IGCSE · Practice questions · The cosine rule

Two sides and the angle between.

Extended. Six original Cambridge-style questions on the cosine rule and choosing the right rule.

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

Side opposite the angle, then substitute carefully.

Side: \(a^2 = b^2 + c^2 - 2bc\cos A\), with \(A\) between \(b\) and \(c\).
Angle: \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\), subtracting the side opposite \(A\).
Obtuse: a negative cosine means an obtuse angle; no ambiguity.
Choose: complete opposite pair, sine rule; two sides and included angle or three sides, cosine rule.
Accuracy: keep full values; sides 3 s.f., angles 1 d.p.

Calculation

[3 marks]

In triangle \(ABC\), \(b = 6.2\) cm, \(c = 8.5\) cm and \(A = 37^\circ\). Calculate \(a\).

\(a^2 = 6.2^2 + 8.5^2 - 2 \times 6.2 \times 8.5 \times \cos 37^\circ\) ✓
\(a^2 = 38.44 + 72.25 - 84.177\ldots = 26.512\ldots\) ✓
\(a = 5.15\) cm (3 s.f.) ✓

Calculation

[3 marks]

A triangle has sides of 7 cm, 8 cm and 12 cm. Calculate the size of its largest angle.

The largest angle faces the 12 cm side. ✓
\(\cos\theta = \dfrac{7^2 + 8^2 - 12^2}{2 \times 7 \times 8} = \dfrac{-31}{112}\) ✓
\(\theta = \cos^{-1}(-0.2767\ldots) = 106.1^\circ\) (1 d.p.) ✓

Calculation

[3 marks]

In triangle \(ABC\), \(b = 4\) cm, \(c = 9\) cm and \(A = 135^\circ\). Calculate \(a\).

\(a^2 = 16 + 81 - 72\cos 135^\circ\) ✓
\(\cos 135^\circ\) is negative, so \(a^2 = 97 + 50.91\ldots = 147.91\ldots\) ✓
\(a = 12.2\) cm (3 s.f.) ✓

Calculation

[4 marks]

In triangle \(ABC\), \(b = 10\) cm, \(c = 13\) cm and \(A = 52^\circ\).

(a) Calculate \(a\).

(b) Calculate angle \(B\).

(a) \(a^2 = 100 + 169 - 260\cos 52^\circ = 108.93\ldots\), so \(a = 10.4\) cm (3 s.f.). ✓
(b) Sine rule with the pair \(a, A\): \(\sin B = \dfrac{10\sin 52^\circ}{10.436\ldots} = 0.7550\ldots\) ✓
\(B = 49.0^\circ\) (1 d.p.); acute, since \(b\) is not the longest side. ✓

Proof

[3 marks]

A triangle has sides 4 cm, 5 cm and 6 cm. Without a calculator, show that the cosine of its largest angle is \(\dfrac{1}{8}\), and explain why the triangle is acute-angled.

The largest angle faces the 6 cm side: \(\cos\theta = \dfrac{4^2 + 5^2 - 6^2}{2 \times 4 \times 5} = \dfrac{5}{40} = \dfrac{1}{8}\). ✓
\(\dfrac{1}{8} > 0\), so the largest angle is acute. ✓
The other two angles are smaller still, so every angle is acute. ✓

Analysis

[5 marks]

Two hikers leave a camp \(C\) at the same time. Asha walks 6 km on a bearing of \(030^\circ\) to point \(A\). Ben walks 8 km on a bearing of \(110^\circ\) to point \(B\).

(a) Calculate the distance \(AB\).

(b) Calculate angle \(CAB\).

(a) Angle \(ACB = 110^\circ - 30^\circ = 80^\circ\). \(AB^2 = 36 + 64 - 96\cos 80^\circ = 83.33\ldots\), so \(AB = 9.13\) km (3 s.f.). ✓
(b) Sine rule: \(\sin(CAB) = \dfrac{8\sin 80^\circ}{9.128\ldots} = 0.8630\ldots\), so angle \(CAB = 59.7^\circ\) (acute, since \(CB = 8\) km is not the longest side). ✓
(c) The bearing of \(C\) from \(A\) is \(030^\circ + 180^\circ = 210^\circ\). \(B\) is \(59.7^\circ\) anticlockwise from that direction. ✓
Bearing of \(B\) from \(A\) \(= 210^\circ - 59.7^\circ = 150.3^\circ\) ✓

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.