Key Idea: This topic is your toolkit for solving any triangle — finding missing sides, angles and areas. It runs through geometry and trig questions on both papers, almost always worked by hand on Paper 1 (the GDC only does the final arithmetic).
📐 Right-angled triangles: SOH-CAH-TOA
- hypotenuse — longest side, opposite the right angle
- side opposite the angle θ you are using
- side next to θ (not the hypotenuse)
Know an angle + a side → form the ratio and rearrange for the missing side. Know two sides → form the ratio and take the inverse (sin⁻¹, cos⁻¹, tan⁻¹) for the angle. Three sides, no angle? Use Pythagoras a² + b² = c² — trig is only for when an angle is involved.
🔺 Any triangle: the sine & cosine rules
- side opposite angle A (same for b/B, c/C)
- flip it the other way up when you want an angle
- angle opposite side a — must match the side on the left
- the two sides enclosing angle A
| What you're given | Rule to use |
|---|---|
| A right angle | SOH-CAH-TOA |
| SAS — two sides + the angle between them | Cosine rule (find the third side) |
| SSS — all three sides | Cosine rule, rearranged (find an angle) |
| A side with its opposite angle (AAS / ASA) | Sine rule |
📏 Area of a triangle
- the two sides you know
- the **included** angle — the one *between* sides a and b
Important: The angle in ½ab·sinC must be the included angle (between the two sides you use), not just any angle. If the included angle is missing, find it first with the cosine rule (SSS) or the sine rule, then apply the area formula.
✏️ IB-style worked examples
IB-style question — right-angled triangle (Paper 1)
A ladder leans against a wall, reaching 6 m up. The ladder makes an angle of 65° with the ground. Find the length of the ladder.
Step by step:
The 6 m is opposite the 65° angle; the ladder is the hypotenuse → use sin.
Rearrange (unknown is on the bottom): multiply up, divide.
The ladder is ≈ 6.62 m long.
IB-style question — sine & cosine rules (Paper 1)
In triangle PQR, p = 9, r = 11 and the included angle Q = 58°. Find side q, then angle P.
Step by step:
Two sides + the angle between (SAS) → cosine rule for q.
Evaluate.
Now there's a pair (q opposite Q) → sine rule, flipped, for P.
Solve.
q ≈ 9.85, P ≈ 50.7°.
IB-style question — area, then a missing side (Paper 1)
A triangular garden has two sides of 8 m and 11 m with an included angle of 40°. Find its area, then the length of the third side.
Step by step:
Included angle given → area straight away.
Third side: SAS again → cosine rule.
Evaluate.
Area ≈ 28.3 m²; third side ≈ 7.09 m.
Important: When you find an angle from a sine value, there are two answers: the acute one and its obtuse partner (180° − it). Your GDC only shows the acute one. e.g. sin C = 0.5 gives C = 30° or 150°. Use the context (the longest side faces the largest angle; or whether the angle is labelled acute/obtuse) to pick the right one — or keep both. cos⁻¹ is safe (one answer in a triangle); sin⁻¹ is the trap.
Tap each card to reveal the answer.
Right triangle: opposite 7, adjacent 24. Find θ. tan θ = 7/24, so θ = tan⁻¹(7/24) ≈ 16.3°.
Triangle: a = 12, A = 35°, B = 80°. Find b. Sine rule: b = 12·sin80° / sin35° ≈ 20.6.
Triangle: sides 5, 6, 7. Find the largest angle. Largest angle faces side 7. cos = (5²+6²−7²)/(2·5·6) = 12/60, so ≈ 78.5°.
Which rule for two sides + the angle between them? SAS → the cosine rule (to find the third side).
Area of a triangle with sides 9, 10 and included angle 50°. ½·9·10·sin50° ≈ 34.5.
sin C = 0.6 in a triangle — what are the possible angles? C ≈ 36.9° or 143.1° (acute and its obtuse partner).
Exam Tips
- First decide the rule: right angle → SOH-CAH-TOA; SAS/SSS → cosine; side-with-opposite-angle → sine.
- In the cosine rule, the side on the left must be opposite the angle in the cos term.
- Finding an angle from sin? Check the obtuse partner 180° − θ — the GDC hides it.
- Area needs the INCLUDED angle (between the two sides); find it first if it's missing.
- Keep full accuracy in working and round only the final answer to 3 s.f.