IGCSE · Practice questions · Area of a triangle: half ab sin C

Two sides and the angle between.

Extended. Six original Cambridge-style questions on the formula area = ½ab sin C.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Two sides, the angle between them, and a half.

Formula: Area \(= \tfrac{1}{2}ab\sin C\), with \(C\) between \(a\) and \(b\).
Obtuse angles: work exactly the same way; \(\sin C\) is still positive.
Backwards: \(\sin C = \dfrac{2 \times \text{Area}}{ab}\); consider \(180^\circ - C\) as well.
Three sides: cosine rule for an angle first.
Segment: sector area minus triangle area, both using the angle at the centre.

Calculation

[2 marks]

Calculate the area of triangle \(ABC\) in which \(a = 7.4\) cm, \(b = 5.9\) cm and \(C = 63^\circ\).

Area \(= \tfrac{1}{2} \times 7.4 \times 5.9 \times \sin 63^\circ\) ✓
\(= 19.5\) cm² (3 s.f.) ✓

Calculation

[2 marks]

Two sides of a triangle are 12 cm and 15 cm, and the angle between them is \(145^\circ\). Calculate the area.

Area \(= \tfrac{1}{2} \times 12 \times 15 \times \sin 145^\circ = 90\sin 145^\circ\) ✓
\(= 51.6\) cm² (3 s.f.) ✓

Calculation

[3 marks]

A triangle has area 45 cm². Two of its sides meet at an angle of \(28^\circ\). One of these sides is 14 cm. Calculate the length of the other.

\(45 = \tfrac{1}{2} \times 14 \times x \times \sin 28^\circ = 7x\sin 28^\circ\) ✓
\(x = \dfrac{45}{7\sin 28^\circ} = 13.7\) cm (3 s.f.) ✓

Calculation

[3 marks]

A triangle has sides 6 cm and 10 cm and an area of 18 cm². Find the two possible sizes of the angle between these two sides.

\(18 = \tfrac{1}{2} \times 6 \times 10 \times \sin C = 30\sin C\), so \(\sin C = 0.6\). ✓
\(C = \sin^{-1}(0.6) = 36.9^\circ\) ✓
or \(C = 180^\circ - 36.9^\circ = 143.1^\circ\) ✓

Proof

[2 marks]

Without a calculator, show that the area of an equilateral triangle of side \(x\) cm is \(\dfrac{\sqrt{3}}{4}x^2\) cm².

Every angle of an equilateral triangle is \(60^\circ\), and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\). ✓
Area \(= \tfrac{1}{2} \times x \times x \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4}x^2\). ✓

Analysis

[4 marks]

A circular table top has radius 50 cm and centre \(O\). A straight cut \(AB\) removes a segment, where angle \(AOB = 80^\circ\).

(a) Calculate the area of triangle \(AOB\).

(b) Calculate the area of the segment removed.

(a) Area of triangle \(= \tfrac{1}{2} \times 50 \times 50 \times \sin 80^\circ = 1231.0\ldots\) so 1230 cm² (3 s.f.) ✓
(b) Sector area \(= \dfrac{80}{360} \times \pi \times 50^2 = 1745.3\ldots\) cm² ✓
Segment \(= 1745.3\ldots - 1231.0\ldots = 514.3\ldots\) ✓
Segment area \(= 514\) cm² (3 s.f.) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.