Pythagoras and Trigonometry in Right-Angled Triangles Revision Notes | GCSE AQA Higher Maths

1. Why Right-Angled Triangles Are Everywhere

A builder checking a roof pitch, a navigator finding a straight-line distance, an engineer calculating a ramp angle — all of them are solving right-angled triangle problems. Two tools handle every such problem at GCSE: Pythagoras' theorem and the trigonometric ratios.

The key decision you face in every question is: which tool do I reach for? By the end of this lesson, that choice will feel automatic. Start with the sides — if no angle is involved, Pythagoras is your tool. If an angle appears (or is needed), trigonometry takes over.

2. Pythagoras' Theorem

Think of the theorem as a balancing act between areas. If you drew a square on each side of the triangle, the two smaller squares together would have exactly the same total area as the large square on the hypotenuse. That relationship is $a^2 + b^2 = c^2$.

Worked example: Finding the Hypotenuse

A right-angled triangle has shorter sides of length 6 cm and 8 cm. Calculate the length of the hypotenuse. [3 marks]

1. $a^2 + b^2 = c^2$ — State Pythagoras' theorem — this earns the method mark [M1].
2. $6^2 + 8^2 = c^2$ — Replace $a$ with 6 and $b$ with 8. Substitution earns [M1].
3. $6 \times 6 = 36$ and $8 \times 8 = 64$, so $36 + 64 = c^2$ — Expand powers longhand to avoid arithmetic errors.
4. $c^2 = 100$, so $c = \sqrt{100} = 10$ — Take the positive square root. Final answer with units [A1].
5. $c = 10 \text{ cm}$ — Always include units — missing units can cost a mark.

Answer: The hypotenuse is 10 cm.

Finding a shorter side works the same way — you rearrange the formula first. If you know the hypotenuse $c$ and one leg $a$, rearrange to get $b^2 = c^2 - a^2$, then square root.

Worked example: Finding a Shorter Side

A right-angled triangle has hypotenuse 13 cm and one shorter side of 5 cm. Calculate the length of the other shorter side. [3 marks]

1. $a^2 + b^2 = c^2 \Rightarrow b^2 = c^2 - a^2$ — Rearrange before substituting — this keeps your working clear [M1].
2. $b^2 = 13^2 - 5^2$ — The hypotenuse is 13, the known leg is 5.
3. $13 \times 13 = 169$ and $5 \times 5 = 25$, so $b^2 = 169 - 25 = 144$ — Expand powers longhand. Subtraction gives 144 [M1].
4. $b = \sqrt{144} = 12$ — Take the positive square root.
5. $b = 12 \text{ cm}$ — Final answer with units [A1].

Answer: The missing shorter side is 12 cm.

3. The Trigonometric Ratios

Pythagoras connects three sides. Trigonometry connects two sides and an angle. Once an angle $\theta$ (other than the right angle) is labelled in your triangle, each side gets a specific name relative to that angle.

With those labels in place, the three trigonometric ratios each pick two of the three sides. A popular memory aid is SOH CAH TOA — it encodes all three ratios in one phrase.

4. Using Trigonometry to Find a Missing Side

When a question gives you one angle and one side and asks for another side, follow a three-step process: label the sides, choose the correct ratio, then rearrange and calculate.

Worked example: Finding a Missing Side Using Trigonometry

In a right-angled triangle, angle $\theta = 40°$. The hypotenuse is 15 cm. Calculate the length of the side opposite $\theta$. Give your answer to 1 decimal place. [3 marks]

1. Known: hypotenuse $= 15$ cm, angle $\theta = 40°$. Need: opposite. — Identifying which sides are involved tells you which ratio to use.
2. $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ (SOH — opposite and hypotenuse involved) — SOH connects opposite and hypotenuse [M1].
3. $\sin 40° = \dfrac{\text{opposite}}{15}$ — Replace $\theta$ with 40° and hypotenuse with 15.
4. $\text{opposite} = 15 \times \sin 40°$ — Multiply both sides by 15 [M1].
5. $\text{opposite} = 15 \times 0.6428... = 9.642...$ — Use the calculator value of $\sin 40°$. Do not round until the final step.
6. $\text{opposite} \approx 9.6 \text{ cm}$ — Round to 1 decimal place as instructed. Include units [A1].

Answer: The opposite side is 9.6 cm (to 1 d.p.).

When the unknown side is in the denominator — for example, finding the hypotenuse when you know the opposite — rearrange before multiplying. Write $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$, then $\text{hypotenuse} = \frac{\text{opposite}}{\sin \theta}$.

Worked example: Finding the Hypotenuse Using Trigonometry

In a right-angled triangle, angle $\theta = 28°$. The side adjacent to $\theta$ is 10 cm. Calculate the length of the hypotenuse. Give your answer to 2 decimal places. [3 marks]

1. Known: adjacent $= 10$ cm, angle $\theta = 28°$. Need: hypotenuse. — Adjacent and hypotenuse — that points to cosine (CAH).
2. $\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ — CAH: cosine links adjacent and hypotenuse [M1].
3. $\cos 28° = \dfrac{10}{\text{hypotenuse}}$, so $\text{hypotenuse} = \dfrac{10}{\cos 28°}$ — The unknown is in the denominator, so divide to isolate it [M1].
4. $\text{hypotenuse} = \dfrac{10}{0.8829...} = 11.326...$ — Use the calculator value of $\cos 28°$. Keep full precision until the final step.
5. $\text{hypotenuse} \approx 11.33 \text{ cm}$ — Round to 2 decimal places as instructed. Include units [A1].

Answer: The hypotenuse is 11.33 cm (to 2 d.p.).

5. Finding a Missing Angle

When two sides are known and you need an angle, the process reverses. You still choose the correct ratio from SOH CAH TOA, but instead of multiplying or dividing, you apply the inverse of that ratio.

The inverse functions are written $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ on your calculator (often labelled as the second function above the sin, cos, tan buttons). They undo the ratio and return the angle.

Worked example: Finding a Missing Angle Using Trigonometry

A right-angled triangle has opposite side 7 cm and hypotenuse 10 cm. Calculate angle $\theta$. Give your answer to 1 decimal place. [3 marks]

1. Known: opposite $= 7$ cm, hypotenuse $= 10$ cm. Need: angle $\theta$. — Opposite and hypotenuse — use sine (SOH).
2. $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{7}{10} = 0.7$ — Calculate the ratio value first [M1].
3. $\theta = \sin^{-1}(0.7)$ — Use the $\sin^{-1}$ function on your calculator to find the angle [M1].
4. $\theta = 44.427...° \approx 44.4°$ — Round to 1 decimal place as instructed [A1].

Answer: $\theta \approx 44.4°$ (to 1 d.p.)

6. Choosing the Right Tool: A Decision Framework

Every right-angled triangle question at GCSE falls into one of two categories. Recognising which category you are in before writing anything is the single most valuable habit you can build.

• Only sides involved (no angle given or needed)? Use Pythagoras' theorem: $a^2 + b^2 = c^2$.
• An angle is given or an angle is needed? Use a trigonometric ratio (SOH CAH TOA).
• Which trig ratio? Label the three sides relative to the given angle, then match the two sides in the question to SOH, CAH, or TOA.
• Finding a side? Substitute and rearrange (multiply or divide).
• Finding an angle? Calculate the ratio value, then apply $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$.

7. Exam Technique and Common Mistakes

The methods above are straightforward when applied carefully. Most marks are lost not through misunderstanding but through avoidable slips. The guidance below targets the errors that appear most frequently in mark schemes.

Together, Pythagoras' theorem ($a^2 + b^2 = c^2$) and the three trigonometric ratios ($\sin \theta$, $\cos \theta$, $\tan \theta$) give you a complete toolkit for every right-angled triangle problem on the GCSE Higher specification. Before calculating, label the hypotenuse, opposite, and adjacent sides clearly. Then decide whether the problem is about sides only, which points to Pythagoras, or about a side-angle relationship, which points to trigonometry. This short pause prevents most method errors and makes it much easier to check whether your answer is sensible once you finish.