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GCSE Maths · EdexcelChapter 20 · Pythagoras and trigonometry
GCSE Maths · Pythagoras and Trigonometry · Higher · Lesson 20.10

GCSE Maths: Area of a triangle: ½ab sin C (Higher)

Part of the GCSE Maths pythagoras and trigonometry chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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20.10

Area of a triangle: ½ab sin C

Edexcel 1MA1: G23
Grade 6-8

You already know that the area of a triangle is ½ × base × height. But sometimes the height is not given. If you know two sides and the angle between them, you can still find the area.

Area = ½ ab sin C

Here a and b are two sides and C is the angle between them.

Why it works. Take a as the base. Draw the height h from the opposite vertex. In the right-angled triangle, the hypotenuse is b and the angle at C is C, so sin C = h ÷ b, which gives h = b sin C. Then area = ½ × base × height = ½ × a × b sin C. That is the formula.

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

The formula can be used to find the area, but it can also be rearranged to find a side or an angle when the area is known. To find an angle, you get sin C. As for the sine rule, there are two angles between 0° and 180° with the same sine, an acute one and an obtuse one (180° minus the acute one). Read the question to see which is wanted, or give both.

The formula is also useful for polygons. A regular polygon with n sides splits into n identical isosceles triangles meeting at the centre. Each has two equal sides (the radius) and an angle of 360° ÷ n between them, so its area is ½ r2 sin(360° ÷ n).

10 cm45°
A regular octagon with radius 10 cm: 8 triangles, each with two sides of 10 cm and an angle of 45° between them.

Key ideas

  • Area = ½ ab sin C, where C is the angle between sides a and b.
  • To find an angle: sin C = 2 × area ÷ (ab). Remember the possible obtuse angle 180° − C.
  • To find a side, rearrange: a = 2 × area ÷ (b sin C).
  • Regular polygons: n triangles, each with area ½ r2 sin(360° ÷ n).
Worked example 20.10.1
Work out the area of the triangle. Give your answer correct to 1 decimal place.
ABC8 cm11 cm40°

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Area of any triangle = ½ × a × b × sin C, where C is the angle between sides a and b.

  1. Spot the two sides and the angle between them. The formula needs two sides and the included angle (the angle between those two sides).
    Sides: 8 cm and 11 cm. Included angle C = 40°.
  2. Write the formula and put the numbers in. Area = half × one side × the other side × sine of the angle between them.
    Area = ½ ab sin C
    Area = ½ × 8 × 11 × sin 40°
  3. Use the calculator. Find the sine first, then multiply everything together.
    sin 40° = 0.64278…
    Area = 0.5 × 8 × 11 × 0.64278… = 28.28265…
  4. Round to 1 d.p. and write the units. Area is measured in square units.
    Area = 28.3 cm2

Check. Another way: take 8 as the base. The height is 11 × sin 40° = 7.07066…. Area = ½ × 8 × 7.07066… = 28.28265…. ✓

Watch out. The angle must be the one between the two sides you use. If it is not between them, find the angle you need first (angles in a triangle add up to 180°).

Worked example 20.10.2
A triangle has two sides of length 8 cm and 11 cm, and an area of 30 cm2. The angle between the two sides is C. Work out the two possible sizes of C, correct to 1 decimal place.
ABC8 cm11 cmC

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Put the area into Area = ½ ab sin C, rearrange to find sin C, then use the inverse sine.

  1. Write the area formula and put the numbers in. The area is ½ × a × b × sin C, and here the angle C is unknown.
    30 = ½ × 8 × 11 × sin C
  2. Work out ½ × a × b. This is a plain number, so we can combine it.
    ½ × 8 × 11 = 44
    30 = 44 sin C
  3. Divide both sides by 44. This gets sin C on its own.
    sin C = 3044 = 0.68181…
  4. Use the inverse sine. The inverse button turns the sine back into an angle.
    C = sin−1(0.68181…) = 42.98588…°
  5. Think about a second (obtuse) angle. Sine of an obtuse angle equals sine of 180° minus it, so both give the same area.
    Second angle = 180° − 42.98588…° = 137.01411…°
  6. Round to 1 d.p. Both angles give the same area, so both are answers.
    C = 43.0° or 137.0°

Check. Work forwards: ½ × 8 × 11 × sin 43.0° = 30.0 cm2, close to 30. ✓

Watch out. The inverse sine only gives the acute angle. An obtuse angle gives the same sine, so check whether the question wants it.

