High School Maths · Year 2 · Unit 11 · Lesson 4 of 5

Area = ½ab sin C, and Choosing the Right Rule: One Formula for Any Triangle

About 15 minutes · the area of a triangle as ½ab sin C · the angle between the two sides · finding a side or an angle from the area · two angles with the same area · parallelograms and regular polygons · choosing the sine rule, the cosine rule or the area formula · calculator allowed · exit quiz

Hook

A triangular field has two sides of 40 m and 55 m, with an angle of 63° between them.

What is its area? You do not know the height, so ½ × base × height seems stuck.

About 980 m². The height is 55 × sin 63°, so the area is ½ × 40 × 55 × sin 63° = 980.1… m². This is the formula Area = ½ab sin C: two sides and the angle between them. No height needed.

Learn

Master the essentials

h = b sin C, so the area is ½ × a × b sin C.

The area formula

Use two sides and the angle between them.

Area = ½ab sin C

Where it comes from

The height from A to side a is b sin C. Then use ½ × base × height.

½ × a × (b sin C)

The included angle

The angle must be between the two sides you use.

½ × 6 × 7 × sin 30° = 10.5 cm²

Reverse: find the angle

Rearrange: sin C = 2 × area ÷ ab. Then use sin⁻¹.

Area 15 cm², sides 6 cm and 10 cm: sin C = 30 ÷ 60 = 0.5

Two angles, same area

sin C = sin(180° − C), so an angle and its supplement give the same area.

sin 130° = sin 50°

Choose the sine rule

When you know a side and the angle opposite it.

Know a, A and B: the sine rule finds b.

Choose the cosine rule

When you know two sides and the angle between them, or all three sides.

Know b, c and A: the cosine rule finds a.

Area from three sides

Find one angle with the cosine rule, then use the area formula.

Two steps: cos C first, then ½ab sin C.

Worked examples

See it done

1. In triangle ABC, AB = 7.8 cm, AC = 12.5 cm and angle BAC = 47°. Find the area of the triangle, correct to 3 significant figures.

  1. Angle A is between AB and AC: use the area formula.
  2. Area = ½ × 7.8 × 12.5 × sin 47°.
  3. = 35.65… = 35.7 cm².

35.7 cm²

2. A triangle has an area of 36 cm². Two of its sides are 9 cm and 11 cm, and the angle between them is acute. Find this angle, correct to 1 decimal place.

  1. ½ × 9 × 11 × sin C = 36.
  2. sin C = 2 × 36 ÷ (9 × 11) = 72 ÷ 99 = 0.7272….
  3. C = sin⁻¹(0.7272…) = 46.7° (acute, as the question says).

46.7°

3. In triangle ABC, AB = 10.6 cm, angle A = 54° and angle B = 63°. Find the area of the triangle, correct to 3 significant figures.

  1. There are not yet two sides. Find one more with the sine rule.
  2. Angle C = 180° − 54° − 63° = 63°. AB = 10.6 cm is opposite C.
  3. BC (opposite A) = 10.6 × sin 54° ÷ sin 63° = 9.624… cm.
  4. Angle B is between AB and BC: area = ½ × 10.6 × 9.624… × sin 63° = 45.45… = 45.5 cm².

45.5 cm²

Common mistake: Using an angle that is not between the two sides. In ½ab sin C, the angle C must be the one where sides a and b meet. If the given angle is somewhere else, find the angle you need first (from the angle sum, or the sine or cosine rule).

PractiseCalculator allowed

10 questions with instant feedback

Choose the correct answer. A calculator is allowed: check it is in degrees mode. Use two sides and the angle between them for the area, and decide which rule each step needs. Give lengths and areas 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.A triangle has sides of 6.3 cm and 9.4 cm with an angle of 58° between them. Find its area, correct to 3 significant figures.

Not quite. This forgets the ½.

Solution

  1. Area = ½ × 6.3 × 9.4 × sin 58°.
  2. = 25.11… = 25.1 cm².

Answer: D. 25.1 cm².

Not quite. This uses cos 58°. The area formula uses sin.

Solution

  1. Area = ½ × 6.3 × 9.4 × sin 58°.
  2. = 25.11… = 25.1 cm².

Answer: D. 25.1 cm².

Not quite. This is what a calculator in radians mode gives. Switch to degrees.

Solution

  1. Area = ½ × 6.3 × 9.4 × sin 58°.
  2. = 25.11… = 25.1 cm².

Answer: D. 25.1 cm².

Correct.

Solution

  1. Area = ½ × 6.3 × 9.4 × sin 58°.
  2. = 25.11… = 25.1 cm².

2.Triangle PQR has PQ = 8 cm, QR = 11 cm and angle PQR = 49°. Which calculation gives its area?

