In triangle ABC, AB = 9 cm , AC = 12 cm , and $\hat B$ is twice the size of ${\hat C}$ .

Find the cosine of ${\hat C}$ .

$\frac{9}{{\sin C}} = \frac{{12}}{{\sin B}}$     (M1)

$\frac{9}{{\sin C}} = \frac{{12}}{{\sin 2C}}$     A1

Using double angle formula $\frac{9}{{\sin C}} = \frac{{12}}{{2\sin C\cos C}}$     M1

$\Rightarrow 9(2\sin C\cos C) = 12\sin C$

$\Rightarrow 6\sin C(3\cos C - 2) = 0\,\,\,\,\,{\text{or equivalent}}$     (A1)

$(\sin C \ne 0)$

$\Rightarrow \cos C = \frac{2}{3}$     A1

[5 marks]

There were many totally correct solutions to this question, but again a significant minority did not make much progress. The most common reasons for this were that candidates immediately assumed that because the question asked for the cosine of ${\hat C}$ that they should use the cosine rule, or they did not draw a diagram and then confused which angles were opposite which sides.