Sine & Cosine Rule – Grade 10 IGCSE Maths

Welcome to Sine & Cosine Rule!

The Sine Rule and Cosine Rule extend trigonometry to any triangle — not just right-angled ones. These are essential tools for solving problems involving non-right triangles, bearings, and areas. Mastering them unlocks the full power of IGCSE trigonometry.

Sine Rule: a/sinA = b/sinB = c/sinC | Cosine Rule: a² = b² + c² − 2bc cosA | Area = ½ab sinC

Sine Rule

AAS and SSA cases, finding sides and angles

Ambiguous Case

When SSA gives two possible triangles

Cosine Rule

SAS and SSS cases, Pythagoras connection

Decision Guide

Which rule to use — quick reference

Area Formula

Area = ½ab sinC for any triangle

Bearings Problems

Derive triangle angles from 3-figure bearings

1. The Sine Rule

In any triangle ABC, where side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C:

a/sinA = b/sinB = c/sinC (to find sides)
sinA/a = sinB/b = sinC/c (to find angles)

When to use the Sine Rule:
• AAS — two angles and one side known (find any other side)
• SSA — two sides and the angle opposite one of them (find the other angle)
Rule of thumb: if you know an angle AND the side opposite it, the Sine Rule will work.

Example — Finding a side (AAS):
In triangle ABC: A = 40°, B = 65°, b = 12 cm. Find a.
a/sin40° = 12/sin65° ⟹ a = 12 × sin40° / sin65° = 12 × 0.6428 / 0.9063 ≈ 8.50 cm

Example — Finding an angle (SSA):
In triangle ABC: a = 8, b = 11, A = 35°. Find B.
sinB/11 = sin35°/8 ⟹ sinB = 11 × sin35°/8 = 11 × 0.5736/8 = 0.7887
B = sin⁻¹(0.7887) ≈ 52.0° (check ambiguous case below)

2. The Ambiguous Case (SSA)

When given two sides and an angle NOT between them (SSA), there can be zero, one, or two valid triangles. This is the ambiguous case.

How to check for the ambiguous case:
Given angle A and sides a, b (where a is opposite A):
• If a ≥ b → only one triangle (A is obtuse or a is long enough)
• If b × sinA < a < b → two triangles possible
• If a = b × sinA → exactly one right-angled triangle
• If a < b × sinA → no triangle possible

Finding both solutions:
If B₁ = sin⁻¹(sinB), then the second angle is B₂ = 180° − B₁.
Check: A + B₂ < 180°? If yes, B₂ gives a valid second triangle.
Triangle 1: angles A, B₁, (180°−A−B₁) Triangle 2: angles A, B₂, (180°−A−B₂)

Exam tip: Cambridge questions will often say "find the two possible values" or "show there are two possible triangles". Always substitute back to check both are valid.

3. The Cosine Rule

Finding a side: a² = b² + c² − 2bc cosA
Finding an angle: cosA = (b² + c² − a²) / (2bc)

When to use the Cosine Rule:
• SAS — two sides and the included angle (find the third side)
• SSS — all three sides known (find any angle)
The cosine rule handles situations where the sine rule cannot (no opposite side/angle pair).

Example — Finding a side (SAS):
In triangle ABC: b = 7, c = 5, A = 60°. Find a.
a² = 7² + 5² − 2(7)(5)cos60° = 49 + 25 − 70 × 0.5 = 74 − 35 = 39
a = √39 ≈ 6.24 cm

Example — Finding an angle (SSS):
Triangle with sides a = 9, b = 7, c = 5. Find angle A.
cosA = (7² + 5² − 9²)/(2 × 7 × 5) = (49 + 25 − 81)/70 = −7/70 = −0.1
A = cos⁻¹(−0.1) ≈ 95.7° (obtuse — negative cosine means obtuse angle ✓)

Connection to Pythagoras: When A = 90°, cosA = 0, so the cosine rule becomes a² = b² + c², which is exactly Pythagoras' theorem. The cosine rule is the generalised version.