Worked example 20.10.3
A triangle has an area of 35 cm2. Two of its sides are a and 10 cm, and the angle between them is 48°. Work out the length of side a, correct to 1 decimal place.
ABCa10 cm48°

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Put the area into Area = ½ ab sin C and rearrange to find the missing side.

  1. Write the area formula and put the numbers in. The area is ½ × a × b × sin C, and here one side is unknown.
    35 = ½ × a × 10 × sin 48°
  2. Work out the known numbers. Combine ½, the side and sin C into one number.
    sin 48° = 0.74314…
    ½ × 10 × 0.74314… = 3.71572…
    35 = 3.71572… a
  3. Divide both sides by 3.71572…. This leaves a on its own.
    a = 35 ÷ 3.71572… = 9.41942…
  4. Round to 1 d.p. Give the unit.
    a = 9.4 cm

Check. Work forwards: ½ × 9.4 × 10 × sin 48° = 34.9 cm2, close to 35. ✓

Watch out. Be careful rearranging: divide by everything that multiplies the unknown side (the ½, the other side and sin C).

Worked example 20.10.4
A regular octagon has a distance of 10 cm from its centre to each corner. Work out its area, correct to 1 decimal place.
10 cm45°

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. A regular octagon is made of 8 identical triangles. Find the area of one triangle with ½ ab sin C and multiply by 8.

  1. Find the angle at the centre. The 8 angles at the centre make a full turn of 360°.
    360° ÷ 8 = 45°
  2. Find the area of one triangle. Two sides are the radius 10 cm, and the angle between them is 45°.
    ½ × 10 × 10 × sin 45° = 50 × 0.70710… = 35.35533… cm2
  3. Multiply by 8. There are 8 identical triangles.
    8 × 35.35533… = 282.84271…
  4. Round.
    Area = 282.8 cm2

Check. Exact version: 8 × 50 × √2÷2 = 200√2 = 282.84271…. ✓

Watch out. Multiply by the number of triangles at the end.

Watch out

The angle in the formula must be the one between the two sides you use. If it is not, you must find the angle you need first, using the fact that angles in a triangle add to 180°. Also remember the units: area is in square units, such as cm2.

Exercise 20L  Area of a triangle using ½ab sin C

Calculator allowed
1
FWork out the area of each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
ABC6 cm9 cm35°
(b)
ABC10.4 cm7.5 cm62°
(c)
ABC5.2 cm8.8 cm118°
(d)
ABC12 cm12 cm25°
2
FWork out the size of the acute angle C between the two sides of each triangle. Give your answers correct to 1 decimal place.
(a)Area = 20 cm2, sides 8 cm and 7 cm. Angle C is acute.
ABC8 cm7 cmC
(b)Area = 30 cm2, sides 8 cm and 11 cm. Angle C is acute.
ABC8 cm11 cmC
3
FWork out the length of the side a in each triangle, correct to 1 decimal place. The angle shown is between the two sides.
(a)Area = 35 cm2, other side 10 cm, angle 48°
ABCa10 cm48°
(b)Area = 22.5 cm2, other side 7.5 cm, angle 90°
ABCa7.5 cm90°
4
FEach regular polygon is split into identical triangles from its centre. Work out the area of each polygon, correct to 1 decimal place.
(a)Regular octagon with centre-to-corner distance 10 cm
10 cm45°
(b)Regular pentagon with centre-to-corner distance 6 cm
6 cm72°
(c)Regular dodecagon with centre-to-corner distance 4 cm
4 cm30°
5
PSA parallelogram has sides of length 9 cm and 12 cm, and the angle between them is 65°. Work out the area of the parallelogram, correct to 1 decimal place.
ABCD12 cm9 cm65°
6
PSA triangular field has two sides of length 120 m and 95 m with an angle of 70° between them. Grass seed costs £0.35 for every square metre. Work out the cost of seeding the whole field, to the nearest pound.
ABC95 m120 m70°
7
PSA triangle has sides of length 7 cm, 8 cm and 9 cm. Work out its area, correct to 1 decimal place.
ABC9 cm8 cm7 cm
8
RSam says that the area of triangle ABC is ½ × 7 × 9 × sin 50°. Explain what is wrong with his method. (AB = 7 cm, BC = 9 cm, angle A = 50°.)
ABC9 cm7 cm50°
9
RShow that the area of an equilateral triangle with sides of length 6 cm is 9√3 cm2.
ABC6 cm6 cm60°60°
10
RChallenge. A triangle has two sides of length 8 cm and 10 cm. Its area is 24 cm2. Work out the two possible sizes of the angle between these sides, correct to 1 decimal place. Explain why both answers are possible.
ABC8 cm10 cmC

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