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]