4. Decision Guide — Which Rule?

Given information	Case	Rule to use
Two angles + any side	AAS or ASA	Sine Rule
Two sides + angle opposite one of them	SSA	Sine Rule (check ambiguous case)
Two sides + included angle	SAS	Cosine Rule
All three sides	SSS	Cosine Rule

Quick check: Do you have a matching angle-side pair? If yes → Sine Rule. If you know the included angle between two sides, or all three sides → Cosine Rule.

5. Area of a Triangle

Area = ½ × a × b × sinC

Where a and b are any two sides and C is the included angle between them. This works for any triangle — you need two sides and the angle between them.

Example: Triangle with sides 6 cm and 9 cm, included angle 48°.
Area = ½ × 6 × 9 × sin48° = 27 × 0.7431 ≈ 20.1 cm²

Connection to base × height:
Area = ½ × base × height ⟹ height = b × sinC (from right-angle trig in the triangle)
So Area = ½ × a × (b sinC) = ½ab sinC ✓

Always make sure C is the angle between sides a and b. Using a non-included angle is the most common error here.

6. Bearings Problems

A bearing is a 3-figure angle measured clockwise from North. Bearing problems often create non-right-angled triangles that require the cosine rule.

Key steps for bearings problems:
1. Draw a clear diagram with North arrows at each point
2. Mark the given bearings (clockwise from North)
3. Use co-interior (allied) or alternate angles to find the interior triangle angle
4. Apply the cosine rule (usually SAS) to find the unknown distance
5. Use sine rule if an angle is needed

Classic example:
A ship sails bearing 070° for 80 km, then bearing 155° for 60 km. Find distance from start.
Interior angle at the turn = 180° − 70° + (180° − 155°) = 180° − 70° + 25° = 135°
Wait — use: interior angle = 180° − (155° − 70°) = 180° − 85° = 95°
d² = 80² + 60² − 2(80)(60)cos95° = 6400 + 3600 − 9600(−0.0872) = 10000 + 837 = 10837
d ≈ 104 km

The interior angle at the turning point: if first bearing is θ₁ and second is θ₂ (θ₂ > θ₁), interior angle = 180° − (θ₂ − θ₁). Draw a North line at the turning point to see this clearly.

Example 1 — Sine Rule: Find a side

Given: Triangle PQR, angle P = 50°, angle Q = 72°, side q = 15 cm. Find p.

Angle R = 180° − 50° − 72° = 58° (so we know all angles)

p/sinP = q/sinQ ⟹ p/sin50° = 15/sin72°

p = 15 × sin50°/sin72° = 15 × 0.7660/0.9511 = 12.1 cm (3 s.f.)

Example 2 — Sine Rule: Find angle, check ambiguous case

Given: Triangle, a = 7 cm, b = 10 cm, A = 42°. Find angle B.

sinB/10 = sin42°/7 ⟹ sinB = 10 × 0.6691/7 = 0.9559

B₁ = sin⁻¹(0.9559) ≈ 72.9° Check A + B₁ = 42 + 72.9 = 114.9° < 180° ✓

B₂ = 180° − 72.9° = 107.1° Check A + B₂ = 42 + 107.1 = 149.1° < 180° ✓

Also: b × sinA = 10 × 0.6691 = 6.69 < a = 7 < b = 10 → ambiguous case confirmed. Two solutions: B ≈ 72.9° or B ≈ 107.1°

Example 3 — Cosine Rule: Find a side (SAS)

Given: b = 8 cm, c = 11 cm, included angle A = 55°. Find side a.

a² = 8² + 11² − 2(8)(11)cos55° = 64 + 121 − 176 × 0.5736

a² = 185 − 100.95 = 84.05

a = √84.05 ≈ 9.17 cm (3 s.f.)

