Key Idea: Topic 3.2 covers the three core tools of non-right-angle trigonometry: the sine rule, the cosine rule, and the triangle area formula. These handle any triangle when you don't have a right angle. The key is recognising which rule to use based on what information you are given.
✅ Right-angle trig: SOH CAH TOA
- SOH CAH TOA
- sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Only works in right-angled triangles — identify the right angle first.
- Finding an angle
- Use the inverse: θ = sin⁻¹(opp/hyp). Example: opposite = 6, hypotenuse = 10 → θ = sin⁻¹(0.6) = 36.9°.
📐 Non-right-angle rules
| Rule | Formula | When to use |
|---|---|---|
| Sine rule | a/sin A = b/sin B = c/sin C | Use when you have angle + opposite side pair. Also for AAS or ASA. |
| Cosine rule (find side) | a² = b² + c² − 2bc cos A | Use when you have two sides + the included angle (SAS). |
| Cosine rule (find angle) | cos A = (b² + c² − a²) / (2bc) | Use when you have all three sides (SSS). |
| Area of triangle | Area = ½ ab sin C | Use when you have two sides and the included angle. |
Example: Sine rule: Triangle with A = 35°, a = 8, B = 72°. Find b. b/sin 72° = 8/sin 35° → b = 8 × sin72°/sin35° = 13.3 (3 s.f.) Cosine rule (SAS): a = 7, b = 10, C = 50°. Find c. c² = 7² + 10² − 2(7)(10)cos50° = 49 + 100 − 140(0.643) = 59.0 → c = 7.68 Area: Two sides 6 and 9, included angle 40°. Area = ½ × 6 × 9 × sin40° = 17.4 cm²
Ambiguous case of the sine rule: when you have SSA (two sides and a non-included angle), there can be two valid triangles. Check whether 180° − θ is also a valid solution. For 3D trig problems: identify the 2D triangle cross-section inside the 3D shape. Then apply the appropriate 2D rule.
Paper 1 (GDC allowed): Label all sides and angles on your diagram before starting. State which rule you're using. Paper 2 (GDC allowed): The GDC handles all calculations — focus on setting up the correct formula and substituting accurately. Round intermediate values carefully (don't round mid-calculation).
IB-style question [6 marks]
A triangular plot of land has corners P, Q and R. PQ = 40 m, QR = 55 m, and angle PQR = 68°. (a) Find the length PR, correct to 3 significant figures. (b) Find the area of the plot, correct to 3 significant figures. (c) Find the size of angle QPR, correct to 3 significant figures.
Step by step:
(a) You have two sides and the angle between them (SAS), so use the cosine rule. Write it, then substitute.
Work out the right-hand side, then take the square root.
(b) Two sides with the included angle → use the area formula.
Evaluate.
(c) Angle QPR is opposite QR, so use the sine rule. Pair each angle with its opposite side.
Substitute QR = 55, PR = 54.6, Q = 68°, then take the inverse sine.
(a) PR ≈ 54.6 m. (b) Area ≈ 1020 m². (c) angle QPR ≈ 69.2°.