Not quite. The area formula uses sin, not cos.

Solution

  1. Angle Q is between PQ and QR.
  2. Area = ½ × PQ × QR × sin Q = ½ × 8 × 11 × sin 49°.

Answer: B. ½ × 8 × 11 × sin 49°.

Correct.

Solution

  1. Angle Q is between PQ and QR.
  2. Area = ½ × PQ × QR × sin Q = ½ × 8 × 11 × sin 49°.

Not quite. This forgets the ½. It gives the area of a parallelogram.

Solution

  1. Angle Q is between PQ and QR.
  2. Area = ½ × PQ × QR × sin Q = ½ × 8 × 11 × sin 49°.

Answer: B. ½ × 8 × 11 × sin 49°.

Not quite. This is the area only when the angle is 90°. Multiply by sin 49°.

Solution

  1. Angle Q is between PQ and QR.
  2. Area = ½ × PQ × QR × sin Q = ½ × 8 × 11 × sin 49°.

Answer: B. ½ × 8 × 11 × sin 49°.

3.You know all three sides of a triangle. Which should you use first to find one of its angles?

Not quite. The sine rule needs an angle and the side opposite it. You know no angles yet.

Solution

  1. With three sides and no angles, no side has a known opposite angle.
  2. The cosine rule for an angle uses three sides: cos A = (b² + c² − a²) ÷ 2bc.

Answer: B. The cosine rule.

Correct.

Solution

  1. With three sides and no angles, no side has a known opposite angle.
  2. The cosine rule for an angle uses three sides: cos A = (b² + c² − a²) ÷ 2bc.

Not quite. This needs an angle already. It finds the area, not an angle.

Solution

  1. With three sides and no angles, no side has a known opposite angle.
  2. The cosine rule for an angle uses three sides: cos A = (b² + c² − a²) ÷ 2bc.

Answer: B. The cosine rule.

Not quite. SOH CAH TOA works only in a right-angled triangle.

Solution

  1. With three sides and no angles, no side has a known opposite angle.
  2. The cosine rule for an angle uses three sides: cos A = (b² + c² − a²) ÷ 2bc.

Answer: B. The cosine rule.

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

4.A parallelogram ABCD has AB = 13 cm, AD = 8.4 cm and angle DAB = 71°. Find its area, correct to 3 significant figures.

Not quite. This is the area of one triangle ABD. The parallelogram is two of these triangles.

Solution

  1. The diagonal BD splits the parallelogram into two equal triangles.
  2. Each triangle: ½ × 13 × 8.4 × sin 71°.
  3. Parallelogram = 13 × 8.4 × sin 71° = 103.2… = 103 cm².

Answer: C. 103 cm².

Not quite. This multiplies the sides, as for a rectangle. Multiply by sin 71° as well.

Solution

  1. The diagonal BD splits the parallelogram into two equal triangles.
  2. Each triangle: ½ × 13 × 8.4 × sin 71°.
  3. Parallelogram = 13 × 8.4 × sin 71° = 103.2… = 103 cm².

Answer: C. 103 cm².

Correct.

Solution

  1. The diagonal BD splits the parallelogram into two equal triangles.
  2. Each triangle: ½ × 13 × 8.4 × sin 71°.
  3. Parallelogram = 13 × 8.4 × sin 71° = 103.2… = 103 cm².

Not quite. This uses cos 71°. Use sin.

Solution

  1. The diagonal BD splits the parallelogram into two equal triangles.
  2. Each triangle: ½ × 13 × 8.4 × sin 71°.
  3. Parallelogram = 13 × 8.4 × sin 71° = 103.2… = 103 cm².

Answer: C. 103 cm².

5.A triangle has an area of 54 cm². One side is 12 cm, and the angle between this side and the side x is 48°. Find x, correct to 3 significant figures.

Correct.

Solution

  1. ½ × 12 × x × sin 48° = 54.
  2. 6 × x × sin 48° = 54.
  3. x = 54 ÷ (6 × sin 48°) = 12.11… = 12.1 cm.

Not quite. This forgets the ½ in the formula: x = 2 × area ÷ (12 sin 48°).

Solution

  1. ½ × 12 × x × sin 48° = 54.
  2. 6 × x × sin 48° = 54.
  3. x = 54 ÷ (6 × sin 48°) = 12.11… = 12.1 cm.

Answer: A. 12.1 cm.

Not quite. This uses cos 48°. Use sin.

Solution

  1. ½ × 12 × x × sin 48° = 54.
  2. 6 × x × sin 48° = 54.
  3. x = 54 ÷ (6 × sin 48°) = 12.11… = 12.1 cm.