Example 4 — Cosine Rule: Find an angle (SSS)

Given: a = 5 cm, b = 8 cm, c = 6 cm. Find angle C.

cosC = (a² + b² − c²)/(2ab) = (25 + 64 − 36)/(2 × 5 × 8) = 53/80 = 0.6625

C = cos⁻¹(0.6625) ≈ 48.5° (3 s.f.)

Example 5 — Area Formula

Given: Triangle with sides 12 cm and 9 cm, included angle 67°. Find the area.

Area = ½ × 12 × 9 × sin67° = 54 × 0.9205 ≈ 49.7 cm² (3 s.f.)

Example 6 — Bearings Problem

Problem: Port A. Ship sails bearing 065° for 50 km to B, then bearing 148° for 35 km to C. Find AC.

Interior angle at B: supplement of the back-bearing difference. North at B points up. Angle ABC (interior) = 180° − (148° − 065°) = 180° − 83° = 97°

AB = 50, BC = 35, included angle B = 97°

AC² = 50² + 35² − 2(50)(35)cos97° = 2500 + 1225 − 3500(−0.1219) = 3725 + 426.7 = 4151.7

AC = √4151.7 ≈ 64.4 km

Common Mistakes to Avoid

Mistake 1 — Using the Sine Rule for SAS

If you know two sides and the included angle (e.g. sides 5, 8 and angle 40° between them), you CANNOT use the sine rule — you have no matching angle-side pair. You must use the cosine rule: a² = b² + c² − 2bc cosA.

Mistake 2 — Ignoring the Ambiguous Case

In SSA situations, always check if a second angle is possible. Compute B₁ = sin⁻¹(value), then test B₂ = 180° − B₁. If A + B₂ < 180°, both solutions are valid. Giving only one answer when two exist loses marks.

Mistake 3 — Cosine Rule Rearrangement Error

When finding an angle: cosA = (b² + c² − a²)/(2bc). Students often write b² + c² + a² or forget to divide by 2bc. The formula is a rearrangement — always check by starting from a² = b² + c² − 2bc cosA and making cosA the subject.

Mistake 4 — Wrong Angle in Area Formula

Area = ½ab sinC only works when C is the angle BETWEEN sides a and b (the included angle). If you use a non-included angle the formula gives the wrong answer. Always identify which two sides enclose the angle.

Mistake 5 — Bearings Interior Angle Without a Diagram

Never try to calculate a bearing problem without drawing a diagram first. Mark North at every point. Use co-interior (allied) angles (sum to 180°) on parallel North lines to find the triangle's interior angle. Without the diagram, sign errors are almost inevitable.

Key Formulas — Sine & Cosine Rule

Sine Rule (sides): a/sinA = b/sinB = c/sinC

Sine Rule (angles): sinA/a = sinB/b = sinC/c

Cosine Rule (side): a² = b² + c² − 2bc cosA

Cosine Rule (angle): cosA = (b² + c² − a²) / (2bc)

Area of triangle: Area = ½ ab sinC

✓ Cambridge IGCSE provides the Sine Rule and Cosine Rule on the formula sheet. The Area formula (½ ab sinC) is also provided. You still need to know WHEN and HOW to apply them.

Labelling convention:
Side a is opposite angle A. Side b is opposite angle B. Side c is opposite angle C.
This is consistent in both rules — always identify your opposite pairs first.

Ambiguous case condition: SSA with b × sinA < a < b → two triangles exist. The two angles are B and (180° − B).

Bearings: 3-figure clock-wise from North. Interior angle at turning point = 180° − (second bearing − first bearing) when second bearing > first bearing.

Triangle Solver & Visualiser

Select the case, enter known values, and see the solved triangle drawn to scale.

Case:

Select a case, enter values, and click Solve.

Exercise 1 — Sine Rule: Find Sides

Use the sine rule to find the unknown side. Give answers to 3 significant figures.

