High School Maths · Year 2 · Unit 11 · Lesson 3 of 5

The Cosine Rule: When the Sine Rule Cannot Start

About 15 minutes · when the sine rule cannot be used · the cosine rule for a side (two sides and the angle between them) · the cosine rule for an angle (three sides) · obtuse angles and negative cosines · order of operations · calculator allowed · exit quiz

Hook

Two straight roads leave a village with an angle of 47° between them. Ana drives 8 km along one road. Ben drives 11 km along the other.

How far apart are they now, in a straight line? Try the sine rule: you will find there is no side with its opposite angle to start from.

About 8.06 km. You know two sides and the angle between them, so use the cosine rule: distance² = 8² + 11² − 2 × 8 × 11 × cos 47° = 64.96…, and the square root is 8.060… km. It works like Pythagoras, with a correction for the angle not being 90°.

Learn

Master the essentials

Angle A is between sides b and c. Side a is opposite it.

When to use it

Use the cosine rule when you know two sides and the angle BETWEEN them, or all three sides.

No side-and-opposite-angle pair? Use the cosine rule.

Finding a side

The side opposite angle A, from the two sides that make angle A.

a² = b² + c² − 2bc cos A

The included angle

Angle A must be the angle between sides b and c.

Sides 6 cm and 9 cm with 50° between them.

Order of operations

Work out 2bc cos A first, then subtract it from b² + c².

b = 4, c = 9, A = 60°: a² = 16 + 81 − 36 = 61

Square root last

The formula gives a². Take the square root to get the side.

a = √61 = 7.81

Finding an angle

Rearrange when you know all three sides. The angle is opposite side a.

cos A = (b² + c² − a²) ÷ 2bc

Obtuse angles

If cos A is negative, A is obtuse. cos⁻¹ gives the right angle automatically.

cos A = −0.25, so A = cos⁻¹(−0.25) = 104.5°

Link to Pythagoras

When A = 90°, cos A = 0, so the rule becomes Pythagoras.

a² = b² + c² − 0

Worked examples

See it done

1. In triangle ABC, AB = 7.2 cm, AC = 9.5 cm and angle BAC = 54°. Find BC, correct to 3 significant figures.

  1. Two sides and the angle between them: use the cosine rule for BC, the side opposite angle A.
  2. BC² = 9.5² + 7.2² − 2 × 9.5 × 7.2 × cos 54°.
  3. = 90.25 + 51.84 − 80.40… = 61.68….
  4. BC = √61.68… = 7.853… = 7.85 cm.

7.85 cm

2. A triangle has sides of 8 cm, 11 cm and 14 cm. Find its largest angle, correct to 1 decimal place.

  1. The largest angle is opposite the longest side, 14 cm.
  2. cos θ = (8² + 11² − 14²) ÷ (2 × 8 × 11) = (64 + 121 − 196) ÷ 176.
  3. = −11 ÷ 176 = −0.0625. The cosine is negative, so the angle is obtuse.
  4. θ = cos⁻¹(−0.0625) = 93.6°.

93.6°

3. A garden path goes from P to Q, then turns to R. PQ = 6.5 m, QR = 4.8 m and angle PQR = 118°. How far is it from P straight to R? Give your answer correct to 3 significant figures.

  1. Two sides and the angle between them at Q: use the cosine rule.
  2. PR² = 6.5² + 4.8² − 2 × 6.5 × 4.8 × cos 118°.
  3. cos 118° is negative, so 2 × 6.5 × 4.8 × cos 118° = −29.29…. Subtracting a negative adds it.
  4. PR² = 42.25 + 23.04 + 29.29… = 94.58…, so PR = 9.725… = 9.73 m.

9.73 m

Common mistake: Working left to right on the calculator. In a² = b² + c² − 2bc cos A, do not work out (b² + c² − 2bc) and then multiply by cos A. Multiply 2 × b × c × cos A first, then take it away from b² + c². And with an obtuse angle, cos A is negative, so you end up ADDING.

PractiseCalculator allowed

10 questions with instant feedback

Choose the correct answer. A calculator is allowed: check it is in degrees mode. Find the angle between the two known sides, or use all three sides. Give lengths correct to 3 significant figures and angles correct to 1 decimal place. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.In triangle ABC, AB = 8.9 cm, AC = 6.4 cm and angle BAC = 41°. Find BC, correct to 3 significant figures.

Not quite. This is BC². Take the square root.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. BC² = 6.4² + 8.9² − 2 × 6.4 × 8.9 × cos 41° = 34.19….
  3. BC = √34.19… = 5.847… = 5.85 cm.

Answer: D. 5.85 cm.

