Trigonometry — Sine Rule and Cosine Rule (Extended)
SOHCAHTOA only works for right-angled triangles. The sine and cosine rules extend trigonometry to any triangle. At Year 4 Advanced, you must choose and justify the appropriate rule for each problem.
What You'll Learn
- Apply the sine rule to find unknown sides and angles in non-right triangles
- Apply the cosine rule when given SAS (two sides and included angle) or SSS (three sides)
- Calculate the area of any triangle using ½ab sin C
- Recognise and handle the ambiguous case of the sine rule
- Choose between sine rule and cosine rule with justification
- Solve real-world problems involving bearings, navigation, and surveying
IB Assessment Focus
Criterion A: Apply sine/cosine rules to unfamiliar problems (bearing, surveying, 3D).
Criterion B: Justify the choice of rule; explain why the cosine rule reduces to Pythagoras when the angle is 90°.
Criterion C: Label diagrams clearly; use correct notation; show logical working.
Criterion D: Round appropriately and justify accuracy in context (nearest degree, 1 d.p., etc.).
Key Vocabulary
| Term | Definition |
|---|---|
| Sine rule | a/sin A = b/sin B = c/sin C — relates sides to their opposite angles |
| Cosine rule | a² = b² + c² − 2bc cos A — a generalisation of Pythagoras |
| Included angle | The angle between two known sides |
| Opposite side | The side directly across from a given angle in a triangle |
| Ambiguous case | When using the sine rule to find an angle, two solutions may exist (sin θ = sin(180° − θ)) |
| Bearing | A direction measured clockwise from north, expressed as a three-figure angle (e.g., 045°) |
| Non-right triangle | A triangle with no 90° angle — SOHCAHTOA does not apply directly |
The Sine Rule
The sine rule connects each side of a triangle with the sine of its opposite angle. Use it when you know a side and its opposite angle plus one more piece of information.
When to Use the Sine Rule
- AAS / ASA: Two angles and one side known → find the remaining side
- SSA: Two sides and an angle opposite one of them → find the opposite angle (watch for ambiguous case)
Finding an Angle
The Cosine Rule
The cosine rule is a generalisation of Pythagoras’ theorem. When the included angle is 90°, the −2bc cos A term vanishes (since cos 90° = 0), giving a² = b² + c².
When to Use the Cosine Rule
- SAS: Two sides and the included angle → find the third side
- SSS: Three sides known → find any angle
Area of Any Triangle
When you know two sides and the included angle, use the sine area formula instead of base × height.
Choosing the Right Formula
| Information Given | Use |
|---|---|
| Base and perpendicular height | Area = ½ × base × height |
| Two sides and included angle | Area = ½ ab sin C |
| Three sides (no angle) | Use cosine rule first to find an angle, then use ½ ab sin C |
The Ambiguous Case
When using the sine rule to find an angle (SSA configuration), there can be two valid triangles. This happens because sin θ = sin(180° − θ).
- sin B = b sin A / a = 10 × sin 40° / 7 = 10 × 0.6428 / 7 = 0.9183.
- B = sin−1(0.9183) = 66.8°.
- Supplementary angle: B’ = 180° − 66.8° = 113.2°.
- Check: A + B’ = 40° + 113.2° = 153.2° < 180° — geometrically valid!
- Both B = 66.8° and B = 113.2° produce valid triangles. This is the ambiguous case.
Worked Examples
Multi-step solutions showing the reasoning expected at Year 4 Advanced.
c² = a² + b² − 2ab cos C = 14² + 9² − 2(14)(9) cos 72°
= 196 + 81 − 252 × 0.3090 = 277 − 77.87 = 199.13.
c = √199.13 = 14.11 cm (2 d.p.).
At the turning point, the angle between the two legs: the ship was heading 060° (measuring from north clockwise). It then turns to 150°. The angle turned = 150° − 60° = 90°. The interior angle of the triangle at this point = 180° − 90° = 90°... Actually, the included angle = 150° − 60° = 90°.
Using cosine rule with included angle 90°:
d² = 15² + 20² − 2(15)(20) cos 90° = 225 + 400 − 0 = 625.
d = 25 km.
(When the included angle is 90°, the cosine rule reduces to Pythagoras.)
Let a = 14, b = 8, c = 11.
cos A = (b² + c² − a²) / 2bc = (64 + 121 − 196) / (2 × 8 × 11) = −11/176 = −0.0625.
A = cos−1(−0.0625) = 93.58°.
Area = ½ × 8 × 11 × sin 93.58° = 44 × 0.9981 = 43.9 cm² (1 d.p.).
When A = 90°: cos 90° = 0.
Therefore: a² = b² + c² − 2bc(0) = b² + c².
This is Pythagoras’ theorem: the square of the hypotenuse equals the sum of the squares of the other two sides. The cosine rule is therefore a generalisation that works for ALL triangles, not just right-angled ones. ∎
E = sin−1(0.8) = 53.1°.
Supplementary: E’ = 180° − 53.1° = 126.9°.
Check E’: D + E’ = 30° + 126.9° = 156.9° < 180° → valid.
Triangle 1: D = 30°, E = 53.1°, F = 96.9°.
Triangle 2: D = 30°, E = 126.9°, F = 23.1°.
Both are valid — this IS the ambiguous case.
Practice Q&A
Attempt each question before revealing the model answer. Justify your choice of rule.
C = cos−1(−0.275) = 106.0° (1 d.p.). The negative cosine confirms C is obtuse.
d² = 12² + 18² − 2(12)(18) cos 85° = 144 + 324 − 432 × 0.0872 = 468 − 37.67 = 430.33.
d = √430.33 = 20.74 km (2 d.p.).
Flashcard Review
Tap each card to reveal the answer. Try to answer from memory first.
Each side divided by the sine of its opposite angle.
Use with SAS (two sides and included angle).
Use with SSS (three sides known).
Requires two sides and the included angle.