1. Triangle: A=50°, B=70°, b=14 cm. Find a (cm).

2. Triangle: P=35°, Q=80°, q=20 cm. Find p (cm).

3. Triangle: A=62°, C=48°, c=9 cm. Find a (cm).

4. Triangle: B=44°, C=76°, c=18 cm. Find b (cm).

5. Triangle: A=30°, B=45°, a=8 cm. Find b (cm).

6. Triangle: X=55°, Y=65°, y=22 cm. Find x (cm).

7. Triangle: A=72°, B=38°, b=6 cm. Find a (cm).

8. Triangle: P=25°, R=110°, r=30 cm. Find p (cm).

Exercise 2 — Sine Rule: Find Angles

Find the unknown angle in degrees. Give answers to 1 decimal place. For the ambiguous case, give the smaller angle.

1. a=6, b=8, A=40°. Find B (degrees).

2. a=5, c=7, A=30°. Find C (degrees).

3. b=10, c=12, B=50°. Find C (degrees).

4. a=9, b=11, A=42°. Find B (degrees, smaller value).

5. p=7, q=9, P=48°. Find Q (degrees).

6. a=13, b=15, A=55°. Find B (degrees).

7. x=8, y=6, X=70°. Find Y (degrees).

8. a=20, b=25, A=45°. Find B (degrees, smaller value).

Exercise 3 — Cosine Rule: Find Sides

Use the cosine rule to find the unknown side. Give answers to 3 s.f.

1. b=5, c=7, A=60°. Find a (cm).

2. a=8, c=10, B=45°. Find b (cm).

3. b=6, c=9, A=35°. Find a (cm).

4. a=12, b=15, C=80°. Find c (cm).

5. b=4, c=7, A=120°. Find a (cm).

6. a=9, c=11, B=55°. Find b (cm).

7. b=3, c=4, A=90°. Find a (cm).

8. a=7, b=7, C=40°. Find c (cm).

Exercise 4 — Cosine Rule: Find Angles

Use the cosine rule to find the unknown angle. Give answers to 1 decimal place.

1. a=7, b=8, c=5. Find angle C (degrees).

2. a=5, b=6, c=8. Find angle A (degrees).

3. a=9, b=11, c=7. Find angle B (degrees).

4. a=4, b=5, c=6. Find angle C (degrees).

5. a=10, b=8, c=6. Find angle A (degrees).

6. a=3, b=4, c=5. Find angle C (degrees).

7. a=12, b=10, c=8. Find angle B (degrees).

8. a=6, b=6, c=6. Find angle A (degrees).

Exercise 5 — Area Formula & Bearings

Mixed application questions. Give areas to 3 s.f. (cm²) and distances to 3 s.f. (km).

1. Triangle, sides 5 and 8 cm, included angle 60°. Area = ? (cm²)

2. Triangle, sides 10 and 12 cm, included angle 45°. Area = ? (cm²)

3. Triangle, sides 7 and 9 cm, included angle 75°. Area = ? (cm²)

4. Triangle, sides 4 and 6 cm, included angle 120°. Area = ? (cm²)

5. Ship: bearing 060° for 40 km, then bearing 140° for 30 km. Distance from start = ? (km)

6. Ship: bearing 080° for 50 km, then bearing 170° for 40 km. Distance from start = ? (km)

7. Triangle PQR, PQ=11, QR=9, angle PQR=50°. Area = ? (cm²)

8. Ship: bearing 030° for 60 km, bearing 120° for 45 km. Distance from start = ? (km)

Practice — 25 Questions

Mixed sine rule, cosine rule, area and bearings. Give sides to 3 s.f. and angles to 1 d.p.

