Triangle ABC has AB = 5 cm, BC = 6 cm and area 10 \({\text{c}}{{\text{m}}^2}\).

(a)     Find \(\sin \hat B\).

(b)     Hence, find the two possible values of AC, giving your answers correct to two decimal places.

(a)     area \( = \frac{1}{2} \times {\text{BC}} \times {\text{AB}} \times \sin B\)     (M1)

\(\left( {10 = \frac{1}{2} \times 5 \times 6 \times \sin B} \right)\)

\(\sin \hat B = \frac{2}{3}\)     A1

(b)     \(\cos B = \pm \frac{{\sqrt 5 }}{3}{\text{ }}( = \pm 0.7453 \ldots ){\text{ or }}B = 41.8 \ldots {\text{ and }}138.1 \ldots \)     (A1)

\({\text{A}}{{\text{C}}^2} = {\text{B}}{{\text{C}}^2} + {\text{A}}{{\text{B}}^2} - 2 \times {\text{BC}} \times {\text{AB}} \times \cos B\)     (M1)

\({\text{AC}} = \sqrt {{5^2} + {6^2} - 2 \times 5 \times 6 \times 0.7453 \ldots } {\text{ or }}\sqrt {{5^2} + {6^2} + 2 \times 5 \times 6 \times 0.7453 \ldots } \)

\({\text{AC}} = 4.03{\text{ or }}10.28\)     A1A1

[6 marks]

Most candidates attempted this question and part (a) was answered correctly by most candidates but in (b), despite the wording of the question, the obtuse angle was often omitted leading to only one solution.

In many cases early rounding led to inaccuracy in the final answers and many candidates failed to round their answers to two decimal places as required.