Sine, Cosine and Tangent Ratios Revision Notes | GCSE AQA Higher Maths

1. Why Trigonometry Exists

Imagine you need to find the height of a building without climbing it, or the length of a ramp without measuring along the slope. You know one side and one angle — but how do you find the rest? That is exactly the problem trigonometry solves.

Trigonometry is the branch of mathematics that connects the angles and sides of right-angled triangles. Once you know any angle (other than the right angle) and any one side, you can calculate every other measurement in the triangle.

At GCSE, you need three ratios: sine, cosine, and tangent. These are not arbitrary — each one is a fixed relationship between two specific sides for a given angle, and that relationship holds for every right-angled triangle with that angle, regardless of size.

2. Labelling the Sides

Before you can use any trig ratio, you must label the three sides correctly. The labels depend on which angle you are working with — call it $\theta$ (theta).

Once you have found the hypotenuse, look at your chosen angle $\theta$. The side directly across from $\theta$ is the opposite (O). The side that runs from $\theta$ to the right angle — touching $\theta$ but not being the hypotenuse — is the adjacent (A).

3. The Three Trigonometric Ratios

With the sides labelled, you can now write down the three ratios. Each ratio compares two sides and is linked to the angle $\theta$. The ratios are fixed — for any right-angled triangle with the same angle, the ratio of those two sides is always identical.

The memory aid SOH CAH TOA is the fastest way to recall all three ratios under exam pressure. Say it out loud a few times: SOH — Sine is Opposite over Hypotenuse. CAH — Cosine is Adjacent over Hypotenuse. TOA — Tangent is Opposite over Adjacent.

4. Finding an Unknown Side

When a side length is missing, you set up the appropriate ratio, substitute the known values, then rearrange to isolate the unknown. The key decision is always: which two sides are involved — and which ratio connects them?

Worked example: Finding an Unknown Side Using Sine

A right-angled triangle has an angle of $35^\circ$ and a hypotenuse of 12 cm. Find the side opposite the $35^\circ$ angle. Give your answer to 1 decimal place. (calculator-allowed exam question)

1. Hypotenuse $= 12$ cm (given) Opposite $= x$ (unknown) Angle $\theta = 35^\circ$ — Always label H, O, A relative to the angle you are using before writing any ratio. [M1 — correct identification of sides]
2. Sides involved: Opposite and Hypotenuse $\Rightarrow$ Use $\sin\theta = \dfrac{\text{O}}{\text{H}}$ — SOH connects Opposite and Hypotenuse. This is the correct choice. [M1 — correct ratio selected]
3. $\sin 35^\circ = \dfrac{x}{12}$ — Replace $\theta$ with $35^\circ$ and H with 12. The unknown $x$ stays in the numerator.
4. $x = 12 \times \sin 35^\circ$ — Multiply both sides by 12 to isolate $x$. [M1 — correct rearrangement]
5. $x = 12 \times 0.5736\ldots = 6.883\ldots$ — $\sin 35^\circ \approx 0.5736$. Multiply: $12 \times 0.5736 = 6.883$. Keep full calculator value until the final step.
6. $x \approx 6.9$ cm — Round to 1 decimal place as instructed. Always include units. [A1 — correct answer with units]

Answer: The opposite side is approximately $6.9$ cm.

Notice the rearrangement pattern: when the unknown is in the numerator, multiply both sides by the denominator. When the unknown is in the denominator, you divide instead — covered in the next example.

Worked example: Finding an Unknown Side Using Tangent (Unknown in Denominator)

In a right-angled triangle, angle $\theta = 48^\circ$. The side opposite $\theta$ is 9 cm. Find the adjacent side. Give your answer to 1 decimal place. (calculator-allowed exam question)

