correct ratio selected]
$\sin 35^\circ = \dfrac{BC}{12}$ — Replace $\theta$ with $35^\circ$ and H with 12. [M1 — correct substitution]
$BC = 12 \times \sin 35^\circ$ — Multiply both sides by 12 to isolate BC. The unknown was in the numerator, so multiply across.
$BC = 12 \times 0.5736\ldots = 6.883\ldots \approx 6.9$ cm — $\sin 35^\circ \approx 0.5736$. Keep the full decimal until the final step, then round. [A1 — correct answer to 1 d.p.]

Answer: $BC = 6.9$ cm (to 1 decimal place)

Worked example: Finding the Adjacent Side Using Cosine

In right-angled triangle $PQR$, the right angle is at $R$. Angle $P = 52^\circ$ and the hypotenuse $PQ = 9$ cm. Find the length of $PR$ (adjacent to angle $P$). Give your answer to 1 decimal place.

$H = PQ = 9$ cm, $\quad A = PR =?$ — PR runs from angle P to the right angle at R, so it is adjacent to angle P.
$\cos\theta = \dfrac{A}{H}$ — we have H and need A, so cosine fits. — CAH: Cosine = Adjacent over Hypotenuse. [M1 — correct ratio]
$\cos 52^\circ = \dfrac{PR}{9}$ — [M1 — correct substitution]
$PR = 9 \times \cos 52^\circ = 9 \times 0.6157\ldots = 5.541\ldots \approx 5.5$ cm — $\cos 52^\circ \approx 0.6157$. Multiply both sides by 9. [A1 — correct answer to 1 d.p.]

Answer: $PR = 5.5$ cm (to 1 decimal place)

Worked example: Finding the Hypotenuse When It Is in the Denominator

A right-angled triangle has an angle of $40^\circ$. The side opposite this angle is $7$ cm. Find the hypotenuse. Give your answer to 2 decimal places.

$O = 7$ cm, $\quad H =?$ — We have the opposite and need the hypotenuse, so sine is the correct ratio.
$\sin 40^\circ = \dfrac{7}{H}$ — [M1 — correct ratio and substitution]
$H \times \sin 40^\circ = 7$ — This clears H from the denominator.
$H = \dfrac{7}{\sin 40^\circ}$ — Isolate H by dividing. Notice: dividing by $\sin 40^\circ$ (which is less than 1) gives a result larger than 7 — the hypotenuse must be the longest side, so this makes sense.
$H = \dfrac{7}{0.6428\ldots} = 10.893\ldots \approx 10.89$ cm — $\sin 40^\circ \approx 0.6428$. Keep full precision until the final rounding step. [A1 — correct answer to 2 d.p.]

Answer: $H = 10.89$ cm (to 2 decimal places)

5. Finding a Missing Angle

When both relevant sides are known and the angle is unknown, you reverse the process using an inverse trig function. On your calculator, these are labelled $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ — sometimes accessed via a SHIFT or 2nd key.

The inverse function undoes the ratio: if $\sin\theta = 0.5$, then $\theta = \sin^{-1}(0.5) = 30^\circ$. You are asking the calculator: "which angle produces this ratio?"

Worked example: Finding an Unknown Angle Using Tangent

A right-angled triangle has an opposite side of $8$ cm and an adjacent side of $5$ cm. Find the angle $\theta$ between the adjacent side and the hypotenuse. Give your answer to 1 decimal place.

$O = 8$ cm, $\quad A = 5$ cm, $\quad H$ not involved — Opposite and adjacent are known, hypotenuse is not needed. Use tangent.
$\tan\theta = \dfrac{O}{A} = \dfrac{8}{5} = 1.6$ — [M1 — correct ratio and substitution]
$\theta = \tan^{-1}(1.6)$ — Use the $\tan^{-1}$ key on your calculator. [M1 — correct use of inverse]
$\theta = 57.994\ldots^\circ \approx 58.0^\circ$ — [A1 — correct answer to 1 d.p.] Always check: the angle should be between $0^\circ$ and $90^\circ$ for a right-angled triangle.

Answer: $\theta = 58.0^\circ$ (to 1 decimal place)

6. Deliberate Practice

Now that all three ratios and both directions (finding sides and finding angles) have been covered, here are four problems to consolidate the method. For each one, start by labelling the sides before picking a ratio.

7. Exam Technique and Common Mistakes

Trigonometry questions are reliable mark-earners once the method is secure — but examiners know exactly where students drop marks. The guidance below targets the most frequent errors seen in GCSE mark schemes.

One final check worth building into every answer: does the size of your answer make sense? The hypotenuse must always be the longest side. An angle must sit between $0^\circ$ and $90^\circ$. If either condition is violated, revisit the labelling before moving on.

Sources and content standards

Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.

AQA Level 2 Certificate in Further Mathematics (8365) specification How we check content Report a correction — include this lesson URL

Continue through the topic

Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.

Sine and Cosine Rules Area of a Triangle Pythagoras in 2D and 3D 2D Trigonometry and Pythagoras' Theorem Geometry topic GCSE AQA Higher pathway

Trigonometry Review Revision Notes | GCSE AQA Higher Further Maths