G22: Sine & Cosine Rules
Know and apply the sine rule and cosine rule to find unknown lengths and angles
📋 Key Concepts
Sine and cosine rules are used to find unknown sides and angles in non-right-angled triangles.
When to use:
- Sine rule: When you know a pair (side + opposite angle)
- Cosine rule: When you have SAS (two sides + included angle) or SSS (three sides)
📝 The Sine Rule
a/sin A = b/sin B = c/sin C
Use when:
- You know two angles and a side (AAS or ASA)
- You know two sides and a non-included angle (SSA)
Alternative form:
sin A/a = sin B/b = sin C/c
📝 Using the Sine Rule to Find a Side
Example 1
In triangle ABC, angle A = 42°, angle B = 65°, side b = 12 cm. Find side a.
Solution:
a/sin A = b/sin B
a/sin 42° = 12/sin 65°
a = 12 × sin 42°/sin 65°
a = 12 × 0.669/0.906
a = 8.86 cm (to 2 d.p.)
📝 Using the Sine Rule to Find an Angle
Example 2
In triangle PQR, side p = 8 cm, side q = 10 cm, angle P = 40°. Find angle Q.
Solution:
p/sin P = q/sin Q
8/sin 40° = 10/sin Q
sin Q = 10 × sin 40°/8
sin Q = 0.8035
Q = sin⁻¹(0.8035) = 53.4°
Or possibly 180° - 53.4° = 126.6° (ambiguous case)
Ambiguous case: When finding an angle using sine rule, there may be two possible answers (θ and 180° - θ).
📝 The Cosine Rule
a² = b² + c² - 2bc cos A
Use when:
- You know two sides and the included angle (SAS)
- You know all three sides (SSS) and want an angle
📝 Using Cosine Rule to Find a Side
Example 3
In triangle ABC, side b = 7 cm, side c = 9 cm, angle A = 50°. Find side a.
Solution:
a² = b² + c² - 2bc cos A
a² = 7² + 9² - 2 × 7 × 9 × cos 50°
a² = 49 + 81 - 126 × 0.643
a² = 130 - 81.0
a² = 49.0
a = 7.00 cm (to 2 d.p.)
📝 Using Cosine Rule to Find an Angle
cos A = (b² + c² - a²)/(2bc)
Example 4
In triangle ABC, sides are a = 6 cm, b = 8 cm, c = 10 cm. Find angle A.
Solution:
cos A = (b² + c² - a²)/(2bc)
cos A = (8² + 10² - 6²)/(2 × 8 × 10)
cos A = (64 + 100 - 36)/160
cos A = 128/160 = 0.8
A = cos⁻¹(0.8) = 36.9°
📝 Choosing the Right Rule
| Given | Rule to Use |
|---|---|
| Two angles + side | Sine rule |
| Two sides + non-included angle | Sine rule |
| Two sides + included angle | Cosine rule |
| Three sides | Cosine rule |
🧮 Worked Examples — Exam Style
Sine rule (pairs of angle-side opposites) or cosine rule (two sides + included angle, or all three sides). Choosing is half the marks.
Find side a in triangle ABC: angle A = 40°, angle B = 65°, b = 12 cm.
Angle C = 75°. Sine rule: a/sin 40° = 12/sin 65°
a = 12 × sin 40°/sin 65° = 12 × 0.6428/0.9063 = 8.51 cm (3 s.f.)
M1 for the sine rule set-up (find the third angle first), A1 for 8.51. Pair each side with the angle OPPOSITE it.
Triangle sides 8 cm and 11 cm with included angle 52°. Find the third side.
Cosine rule: c² = 8² + 11² − 2 × 8 × 11 × cos 52° = 64 + 121 − 108.1 = 76.9
c = 8.77 cm (3 s.f.)
M1 for the cosine rule, A1 for 8.77. Two sides + INCLUDED angle = cosine rule, always.
Exam wording to expect: non-right-angled triangles: "calculate the length of …", "calculate the size of angle …" (3 s.f.).
❓ Practice Questions
Q1: In triangle ABC: A = 35°, B = 70°, b = 15 cm. Find side a.
Q2: In triangle PQR: p = 9 cm, q = 7 cm, r = 11 cm. Find angle R.
Q3: In triangle XYZ: x = 12 cm, y = 8 cm, angle Z = 60°. Find side z.
Q4: In triangle DEF: d = 10 cm, e = 14 cm, angle F = 45°. Find side f.
Q5: In triangle ABC: a = 7 cm, b = 9 cm, A = 48°. Find angle B.
Q6: In triangle ABC: AB = 11 cm, BC = 13 cm, angle ABC = 74°. Find AC
Q7: In triangle PQR: PQ = 6 cm, PR = 9 cm, angle P = 40°. Find QR (3 s.f.).
✅ Answers
- a = 15 × sin 35°/sin 70° = 9.08 cm
- cos R = (9² + 7² - 11²)/(2 × 9 × 7) = -7/126, R = 93.2°
- z² = 12² + 8² - 2 × 12 × 8 × cos 60° = 112, z = 10.6 cm
- f² = 10² + 14² - 2 × 10 × 14 × cos 45° = 104.3, f = 10.2 cm
- sin B = 9 × sin 48°/7, B = sin⁻¹(0.954) = 72.6° or 107.4°
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. a = 15 × sin 35°/sin 70° = 15 × 0.5736/0.9397 = 9.15 cm (the page shows 9.08 — recompute: 15 × 0.5736 = 8.604; ÷ 0.9397 = 9.155 → 9.15 cm, 3 s.f.)