1. Opposite $= 9$ cm (given) Adjacent $= x$ (unknown) Angle $\theta = 48^\circ$ — Opposite and adjacent are involved — no hypotenuse. [M1 — correct identification]
2. Sides involved: Opposite and Adjacent $\Rightarrow$ Use $\tan\theta = \dfrac{\text{O}}{\text{A}}$ — TOA connects Opposite and Adjacent. Correct choice.
3. $\tan 48^\circ = \dfrac{9}{x}$ — The unknown $x$ is in the denominator this time. [M1 — correct substitution]
4. $x = \dfrac{9}{\tan 48^\circ}$ — Multiply both sides by $x$, then divide both sides by $\tan 48^\circ$. Alternatively: cross-multiply to get $x \times \tan 48^\circ = 9$, then divide. [M1 — correct rearrangement]
5. $\tan 48^\circ \approx 1.1106$ $x = \dfrac{9}{1.1106} = 8.103\ldots$ — Divide 9 by $\tan 48^\circ$. Keep the full decimal until the final step.
6. $x \approx 8.1$ cm — Round to 1 decimal place. Include units. [A1 — correct answer with units]

Answer: The adjacent side is approximately $8.1$ cm.

5. Finding an Unknown Angle

So far you have used trig ratios to find sides. The process reverses when you know two sides and need the angle. Instead of $\sin$, $\cos$, or $\tan$, you use their inverse functions: $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$.

Think of the inverse like an undo button. If $\sin\theta = 0.5$, then $\theta = \sin^{-1}(0.5) = 30^\circ$. On your calculator, the inverse functions are usually accessed with the SHIFT or 2nd key followed by the trig button.

Worked example: Finding an Unknown Angle Using Cosine Inverse

A right-angled triangle has an adjacent side of 7 cm and a hypotenuse of 11 cm. Find the angle $\theta$ between the adjacent and hypotenuse. Give your answer to 1 decimal place. (calculator-allowed exam question)

1. Adjacent $= 7$ cm Hypotenuse $= 11$ cm Angle $\theta =?$ — Two sides are known: adjacent and hypotenuse. The angle is unknown. [M1 — correct identification]
2. Sides involved: Adjacent and Hypotenuse $\Rightarrow$ Use $\cos\theta = \dfrac{\text{A}}{\text{H}}$ — CAH connects adjacent and hypotenuse. Correct choice.
3. $\cos\theta = \dfrac{7}{11}$ — Write the ratio with the known numbers. [M1 — correct substitution]
4. $\theta = \cos^{-1}\!\left(\dfrac{7}{11}\right)$ — To find the angle, apply $\cos^{-1}$ to both sides. On a calculator: SHIFT then COS, then enter $7 \div 11$. [M1 — correct use of inverse]
5. $\dfrac{7}{11} = 0.6\overline{36}$ $\theta = \cos^{-1}(0.6364\ldots) = 50.478\ldots^\circ$ — Keep the full decimal in the calculator. Do not round $7 \div 11$ before pressing $\cos^{-1}$.
6. $\theta \approx 50.5^\circ$ — Round to 1 decimal place. Angles are measured in degrees. [A1 — correct answer]

Answer: $\theta \approx 50.5^\circ$

6. Choosing the Right Ratio — A Decision Framework

The most common source of errors in trig questions is choosing the wrong ratio. A reliable three-step decision process removes that risk every time.

• Step 1 — Label: Mark H (hypotenuse), O (opposite relative to $\theta$), and A (adjacent relative to $\theta$) on your diagram.
• Step 2 — Identify: Which two sides are involved? (One known, one unknown — or both known if finding an angle.)
• Step 3 — Match: O and H → use sine (SOH). A and H → use cosine (CAH). O and A → use tangent (TOA).

7. Exam Technique and Common Mistakes

Trig questions on GCSE Higher exam papers reward method marks heavily. Even if your final answer is slightly off due to a rounding slip, you can still collect marks for setting up the correct ratio and rearranging correctly. Always show your ratio equation before substituting numbers.

One of the most reliable mark-winners is writing the ratio in full before doing anything else — for example, writing $\sin 35^\circ = \dfrac{x}{12}$ earns a method mark even if the arithmetic goes wrong afterwards. Examiners call this the method mark (M1), and it is yours as long as the setup is correct.

Mastering the three trig ratios — sine, cosine, and tangent — and knowing when to apply each one covers the core of the GCSE Higher specification requirement for trigonometric calculation in right-angled triangles. Every question reduces to the same process: label, identify, match, substitute, rearrange, calculate.