Not quite. The rule SUBTRACTS 2bc cos A. This adds it.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. BC² = 6.4² + 8.9² − 2 × 6.4 × 8.9 × cos 41° = 34.19….
  3. BC = √34.19… = 5.847… = 5.85 cm.

Answer: D. 5.85 cm.

Not quite. This works out (b² + c² − 2bc) first, then multiplies by cos 41°. Multiply 2bc cos A first.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. BC² = 6.4² + 8.9² − 2 × 6.4 × 8.9 × cos 41° = 34.19….
  3. BC = √34.19… = 5.847… = 5.85 cm.

Answer: D. 5.85 cm.

Correct.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. BC² = 6.4² + 8.9² − 2 × 6.4 × 8.9 × cos 41° = 34.19….
  3. BC = √34.19… = 5.847… = 5.85 cm.

2.A triangle has sides of 5.5 cm, 7.3 cm and 9.8 cm. Find the angle opposite the 9.8 cm side, correct to 1 decimal place.

Correct.

Solution

  1. cos θ = (5.5² + 7.3² − 9.8²) ÷ (2 × 5.5 × 7.3).
  2. = (30.25 + 53.29 − 96.04) ÷ 80.3 = −0.1556….
  3. θ = cos⁻¹(−0.1556…) = 99.0° (obtuse, as cos θ is negative).

Not quite. This is cos θ, not the angle. Use cos⁻¹.

Solution

  1. cos θ = (5.5² + 7.3² − 9.8²) ÷ (2 × 5.5 × 7.3).
  2. = (30.25 + 53.29 − 96.04) ÷ 80.3 = −0.1556….
  3. θ = cos⁻¹(−0.1556…) = 99.0° (obtuse, as cos θ is negative).

Answer: A. 99.0°.

Not quite. This is the angle opposite the 5.5 cm side. The side opposite θ goes last, after the minus sign.

Solution

  1. cos θ = (5.5² + 7.3² − 9.8²) ÷ (2 × 5.5 × 7.3).
  2. = (30.25 + 53.29 − 96.04) ÷ 80.3 = −0.1556….
  3. θ = cos⁻¹(−0.1556…) = 99.0° (obtuse, as cos θ is negative).

Answer: A. 99.0°.

Not quite. This is the angle opposite the 7.3 cm side. Subtract 9.8², the side opposite θ.

Solution

  1. cos θ = (5.5² + 7.3² − 9.8²) ÷ (2 × 5.5 × 7.3).
  2. = (30.25 + 53.29 − 96.04) ÷ 80.3 = −0.1556….
  3. θ = cos⁻¹(−0.1556…) = 99.0° (obtuse, as cos θ is negative).

Answer: A. 99.0°.

3.In triangle PQR, PQ = 7 cm, PR = 12 cm and angle QPR = 38°. Which equation finds QR?

Not quite. The cosine rule subtracts 2bc cos A.

Solution

  1. QR is opposite angle P, which is between the 7 cm and 12 cm sides.
  2. Cosine rule: QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Answer: C. QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Not quite. The two sides must be squared, and the result is QR².

Solution

  1. QR is opposite angle P, which is between the 7 cm and 12 cm sides.
  2. Cosine rule: QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Answer: C. QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Correct.

Solution

  1. QR is opposite angle P, which is between the 7 cm and 12 cm sides.
  2. Cosine rule: QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Not quite. The cosine rule uses cos, not sin.

Solution

  1. QR is opposite angle P, which is between the 7 cm and 12 cm sides.
  2. Cosine rule: QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Answer: C. QR² = 7² + 12² − 2 × 7 × 12 × cos 38°.

Exam standard (Several steps, like real exam questions.)

4.Two hikers leave a camp along straight paths with an angle of 124° between them. One stops 3.6 km from the camp and the other stops 5.3 km from the camp. How far apart are they? Give your answer correct to 3 significant figures.

Correct.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. d² = 3.6² + 5.3² − 2 × 3.6 × 5.3 × cos 124°, and cos 124° = −0.5591….
  3. d² = 12.96 + 28.09 + 21.33… = 62.38….
  4. d = 7.898… = 7.90 km.

Not quite. cos 124° is negative. Subtracting a negative means adding: the distance must be more than with a right angle.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. d² = 3.6² + 5.3² − 2 × 3.6 × 5.3 × cos 124°, and cos 124° = −0.5591….
  3. d² = 12.96 + 28.09 + 21.33… = 62.38….
  4. d = 7.898… = 7.90 km.

Answer: A. 7.90 km.

