Special Triangles and Their Ratios Revision Notes | GCSE AQA Higher Further Maths

1. Why Special Triangles Matter

On Paper 1 (non-calculator), you cannot reach for a calculator to find $\sin 30°$ or $\cos 45°$. Instead, examiners expect you to recall exact values instantly — and the two special triangles are the engine behind all of them.

These two triangles — the 30-60-90 and the 45-45-90 — each have fixed side ratios. Once you know those ratios, you can read off exact trig values, find missing lengths, and handle a whole family of non-calculator geometry questions with confidence.

2. The 45-45-90 Triangle and the Ratio 1: 1: √2

Start with a square of side length 1. Draw one diagonal. That diagonal cuts the square into two identical right-angled triangles, each with two legs of length 1 and angles of 45°, 45°, and 90°.

By Pythagoras, the hypotenuse satisfies $h^2 = 1^2 + 1^2 = 2$, so $h = \sqrt{2}$. The three sides are therefore in the ratio $1: 1: \sqrt{2}$, where $\sqrt{2}$ is always the hypotenuse.

This ratio scales. If the legs are each 5 cm, the hypotenuse is $5\sqrt{2}$ cm. If the hypotenuse is $8\sqrt{2}$, each leg is 8. The multiplier stays consistent throughout.

3. The 30-60-90 Triangle and the Ratio 1: √3: 2

Now picture an equilateral triangle with all sides equal to 2. Every interior angle is 60°. Draw the perpendicular from one vertex to the opposite side — this bisects both the base and the top angle, creating two congruent right-angled triangles.

Each new triangle has a hypotenuse of 2, a base of 1, and angles of 30°, 60°, and 90°. By Pythagoras, the remaining side satisfies $h^2 = 2^2 - 1^2 = 3$, giving $h = \sqrt{3}$. The ratio of sides is therefore $1: \sqrt{3}: 2$.

The key anchor here: the side opposite 30° is always the shortest, and the hypotenuse is exactly double it. The side opposite 60° sits in between, at $\sqrt{3}$ times the shortest side.

4. Reading Off Exact Trigonometric Values

These two triangles are the source of every exact trig value you need at GCSE. Once the ratios are fixed, you can read $\sin$, $\cos$, and $\tan$ directly from the sides using SOH CAH TOA.

Notice the symmetry: $\sin 30° = \cos 60° = \frac{1}{2}$ and $\sin 60° = \cos 30° = \frac{\sqrt{3}}{2}$. This happens because 30° and 60° are complementary angles — swapping opposite and adjacent swaps sine and cosine.

5. Worked Examples: Finding Missing Lengths

Knowing the ratios is one thing; applying them under exam conditions is another. The method is always the same: identify which triangle you have, label the known side against its ratio position, find the scale factor, then multiply every ratio value by that scale factor.

Worked example: Finding a Missing Side in a 30-60-90 Triangle

A right-angled triangle has angles 30°, 60°, and 90°. The side opposite 60° has length $6\sqrt{3}$ cm. Find the exact length of the hypotenuse.

1. Ratio for 30-60-90: $1: \sqrt{3}: 2$ (opposite 30°): (opposite 60°): (hypotenuse) — Always write the ratio with labels so you know which position each side occupies. [M1]
2. Side opposite 60° $= 6\sqrt{3}$ cm, which corresponds to $\sqrt{3}$ in the ratio. — Match the given side to its ratio position before finding the scale factor.
3. $\text{Scale factor} = \dfrac{6\sqrt{3}}{\sqrt{3}} = 6$ — Divide the actual length by the ratio value for that side. [M1]
4. Hypotenuse $= 2 \times 6 = 12$ cm — Multiply the hypotenuse ratio value (2) by the scale factor (6). [A1]
5. Shortest side $= 1 \times 6 = 6$ cm. Check: hypotenuse should be double the shortest side. $2 \times 6 = 12$ ✓ — In a 30-60-90 triangle, the hypotenuse is always exactly twice the shortest side.

Answer: Hypotenuse $= 12$ cm

Worked example: Finding a Missing Side in a 45-45-90 Triangle

An isosceles right-angled triangle has a hypotenuse of length $9\sqrt{2}$ cm. Find the exact length of each leg.

1. Ratio for 45-45-90: $1: 1: \sqrt{2}$ (leg): (leg): (hypotenuse) — The hypotenuse corresponds to $\sqrt{2}$ in the ratio. [M1]
2. $\text{Scale factor} = \dfrac{9\sqrt{2}}{\sqrt{2}} = 9$ — Divide the actual hypotenuse by $\sqrt{2}$. The $\sqrt{2}$ terms cancel cleanly. [M1]
3. Each leg $= 1 \times 9 = 9$ cm — Both legs share ratio value 1, so each equals the scale factor. [A1]
4. $9^2 + 9^2 = 81 + 81 = 162$ $\sqrt{162} = \sqrt{81 \times 2} = 9\sqrt{2}$ ✓ — $9 \times 9 = 81$ and $9 \times 9 = 81$, so $81 + 81 = 162$. Then $\sqrt{162} = 9\sqrt{2}$, confirming the hypotenuse.

Answer: Each leg $= 9$ cm

6. Practice: Apply the Ratios

The two worked examples above covered one triangle each. Now try a short set that mixes both types — the skill is recognising which ratio to reach for before you start calculating.

7. Exam Technique and Common Mistakes

These triangles appear across a wide range of question types — from straightforward "find the missing side" to multi-step problems involving area, perimeter, or trigonometric equations. The mistakes below account for a large proportion of lost marks.

One final point on calculator vs non-calculator: on Papers 2 and 3 you can verify your surd answer numerically, but the method marks still require the ratio and scale-factor working to be shown. A bare decimal with no working scores zero method marks.

Mastering these two ratios means you can handle any non-calculator question involving 30°, 45°, or 60° — from finding exact side lengths to reading off trig values — which directly satisfies both specification requirements for this topic.