Trigonometry is one of the most consistently examined topics across all GCSE Maths papers. For Foundation students, mastering SOH-CAH-TOA is essential for multiple guaranteed marks. For Higher students, the sine and cosine rules extend this to non-right-angled triangles. This guide covers everything from first principles to Higher tier in one place.
Part 1: SOH-CAH-TOA (Foundation and Higher)
What Is SOH-CAH-TOA?
SOH-CAH-TOA is a mnemonic for the three trigonometric ratios used in right-angled triangles:
SOH: sin θ = Opposite / Hypotenuse
CAH: cos θ = Adjacent / Hypotenuse
TOA: tan θ = Opposite / Adjacent
These ratios relate the angles of a right-angled triangle to the ratios of its sides. Before using any of them, you must correctly identify the three sides relative to the angle you are working with.
Labelling the Sides
Given angle θ in a right-angled triangle:
- Hypotenuse (H): Always the longest side, always opposite the right angle
- Opposite (O): The side directly opposite angle θ
- Adjacent (A): The side next to angle θ (not the hypotenuse)
Critical step: Label the three sides before writing any formula. Mislabelling is the most common error in trigonometry questions.
Finding a Missing Side
When to use which ratio
- If you are using or finding the hypotenuse → use sin or cos
- If the hypotenuse is not involved → use tan
- If the opposite side is involved → use sin or tan
- If the adjacent side is involved → use cos or tan
Method: Finding a missing side
Step 1: Label O, H, A relative to the given angle.
Step 2: Identify which sides are involved (given + required) and choose the appropriate ratio.
Step 3: Write the formula and substitute the known values.
Step 4: Solve for the unknown side.
Worked Example 1 — Find the opposite side
In a right-angled triangle, the hypotenuse is 15 cm and the angle is 32°. Find the opposite side.
Step 1: H = 15, θ = 32°, need O.
Step 2: Opposite and Hypotenuse → use sin.
Step 3: sin 32° = O / 15
Step 4: O = 15 × sin 32° = 15 × 0.5299 = 7.95 cm (to 3 s.f.)
Worked Example 2 — Find the adjacent side
A right-angled triangle has a hypotenuse of 12 cm and an angle of 47°. Find the adjacent side.
cos 47° = A / 12 → A = 12 × cos 47° = 12 × 0.6820 = 8.18 cm (to 3 s.f.)
Worked Example 3 — Use tan (no hypotenuse involved)
A right-angled triangle has an opposite side of 9 cm and an angle of 55°. Find the adjacent side.
tan 55° = 9 / A → A = 9 / tan 55° = 9 / 1.428 = 6.30 cm (to 3 s.f.)
Finding a Missing Angle
When all three sides are known but an angle is missing, use the inverse trigonometric functions: sin⁻¹, cos⁻¹, or tan⁻¹.
Method: Finding a missing angle
Step 1: Label O, H, A.
Step 2: Choose the ratio using the two sides you know.
Step 3: Write the ratio as a decimal (e.g. sin θ = 5/13 = 0.385).
Step 4: Apply the inverse function: θ = sin⁻¹(0.385) on your calculator.
Worked Example 4 — Find an angle using sin
A right-angled triangle has opposite = 5 cm and hypotenuse = 13 cm. Find the angle.
sin θ = 5/13 = 0.3846
θ = sin⁻¹(0.3846) = 22.6° (to 1 d.p.)
Worked Example 5 — Find an angle using cos
Adjacent = 7 cm, hypotenuse = 11 cm. Find the angle.
cos θ = 7/11
θ = cos⁻¹(7/11) = 50.5° (to 1 d.p.)
Worked Example 6 — Real-world context (typical AQA/Edexcel style)
A ramp rises 1.2 m over a horizontal distance of 5 m. Find the angle of inclination.
The triangle has opposite = 1.2, adjacent = 5.
tan θ = 1.2 / 5 = 0.24
θ = tan⁻¹(0.24) = 13.5° (to 1 d.p.)
Trigonometry in Non-Right-Angled Triangles (Higher)
Standard SOH-CAH-TOA only works in right-angled triangles. For triangles without a right angle, Higher tier students need two additional tools: the sine rule and the cosine rule.
The standard labelling convention for these rules:
- Sides are labelled a, b, c (lowercase)
- Angles opposite those sides are labelled A, B, C (uppercase)
- So side a is opposite angle A, side b opposite angle B, etc.
