Triangle \(ABC\) has area \({\text{21 c}}{{\text{m}}^{\text{2}}}\). The sides \(AB\) and \(AC\) have lengths \(6\) cm and \(11\) cm respectively. Find the two possible lengths of the side \(BC\).

\(21 = \frac{1}{2} \bullet 6 \bullet 11 \bullet \sin A\)     (M1)

\(\sin A = \frac{7}{{11}}\)     (A1)

EITHER

\(\hat A = 0.6897 \ldots ,{\text{ }}2.452 \ldots \left( {\hat A = \arcsin \frac{7}{{11}},{\text{ }}\pi  - \arcsin \frac{7}{{11}} = 39.521 \ldots ^\circ ,{\text{ }}140.478 \ldots ^\circ } \right)\)     (A1)

OR

\(\cos A =  \pm \frac{{6\sqrt 2 }}{{11}}\;\;\;( =  \pm 0.771 \ldots )\)     (A1)

THEN

\({\text{B}}{{\text{C}}^2} = {6^2} + {11^2} - 2 \bullet 6 \bullet 11\cos A\)     (M1)

\({\text{BC}} = 16.1\) or \(7.43\)     A1A1

Note:     Award M1A1A0M1A1A0 if only one correct solution is given.

[6 marks]