Q2. cos R = (81 + 49 − 121)/126 = 9/126 = 0.0714 → R = 85.9° (1 d.p.)
Q3. z² = 144 + 64 − 192 × cos 60° = 208 − 96 = 112 → z = 10.6 cm
Q4. f² = 100 + 196 − 280 × cos 45° = 296 − 197.99 = 98.0 → f = 9.90 cm (3 s.f.)
Q5. sin B = 9 × sin 48°/7 = 0.9554 → B = 72.7° or 107.3° (ambiguous case — both valid)
Q6. Cosine rule: AC² = 11² + 13² − 2 × 11 × 13 × cos 74° = 121 + 169 − 286 × 0.2756 = 290 − 78.8 = 211.2 → AC = 14.5 cm (3 s.f.)
Q7. QR² = 36 + 81 − 108 × cos 40° = 117 − 82.7 = 34.3 → QR = 5.86 cm (3 s.f.)
📌 Key Facts to Memorise
- Sine rule: a/sin A = b/sin B = c/sin C — side with its OPPOSITE angle
- Cosine rule (side): c² = a² + b² − 2ab cos C
- Cosine rule (angle): cos C = (a² + b² − c²)/(2ab)
- Choose: two sides + included angle → cosine · angle-side opposite pairs → sine · three sides → cosine for an angle
- Answers to 3 s.f. unless told otherwise; ambiguous sine case gives two angles
🗺️ In the Exam
Non-right triangle trigonometry: Higher-boundary content in this folder — 4–6 marks on Higher papers, often as multi-step journeys (angle then side, or both rules).
Links to: ½ab sin C (G23), Pythagoras (G20), bearings (G15).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Draw and label the triangle clearly
- Check which rule to use based on what you know
- For sine rule finding an angle, check for the ambiguous case
- Always check your calculator is in degrees mode
- Round appropriately at the end, not during calculations
- For cosine rule, make sure you know which angle is opposite which side
🧠 Problem-Solving Strategies
Problem-Solving
Sine rule: a/sinA = b/sinB = c/sinC. Use when you have a side-angle pair or two angles and one side. Cosine rule: a squared = b squared + c squared - 2bc cosA. Use when you have two sides and the included angle, or all three sides. The cosine rule is a generalisation of Pythagoras for non-right-angled triangles.
Multi-Step Problem
In triangle ABC, a = 8 cm, b = 6 cm, C = 50 degrees. Find side c and angle A.
Solution: Using cosine rule: c squared = 8 squared + 6 squared - 2(8)(6)cos(50) = 64 + 36 - 96(0.6428) = 100 - 61.7 = 38.3. c = 6.19 cm. Then sine rule: sinA/8 = sin(50)/6.19. sinA = 8 x 0.766/6.19 = 0.990. A = 81.9 degrees.
⚠️ Common Errors
Watch Out!
1. Wrong: Using the sine rule when you have two sides and the included angle Correct: Two sides and the included angle requires the COSINE rule. The sine rule works best with a side-angle pair or two angles and any side.
2. Wrong: The ambiguous case of the sine rule: forgetting that sin(x) = sin(180-x) gives two possible angles Correct: When finding an angle using the sine rule, check if the supplementary angle (180 minus your answer) is also valid. The ambiguous case arises when the given angle is acute and the opposite side is shorter.
3. Wrong: In the cosine rule, using the wrong side as angle A opposite side a Correct: In a squared = b squared + c squared - 2bc cosA, side a must be OPPOSITE angle A. Always label consistently.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: In triangle ABC, a = 7 cm, b = 9 cm and B = 55 degrees. (a) Find angle A. (b) Find side c. (c) Find the area of the triangle.
(a) Sine rule: sinA/7 = sin(55)/9. sinA = 7 x sin(55)/9 = 7 x 0.8192/9 = 0.6372. A = 39.6 degrees. Check: 180 - 55 - 39.6 = 85.4 for angle C. Also check ambiguous case: A could be 180 - 39.6 = 140.4 but 140.4 + 55 = 195.4 which exceeds 180, so only 39.6 is valid.
(b) Angle C = 85.4 degrees. Sine rule: c/sin(85.4) = 9/sin(55). c = 9 x sin(85.4)/sin(55) = 9 x 0.9966/0.8192 = 10.95 cm.
(c) Area = 1/2 x 7 x 9 x sin(85.4) = 31.5 x 0.9966 = 31.4 cm squared.
Mark scheme: M1 sine rule, A1 angle A, M1 check ambiguous case, A1 side c, M1 area formula, A1 area
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Three towns A, B, C form a triangle. AB = 8 km, AC = 6 km and angle A = 40 degrees.
(a) Calculate BC.
(b) A student says "Since I know all three sides, I can only use the cosine rule now." Is this true?
(c) Find the largest angle in the triangle and explain why it must be opposite the longest side.
Answers: (a) Cosine rule: BC squared = 64 + 36 - 2(8)(6)cos(40) = 100 - 96(0.766) = 100 - 73.5 = 26.5. BC = 5.15 km. (b) Not necessarily — you could also use the sine rule to find other angles first since you have a side-angle pair. But the cosine rule is the most direct method. (c) The longest side is AB = 8 km. The largest angle is opposite AB, which is angle C. Using sine rule: sinC/8 = sin(40)/5.15. sinC = 8 x 0.6428/5.15 = 0.999. C = 87.5 degrees. The largest angle is always opposite the longest side because in any triangle, bigger sides face bigger angles.
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