1. Sine rule: A=35°, B=75°, b=12. Find a.

2. Sine rule: P=60°, Q=55°, q=10. Find p.

3. Sine rule angle: a=5, b=7, A=30°. Find B (degrees).

4. Cosine rule side: b=6, c=8, A=50°. Find a.

5. Cosine rule side: a=7, b=9, C=70°. Find c.

6. Cosine rule angle: a=6, b=9, c=11. Find A (degrees).

7. Cosine rule angle: a=8, b=5, c=7. Find B (degrees).

8. Area: sides 6 and 10, included angle 40°. Area (cm²).

9. Area: sides 8 and 12, included angle 55°. Area (cm²).

10. Cosine rule side: b=5, c=5, A=80°. Find a.

11. Sine rule: A=42°, C=68°, c=15. Find a.

12. Sine rule angle: b=9, c=12, B=44°. Find C (degrees).

13. Cosine rule angle: a=10, b=7, c=8. Find C (degrees).

14. Area: sides 15 and 9, included angle 100°. Area (cm²).

15. Bearing: 050° for 30 km, then 130° for 20 km. Distance (km).

16. Cosine rule side: a=11, b=13, C=35°. Find c.

17. Sine rule: B=80°, C=40°, c=9. Find b.

18. Cosine rule angle: a=4, b=6, c=7. Find A (degrees).

19. Area: sides 20 and 14, included angle 65°. Area (cm²).

20. Cosine rule side: b=10, c=14, A=45°. Find a.

21. Sine rule angle: a=12, b=15, A=50°. Find B (degrees, smaller).

22. Area: sides 7 and 11, included angle 30°. Area (cm²).

23. Cosine rule angle: a=9, b=9, c=9. Find A (degrees).

24. Bearing: 090° for 60 km, bearing 200° for 50 km. Distance (km).

25. Sine rule: A=25°, B=65°, b=18. Find a.

Challenge — 12 Questions

Harder multi-step problems. Give answers to 3 s.f. or 1 d.p. as appropriate.

1. Triangle ABC: AB=10, BC=8, CA=7. Find the area of the triangle (cm²). Hint: find an angle first.

2. a=14, b=10, c=9. Find the largest angle (degrees).

3. Two sides 8 and 11, area = 30 cm². Find the included angle (degrees).

4. In triangle PQR, PQ=12, angle P=48°, angle Q=64°. Find QR.

5. a=7, b=10, A=38°. How many valid triangles exist? Enter 1 or 2.

6. Cosine rule: a=√3, b=2, c=1. Find angle A (degrees). (Hint: it's a standard angle.)

7. Ship travels bearing 315° for 80 km, then bearing 045° for 60 km. Distance from start (km).

8. Triangle area = 40 cm², two sides are 10 and 12 cm. Find sin of included angle (×100, round to nearest integer).

9. In △ABC, A=110°, AB=7, AC=5. Find BC.

10. Equilateral triangle, side 8 cm. Area using ½ab sinC (cm²). Round to 1 d.p.

11. a=6, b=9, A=30°. Find the larger possible value of B (degrees).

12. Triangle sides 5, 7, 9. Find the area (cm²) using cosine rule then area formula. Round to 1 d.p.

Exam Style Questions — 5 Questions

Cambridge IGCSE style. Show full working on paper. Enter your final answer.

Q1. In triangle ABC, AB = 9 cm, BC = 12 cm, angle ABC = 74°.
(a) Calculate AC. Give your answer correct to 3 significant figures. (3 marks)
Enter AC in cm:

Q2. Triangle PQR has PQ = 11 cm, PR = 8 cm, QR = 7 cm.
Calculate angle QPR. Give your answer in degrees to 1 decimal place. (3 marks)
Enter angle QPR:

Q3. In triangle XYZ, angle X = 46°, angle Y = 83°, XY = 14 cm.
Calculate XZ. Give your answer correct to 3 significant figures. (3 marks)
Enter XZ in cm:

Q4. A port A. A ship sails on a bearing of 055° for 70 km to reach port B. It then sails on a bearing of 145° for 50 km to reach port C. Calculate the direct distance AC. (4 marks)
Enter AC in km:

Q5. Triangle ABC has AB = 7 cm, BC = 10 cm, angle ABC = 62°.
Calculate the area of the triangle. Give your answer correct to 3 significant figures. (2 marks)
Enter area in cm²: