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.