Home / IGCSE Maths 0580 / Trigonometry / Area of a triangle: half ab sin C
Cambridge IGCSE · Topic 6.5
IGCSE Maths 0580 / Topic 6 / IGCSE

Half base times height, without the height.

Extended. You know that area is half base times height. When the height is not given, trigonometry supplies it: two sides and the angle between them are all you need.

Full-screen labSine and cosine rules lab → Topic 6All trigonometry lessons →
The key idea

Area \(= \tfrac{1}{2}ab\sin C\), where \(C\) is the angle BETWEEN the sides \(a\) and \(b\). It works because the perpendicular height onto side \(a\) is \(b\sin C\).

40° A B C a = 8 cm b = 6 cm h = 6 sin 40°
Fig. 1: drawn to scale with a = 8 cm, b = 6 cm and C = 40°. The height is h = 6 sin 40° = 3.86 cm, so the area is ½ × 8 × 6 sin 40° = 15.4 cm²
Section 01

Where the formula comes from.

In the lab, fix two sides and swing the angle between them. Watch the height \(b\sin C\) rise and fall, and the area follow it: largest at \(90^\circ\), and the same for \(C\) and \(180^\circ - C\).

Worked example Finding the area
In Fig. 1, \(a = 8\) cm, \(b = 6\) cm and \(C = 40^\circ\). Find the area of the triangle.
  1. \(C\) is the angle between \(a\) and \(b\), so the formula applies directly.
  2. Area \(= \tfrac{1}{2} \times 8 \times 6 \times \sin 40^\circ = 24\sin 40^\circ = 15.426\ldots\)
Answer: \(15.4\) cm2 (3 s.f.)
Worked example An obtuse angle
Two sides of a triangle are 9 cm and 12 cm and the angle between them is \(130^\circ\). Find the area.
  1. Area \(= \tfrac{1}{2} \times 9 \times 12 \times \sin 130^\circ\).
  2. \(= 54\sin 130^\circ = 54 \times 0.7660\ldots = 41.366\ldots\)
Answer: \(41.4\) cm2 (3 s.f.)
Examiner trap

Using an angle that is not between the two sides. If you know sides 9 cm and 12 cm and an angle opposite one of them, first find the included angle (with the sine rule), then use the formula.

Section 02

Working backwards.

If the area is given, the formula becomes an equation. Substitute what you know and solve for the missing side or angle.

Worked example Finding the angle
A triangle has sides 8 cm and 10 cm and an area of 30 cm2. Find the angle between the two sides.
  1. \(30 = \tfrac{1}{2} \times 8 \times 10 \times \sin C = 40\sin C\).
  2. \(\sin C = \dfrac{30}{40} = 0.75\), so \(C = \sin^{-1}(0.75) = 48.59\ldots^\circ\).
  3. Since \(\sin(180^\circ - C) = \sin C\), \(C = 180^\circ - 48.59^\circ = 131.41^\circ\) also works.
Answer: \(C = 48.6^\circ\) or \(C = 131.4^\circ\) (two different triangles with the same area)
Worked example Finding a side
A triangle has area 50 cm2. Two of its sides meet at an angle of \(35^\circ\); one of them is 12 cm. Find the other.
  1. \(50 = \tfrac{1}{2} \times 12 \times x \times \sin 35^\circ = 6x\sin 35^\circ\).
  2. \(x = \dfrac{50}{6\sin 35^\circ} = 14.528\ldots\)
Answer: \(14.5\) cm (3 s.f.)
Your turn
A triangle has sides 6 cm and 10 cm and an area of 18 cm2. Find the two possible values of the angle between those sides.
Examiner trap

Forgetting the \(\tfrac{1}{2}\) when working backwards. With area 30 cm2 and sides 8 and 10, \(\sin C = \dfrac{30}{80}\) gives \(22.0^\circ\), which is wrong. The half becomes a 2 when you rearrange: \(\sin C = \dfrac{2 \times \text{Area}}{ab}\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

In the formula Area \(= \tfrac{1}{2}ab\sin C\), what is \(C\)?

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Find the area of a triangle with sides 5 cm and 8 cm and an angle of \(30^\circ\) between them.

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A triangle has sides 7 cm and 9 cm with \(50^\circ\) between them, and a third side of 7.0 cm. Which calculation gives its area?

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Two triangles have sides 6 cm and 10 cm. In one, the angle between them is \(30^\circ\); in the other, \(150^\circ\). How do their areas compare?

Section 03

Polygons, segments and three sides.

Many area problems split into triangles. A regular polygon is a ring of identical isosceles triangles; a segment of a circle is a sector minus a triangle; a triangle given by three sides needs the cosine rule for an angle first.

Worked example A regular hexagon
Find the area of a regular hexagon of side 4 cm.
  1. Join the centre to each vertex: 6 equilateral triangles of side 4 cm, each with a \(60^\circ\) angle at the centre.
  2. One triangle: \(\tfrac{1}{2} \times 4 \times 4 \times \sin 60^\circ = 8\sin 60^\circ = 6.928\ldots\) cm2.
  3. Hexagon: \(6 \times 6.928\ldots = 41.569\ldots\)
Answer: \(41.6\) cm2 (3 s.f.)
Worked example Three sides given
Find the area of the triangle with sides 5 cm, 7 cm and 9 cm.
  1. Cosine rule for the angle between the 5 cm and 7 cm sides: \(\cos C = \dfrac{25 + 49 - 81}{70} = -0.1\), so \(C = 95.739\ldots^\circ\).
  2. Area \(= \tfrac{1}{2} \times 5 \times 7 \times \sin 95.739\ldots^\circ = 17.412\ldots\)
Answer: \(17.4\) cm2 (3 s.f.)
Examiner trap

In a segment problem, using the chord length as a side in \(\tfrac{1}{2}ab\sin C\). The two sides of the triangle are the RADII, and \(C\) is the angle at the centre.

Stage 2 · Exam

Exam-style questions

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In triangle \(PQR\), \(PQ = 11\) cm, \(QR = 7\) cm and angle \(PQR = 118^\circ\). Find the area of the triangle.

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A triangle has sides 8 cm and 9 cm, and an area of 24 cm2. The angle between these sides is obtuse. Find it.

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Find the area of a regular hexagon with side 6 cm.

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Find the area of a triangle with sides 7 cm, 8 cm and 9 cm.

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Area of a triangle, sorted.

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Go deeper · practice
Six original Cambridge-style questions
Areas with acute and obtuse angles, a missing side and a missing angle, an exact equilateral area, and the segment cut off a circular table.