The Sine Rule
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Or equivalently (for finding angles):
$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
When to use the sine rule
Use the sine rule when you know:
- Two angles and one side (AAS or ASA) — to find another side
- Two sides and a non-included angle (SSA) — to find another angle
Worked Example 7 — Find a missing side using sine rule
In triangle ABC: angle A = 48°, angle B = 65°, side a = 12 cm. Find side b.
$$\frac{b}{\sin 65°} = \frac{12}{\sin 48°}$$
$$b = \frac{12 \times \sin 65°}{\sin 48°} = \frac{12 \times 0.9063}{0.7431} = \frac{10.876}{0.7431} = \textbf{14.6 cm (to 3 s.f.)}$$
Worked Example 8 — Find a missing angle using sine rule
In triangle ABC: angle A = 35°, side a = 8 cm, side b = 12 cm. Find angle B.
$$\frac{\sin B}{12} = \frac{\sin 35°}{8}$$
$$\sin B = \frac{12 \times \sin 35°}{8} = \frac{12 \times 0.5736}{8} = \frac{6.883}{8} = 0.8604$$
$$B = \sin^{-1}(0.8604) = \textbf{59.4° (to 1 d.p.)}$$
Note on the ambiguous case: If sin B < 1 but greater than the value needed for 90°, there may be two possible triangles. The exam will usually indicate which is required, or the context makes it obvious.
The Cosine Rule
To find a missing side: $$a^2 = b^2 + c^2 - 2bc\cos A$$
To find a missing angle: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
When to use the cosine rule
Use the cosine rule when you know:
- Two sides and the included angle (SAS) — to find the third side
- All three sides (SSS) — to find any angle
Worked Example 9 — Find a missing side using cosine rule
In triangle ABC: b = 9 cm, c = 7 cm, angle A = 110°. Find side a.
$$a^2 = 9^2 + 7^2 - 2(9)(7)\cos 110°$$
$$a^2 = 81 + 49 - 126 \times (-0.342)$$
$$a^2 = 130 + 43.1 = 173.1$$
$$a = \sqrt{173.1} = \textbf{13.2 cm (to 3 s.f.)}$$
Note: cos 110° is negative (110° is obtuse). This increases a², giving a longer side — which makes sense geometrically.
Worked Example 10 — Find a missing angle using cosine rule
A triangle has sides a = 5, b = 7, c = 8. Find angle A.
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{49 + 64 - 25}{2 \times 7 \times 8} = \frac{88}{112} = 0.7857$$
$$A = \cos^{-1}(0.7857) = \textbf{38.2° (to 1 d.p.)}$$
Area of a Triangle Using Trigonometry (Higher)
When you know two sides and the included angle, you can find the area without the height:
$$\text{Area} = \frac{1}{2}ab\sin C$$
Worked Example 11
A triangle has sides 8 cm and 11 cm with an included angle of 67°. Find its area.
$$\text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin 67° = \frac{1}{2} \times 88 \times 0.9205 = \textbf{40.5 cm}^2 \text{ (to 3 s.f.)}$$
Which Rule to Use — Decision Guide
| Information given | Find | Rule |
|---|---|---|
| Two sides + included angle | Third side | Cosine rule |
| All three sides | Any angle | Cosine rule |
| Two angles + any side | A side | Sine rule |
| Two sides + non-included angle | An angle | Sine rule |
| Right-angled triangle | Side or angle | SOH-CAH-TOA |
| Two sides + included angle | Area | ½ab sin C |
Common Mistakes in Trigonometry Questions
Mistake 1: Using SOH-CAH-TOA on a non-right-angled triangle. Always check first: is there a right angle? If not, you need sine or cosine rule.
Mistake 2: Mislabelling opposite and adjacent. Always label sides relative to the angle you are working with — not the right angle.
Mistake 3: Forgetting that cos is negative for obtuse angles. When applying the cosine rule with an obtuse angle, the −2bc cos A term becomes + (since cos A is negative). Students often forget this and get the wrong answer.
Mistake 4: Not rounding at the final step. Keep full calculator precision throughout multi-step calculations and only round at the end.
Mistake 5: Not giving the angle as a degree. Your calculator should be in degrees (DEG) mode, not radians (RAD). Check this at the start of every paper.
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