Not quite. This is the distance squared. Take the square root.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. d² = 3.6² + 5.3² − 2 × 3.6 × 5.3 × cos 124°, and cos 124° = −0.5591….
  3. d² = 12.96 + 28.09 + 21.33… = 62.38….
  4. d = 7.898… = 7.90 km.

Answer: A. 7.90 km.

Not quite. This uses Pythagoras, as if the angle were 90°. Use the cosine rule with 124°.

Solution

  1. Two sides and the angle between them: use the cosine rule.
  2. d² = 3.6² + 5.3² − 2 × 3.6 × 5.3 × cos 124°, and cos 124° = −0.5591….
  3. d² = 12.96 + 28.09 + 21.33… = 62.38….
  4. d = 7.898… = 7.90 km.

Answer: A. 7.90 km.

5.Two ships leave a port at the same time. Ship S sails on a bearing of 040° at 12 km/h. Ship T sails on a bearing of 115° at 9 km/h. How far apart are they after 2 hours? Give your answer correct to 3 significant figures.

Not quite. This uses the speeds as distances. After 2 hours the ships are 24 km and 18 km from port.

Solution

  1. After 2 hours: S is 24 km from port and T is 18 km from port.
  2. The angle between their courses is 115° − 40° = 75°.
  3. d² = 24² + 18² − 2 × 24 × 18 × cos 75° = 676.3….
  4. d = 26.00… = 26.0 km.

Answer: B. 26.0 km.

Correct.

Solution

  1. After 2 hours: S is 24 km from port and T is 18 km from port.
  2. The angle between their courses is 115° − 40° = 75°.
  3. d² = 24² + 18² − 2 × 24 × 18 × cos 75° = 676.3….
  4. d = 26.00… = 26.0 km.

Not quite. This adds the bearings. The angle between the two courses is 115° − 40° = 75°.

Solution

  1. After 2 hours: S is 24 km from port and T is 18 km from port.
  2. The angle between their courses is 115° − 40° = 75°.
  3. d² = 24² + 18² − 2 × 24 × 18 × cos 75° = 676.3….
  4. d = 26.00… = 26.0 km.

Answer: B. 26.0 km.

Not quite. This is the distance squared. Take the square root.

Solution

  1. After 2 hours: S is 24 km from port and T is 18 km from port.
  2. The angle between their courses is 115° − 40° = 75°.
  3. d² = 24² + 18² − 2 × 24 × 18 × cos 75° = 676.3….
  4. d = 26.00… = 26.0 km.

Answer: B. 26.0 km.

6.The table gives the side lengths of three triangular frames. Find the largest angle of frame B, correct to 1 decimal place.

FrameSide 1Side 2Side 3
A4.5 cm6.2 cm8.8 cm
B7.4 cm5.9 cm10.3 cm
C6.6 cm8.1 cm9.5 cm

Each frame is a triangle made of three straight rods.

Not quite. This uses frame A. Read the row for frame B.

Solution

  1. Frame B: 7.4 cm, 5.9 cm and 10.3 cm. The largest angle is opposite 10.3 cm.
  2. cos θ = (7.4² + 5.9² − 10.3²) ÷ (2 × 7.4 × 5.9) = −0.1891….
  3. θ = cos⁻¹(−0.1891…) = 100.9°.

Answer: C. 100.9°.

Not quite. This is the angle opposite 7.4 cm. The largest angle is opposite the longest side, 10.3 cm.

Solution

  1. Frame B: 7.4 cm, 5.9 cm and 10.3 cm. The largest angle is opposite 10.3 cm.
  2. cos θ = (7.4² + 5.9² − 10.3²) ÷ (2 × 7.4 × 5.9) = −0.1891….
  3. θ = cos⁻¹(−0.1891…) = 100.9°.

Answer: C. 100.9°.

Correct.

Solution

  1. Frame B: 7.4 cm, 5.9 cm and 10.3 cm. The largest angle is opposite 10.3 cm.
  2. cos θ = (7.4² + 5.9² − 10.3²) ÷ (2 × 7.4 × 5.9) = −0.1891….
  3. θ = cos⁻¹(−0.1891…) = 100.9°.

Not quite. This is the cosine. Use cos⁻¹ to find the angle.

Solution

  1. Frame B: 7.4 cm, 5.9 cm and 10.3 cm. The largest angle is opposite 10.3 cm.
  2. cos θ = (7.4² + 5.9² − 10.3²) ÷ (2 × 7.4 × 5.9) = −0.1891….
  3. θ = cos⁻¹(−0.1891…) = 100.9°.

Answer: C. 100.9°.

7.A triangle has sides of 5 cm, 7 cm and 8 cm. Which working shows that the angle opposite the 7 cm side is exactly 60°?