Answer: A. 12.1 cm.

Not quite. This leaves out sin 48°.

Solution

  1. ½ × 12 × x × sin 48° = 54.
  2. 6 × x × sin 48° = 54.
  3. x = 54 ÷ (6 × sin 48°) = 12.11… = 12.1 cm.

Answer: A. 12.1 cm.

6.The table gives two sides and the angle between them for three triangular plots. Find the area of plot B, correct to 3 significant figures.

PlotSide 1Side 2Angle between them
A24 m31 m52°
B18.5 m27 m64°
C35 m22 m118°

Each plot is a triangle.

Correct.

Solution

  1. Plot B: sides 18.5 m and 27 m, with 64° between them.
  2. Area = ½ × 18.5 × 27 × sin 64°.
  3. = 224.4… = 224 m².

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

Solution

  1. Plot B: sides 18.5 m and 27 m, with 64° between them.
  2. Area = ½ × 18.5 × 27 × sin 64°.
  3. = 224.4… = 224 m².

Answer: A. 224 m².

Not quite. This forgets the ½.

Solution

  1. Plot B: sides 18.5 m and 27 m, with 64° between them.
  2. Area = ½ × 18.5 × 27 × sin 64°.
  3. = 224.4… = 224 m².

Answer: A. 224 m².

Not quite. This uses cos 64°. Use sin.

Solution

  1. Plot B: sides 18.5 m and 27 m, with 64° between them.
  2. Area = ½ × 18.5 × 27 × sin 64°.
  3. = 224.4… = 224 m².

Answer: A. 224 m².

7.In triangle ABC, AB = 8.2 cm, AC = 7.5 cm and angle BAC = 56°. Which working shows that the area is 25.5 cm², correct to 3 significant figures?

Not quite. This forgets the ½, and does not give 25.5.

Solution

  1. Angle A is between AB and AC.
  2. Area = ½ × 8.2 × 7.5 × sin 56° = 25.49….
  3. To 3 significant figures this is 25.5 cm² ✓.

Answer: C. ½ × 8.2 × 7.5 × sin 56° = 25.49….

Not quite. The area formula uses sin, not cos.

Solution

  1. Angle A is between AB and AC.
  2. Area = ½ × 8.2 × 7.5 × sin 56° = 25.49….
  3. To 3 significant figures this is 25.5 cm² ✓.

Answer: C. ½ × 8.2 × 7.5 × sin 56° = 25.49….

Correct.

Solution

  1. Angle A is between AB and AC.
  2. Area = ½ × 8.2 × 7.5 × sin 56° = 25.49….
  3. To 3 significant figures this is 25.5 cm² ✓.

Not quite. The area formula uses sin, not tan.

Solution

  1. Angle A is between AB and AC.
  2. Area = ½ × 8.2 × 7.5 × sin 56° = 25.49….
  3. To 3 significant figures this is 25.5 cm² ✓.

Answer: C. ½ × 8.2 × 7.5 × sin 56° = 25.49….

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

8.A triangle has sides of 7 cm, 9 cm and 12 cm. Find its area, correct to 3 significant figures.

Correct.

Solution

  1. Find an angle first. The angle C between the 7 cm and 9 cm sides is opposite 12 cm.
  2. cos C = (7² + 9² − 12²) ÷ (2 × 7 × 9) = −14 ÷ 126, so C = 96.4°.
  3. Area = ½ × 7 × 9 × sin 96.37…° = 31.30… = 31.3 cm².

Not quite. This forgets the ½.

Solution

  1. Find an angle first. The angle C between the 7 cm and 9 cm sides is opposite 12 cm.
  2. cos C = (7² + 9² − 12²) ÷ (2 × 7 × 9) = −14 ÷ 126, so C = 96.4°.
  3. Area = ½ × 7 × 9 × sin 96.37…° = 31.30… = 31.3 cm².

Answer: A. 31.3 cm².

Not quite. This uses the angle opposite the 7 cm side, which is not between the 7 cm and 9 cm sides.

Solution

  1. Find an angle first. The angle C between the 7 cm and 9 cm sides is opposite 12 cm.
  2. cos C = (7² + 9² − 12²) ÷ (2 × 7 × 9) = −14 ÷ 126, so C = 96.4°.
  3. Area = ½ × 7 × 9 × sin 96.37…° = 31.30… = 31.3 cm².

Answer: A. 31.3 cm².

Not quite. The angle opposite 12 cm is between the 7 cm and 9 cm sides, not the 7 cm and 12 cm sides.