Not quite. This puts 8 cm opposite the angle. The side opposite θ is 7 cm, so subtract 7² = 49.

Solution

  1. The angle θ is opposite 7 cm, between the 5 cm and 8 cm sides.
  2. cos θ = (5² + 8² − 7²) ÷ (2 × 5 × 8) = (25 + 64 − 49) ÷ 80.
  3. = 40 ÷ 80 = 0.5, and cos 60° = 0.5, so θ = 60° ✓.

Answer: B. (25 + 64 − 49) ÷ 80 = 0.5, and cos 60° = 0.5.

Correct.

Solution

  1. The angle θ is opposite 7 cm, between the 5 cm and 8 cm sides.
  2. cos θ = (5² + 8² − 7²) ÷ (2 × 5 × 8) = (25 + 64 − 49) ÷ 80.
  3. = 40 ÷ 80 = 0.5, and cos 60° = 0.5, so θ = 60° ✓.

Not quite. The side opposite the angle is subtracted, not added. And a cosine is never more than 1.

Solution

  1. The angle θ is opposite 7 cm, between the 5 cm and 8 cm sides.
  2. cos θ = (5² + 8² − 7²) ÷ (2 × 5 × 8) = (25 + 64 − 49) ÷ 80.
  3. = 40 ÷ 80 = 0.5, and cos 60° = 0.5, so θ = 60° ✓.

Answer: B. (25 + 64 − 49) ÷ 80 = 0.5, and cos 60° = 0.5.

Not quite. The sides must be squared.

Solution

  1. The angle θ is opposite 7 cm, between the 5 cm and 8 cm sides.
  2. cos θ = (5² + 8² − 7²) ÷ (2 × 5 × 8) = (25 + 64 − 49) ÷ 80.
  3. = 40 ÷ 80 = 0.5, and cos 60° = 0.5, so θ = 60° ✓.

Answer: B. (25 + 64 − 49) ÷ 80 = 0.5, and cos 60° = 0.5.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A parallelogram ABCD has AB = 11 cm, AD = 6.5 cm and angle DAB = 65°. Find the length of its longer diagonal, correct to 3 significant figures.

Not quite. This is the shorter diagonal BD, opposite the 65° angle. The longer one is opposite the 115° angle.

Solution

  1. Next to angle A = 65°, angle ABC = 180° − 65° = 115° (co-interior angles).
  2. The longer diagonal AC is opposite the 115° angle in triangle ABC, with sides 11 cm and 6.5 cm.
  3. AC² = 11² + 6.5² − 2 × 11 × 6.5 × cos 115° = 223.6….
  4. AC = 14.95… = 15.0 cm.

Answer: B. 15.0 cm.

Correct.

Solution

  1. Next to angle A = 65°, angle ABC = 180° − 65° = 115° (co-interior angles).
  2. The longer diagonal AC is opposite the 115° angle in triangle ABC, with sides 11 cm and 6.5 cm.
  3. AC² = 11² + 6.5² − 2 × 11 × 6.5 × cos 115° = 223.6….
  4. AC = 14.95… = 15.0 cm.

Not quite. This is the diagonal squared. Take the square root.

Solution

  1. Next to angle A = 65°, angle ABC = 180° − 65° = 115° (co-interior angles).
  2. The longer diagonal AC is opposite the 115° angle in triangle ABC, with sides 11 cm and 6.5 cm.
  3. AC² = 11² + 6.5² − 2 × 11 × 6.5 × cos 115° = 223.6….
  4. AC = 14.95… = 15.0 cm.

Answer: B. 15.0 cm.

Not quite. This uses Pythagoras. The angles of the parallelogram are not 90°.

Solution

  1. Next to angle A = 65°, angle ABC = 180° − 65° = 115° (co-interior angles).
  2. The longer diagonal AC is opposite the 115° angle in triangle ABC, with sides 11 cm and 6.5 cm.
  3. AC² = 11² + 6.5² − 2 × 11 × 6.5 × cos 115° = 223.6….
  4. AC = 14.95… = 15.0 cm.

Answer: B. 15.0 cm.

9.Two sides of a triangle are x cm and (x + 5) cm, and the angle between them is 60°. The third side is 19 cm. Find x.

Not quite. This is x + 5, the longer side. The question asks for x.

Solution

  1. 19² = x² + (x + 5)² − 2 × x × (x + 5) × cos 60°, and cos 60° = 0.5.
  2. 361 = x² + x² + 10x + 25 − x² − 5x = x² + 5x + 25.
  3. x² + 5x − 336 = 0, so (x + 21)(x − 16) = 0.
  4. x is a length, so x = 16. Check: 16² + 21² − 16 × 21 = 361 = 19² ✓.

Answer: C. 16 cm.

Not quite. This uses Pythagoras and leaves out the − 2bc cos A term.

Solution

  1. 19² = x² + (x + 5)² − 2 × x × (x + 5) × cos 60°, and cos 60° = 0.5.
  2. 361 = x² + x² + 10x + 25 − x² − 5x = x² + 5x + 25.
  3. x² + 5x − 336 = 0, so (x + 21)(x − 16) = 0.
  4. x is a length, so x = 16. Check: 16² + 21² − 16 × 21 = 361 = 19² ✓.

Answer: C. 16 cm.

Correct.

Solution

  1. 19² = x² + (x + 5)² − 2 × x × (x + 5) × cos 60°, and cos 60° = 0.5.
  2. 361 = x² + x² + 10x + 25 − x² − 5x = x² + 5x + 25.
  3. x² + 5x − 336 = 0, so (x + 21)(x − 16) = 0.
  4. x is a length, so x = 16. Check: 16² + 21² − 16 × 21 = 361 = 19² ✓.

Not quite. This uses cos 60° = −0.5. In fact cos 60° = 0.5, so the term is − x(x + 5).

Solution

  1. 19² = x² + (x + 5)² − 2 × x × (x + 5) × cos 60°, and cos 60° = 0.5.
  2. 361 = x² + x² + 10x + 25 − x² − 5x = x² + 5x + 25.
  3. x² + 5x − 336 = 0, so (x + 21)(x − 16) = 0.
  4. x is a length, so x = 16. Check: 16² + 21² − 16 × 21 = 361 = 19² ✓.

Answer: C. 16 cm.

10.Quadrilateral ABCD has AB = 7 cm, AD = 9 cm and angle DAB = 72°. Also BC = 8 cm and CD = 6 cm. Find angle BCD, correct to 1 decimal place.

Not quite. This is angle CBD. Angle BCD is opposite the diagonal BD.

Solution

  1. Triangle ABD: BD² = 7² + 9² − 2 × 7 × 9 × cos 72°, so BD = 9.542… cm.
  2. Triangle BCD: angle BCD is opposite BD.
  3. cos BCD = (8² + 6² − BD²) ÷ (2 × 8 × 6) = 0.09308….
  4. Angle BCD = 84.7°.

Answer: D. 84.7°.

Not quite. This is cos BCD. Use cos⁻¹ to find the angle.

Solution

  1. Triangle ABD: BD² = 7² + 9² − 2 × 7 × 9 × cos 72°, so BD = 9.542… cm.
  2. Triangle BCD: angle BCD is opposite BD.
  3. cos BCD = (8² + 6² − BD²) ÷ (2 × 8 × 6) = 0.09308….
  4. Angle BCD = 84.7°.

Answer: D. 84.7°.

Not quite. This finds BD by Pythagoras. Angle DAB is 72°, not 90°: use the cosine rule.

Solution

  1. Triangle ABD: BD² = 7² + 9² − 2 × 7 × 9 × cos 72°, so BD = 9.542… cm.
  2. Triangle BCD: angle BCD is opposite BD.
  3. cos BCD = (8² + 6² − BD²) ÷ (2 × 8 × 6) = 0.09308….
  4. Angle BCD = 84.7°.

Answer: D. 84.7°.

Correct.

Solution

  1. Triangle ABD: BD² = 7² + 9² − 2 × 7 × 9 × cos 72°, so BD = 9.542… cm.
  2. Triangle BCD: angle BCD is opposite BD.
  3. cos BCD = (8² + 6² − BD²) ÷ (2 × 8 × 6) = 0.09308….
  4. Angle BCD = 84.7°.

Apply

Off the line on the golf course

A golfer hits the ball from the tee T. It lands at B, 220 m from T. The hole H is 245 m from T, and the ball went 9° off the line TH. A calculator is allowed.

1. How far is the ball from the hole? Give 3 significant figures.

2. Find angle TBH, the angle at the ball between the tee and the hole, correct to 1 decimal place.

3. Explain how you can tell from your working that angle TBH is obtuse.

  1. Two sides and the angle between them: BH² = 220² + 245² − 2 × 220 × 245 × cos 9° = 1,952…, so BH = 44.18… = 44.2 m.
  2. Angle TBH is opposite TH = 245 m: cos TBH = (220² + BH² − 245²) ÷ (2 × 220 × BH) = −0.4975…, so TBH = 119.8°.
  3. The cosine is negative, and only an obtuse angle has a negative cosine. (Also, 245² is bigger than 220² + BH².)

Next: the Lesson 3 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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