Solution

  1. Find an angle first. The angle C between the 7 cm and 9 cm sides is opposite 12 cm.
  2. cos C = (7² + 9² − 12²) ÷ (2 × 7 × 9) = −14 ÷ 126, so C = 96.4°.
  3. Area = ½ × 7 × 9 × sin 96.37…° = 31.30… = 31.3 cm².

Answer: A. 31.3 cm².

9.A triangle has an area of 30 cm². Two of its sides are 8 cm and 10 cm, and the angle between them is obtuse. Find this angle, correct to 1 decimal place.

Not quite. This is the acute angle with the same sine. The question says the angle is obtuse.

Solution

  1. ½ × 8 × 10 × sin C = 30, so sin C = 60 ÷ 80 = 0.75.
  2. sin⁻¹(0.75) = 48.6°, which is acute.
  3. The obtuse angle is 180° − 48.6° = 131.4°.

Answer: B. 131.4°.

Correct.

Solution

  1. ½ × 8 × 10 × sin C = 30, so sin C = 60 ÷ 80 = 0.75.
  2. sin⁻¹(0.75) = 48.6°, which is acute.
  3. The obtuse angle is 180° − 48.6° = 131.4°.

Not quite. This is sin C. Use sin⁻¹, then 180° minus the result.

Solution

  1. ½ × 8 × 10 × sin C = 30, so sin C = 60 ÷ 80 = 0.75.
  2. sin⁻¹(0.75) = 48.6°, which is acute.
  3. The obtuse angle is 180° − 48.6° = 131.4°.

Answer: B. 131.4°.

Not quite. This forgets the ½: sin C = 2 × 30 ÷ 80, not 30 ÷ 80.

Solution

  1. ½ × 8 × 10 × sin C = 30, so sin C = 60 ÷ 80 = 0.75.
  2. sin⁻¹(0.75) = 48.6°, which is acute.
  3. The obtuse angle is 180° − 48.6° = 131.4°.

Answer: B. 131.4°.

10.In triangle ABC, AB = 15 cm, angle A = 38° and angle B = 76°. Find the area of the triangle, correct to 3 significant figures.

Not quite. This uses angle A with sides BC and AB. Angle A is not between them: use angle B.

Solution

  1. Angle C = 180° − 38° − 76° = 66°. AB = 15 cm is opposite C.
  2. BC (opposite A) = 15 × sin 38° ÷ sin 66° = 10.10… cm.
  3. Angle B is between AB and BC: area = ½ × 15 × 10.10… × sin 76° = 73.56… = 73.6 cm².

Answer: D. 73.6 cm².

Not quite. This forgets the ½.

Solution

  1. Angle C = 180° − 38° − 76° = 66°. AB = 15 cm is opposite C.
  2. BC (opposite A) = 15 × sin 38° ÷ sin 66° = 10.10… cm.
  3. Angle B is between AB and BC: area = ½ × 15 × 10.10… × sin 76° = 73.56… = 73.6 cm².

Answer: D. 73.6 cm².

Not quite. This uses 15 cm for both sides. Find a second side with the sine rule first.

Solution

  1. Angle C = 180° − 38° − 76° = 66°. AB = 15 cm is opposite C.
  2. BC (opposite A) = 15 × sin 38° ÷ sin 66° = 10.10… cm.
  3. Angle B is between AB and BC: area = ½ × 15 × 10.10… × sin 76° = 73.56… = 73.6 cm².

Answer: D. 73.6 cm².

Correct.

Solution

  1. Angle C = 180° − 38° − 76° = 66°. AB = 15 cm is opposite C.
  2. BC (opposite A) = 15 × sin 38° ÷ sin 66° = 10.10… cm.
  3. Angle B is between AB and BC: area = ½ × 15 × 10.10… × sin 76° = 73.56… = 73.6 cm².

Apply

Fencing a field: one triangle, three rules

A farmer has a triangular field ABC with AB = 46 m, AC = 38 m and angle BAC = 72°. A calculator is allowed.

1. Find the area of the field, correct to 3 significant figures.

2. Find the length of the third side BC.

3. Find angle ABC, correct to 1 decimal place. Say which rule you used at each step, and why.

  1. Two sides and the angle between them: area = ½ × 46 × 38 × sin 72° = 831.2… = 831 m².
  2. Two sides and the angle between them again, so the cosine rule: BC² = 46² + 38² − 2 × 46 × 38 × cos 72°, so BC = 49.79… = 49.8 m.
  3. Now BC and angle A are a pair, so the sine rule: sin B = 38 × sin 72° ÷ 49.79…, so B = 46.5° (acute, as AC is shorter than BC).

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

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