Write down the length of ${\rm{MB}}$.

Find the size of angle ${\rm{C\hat MB}}$.

Find the length of ${\rm{CB}}$.

$10$ m     (A1)(C1)

${\rm{A\hat MC}} = 70^\circ \;\;\;$OR$\;\;\;{\rm{A\hat CM}} = 55^\circ$     (A1)

${\rm{C\hat MB}} = 110^\circ$     (A1)     (C2)

${\text{C}}{{\text{B}}^2} = {10^2} + {10^2} - 2 \times 10 \times 10 \times \cos 110^\circ$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the cosine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

$\frac{{{\text{CB}}}}{{\sin 110^\circ }} = \frac{{10}}{{\sin 35^\circ }}$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the sine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

${\rm{A\hat CB}} = 90^\circ$     (A1)

$\sin 55^\circ  = \frac{{{\text{CB}}}}{{55}}\;\;\;$OR$\;\;\;\cos 35^\circ  = \frac{{{\text{CB}}}}{{20}}$     (M1)

Note: Award (A1) for some indication that ${\rm{A\hat CB}} = 90^\circ$, (M1) for correct trigonometric equation.

OR

Perpendicular ${\rm{MN}}$ is drawn from ${\rm{M}}$ to ${\rm{CB}}$.     (A1)

$\frac{{\frac{1}{2}{\text{CB}}}}{{10}} = \cos 35^\circ$     (M1)

Note: Award (A1) for some indication of the perpendicular bisector of ${\rm{BC}}$, (M1) for correct trigonometric equation.

${\text{CB}} = 16.4{\text{ (m)}}\;\;\;\left( {16.3830 \ldots {\text{ (m)}}} \right)$     (A1)(ft)(C3)

Notes: Where a candidate uses ${\rm{C\hat MB}} = 90^\circ$ and finds ${\text{CB}} = 14.1{\text{ (m)}}$ award, at most, (M1)(A1)(A0).

Where a candidate uses ${\rm{C\hat MB}} = 60^\circ$ and finds ${\text{CB}} = 10{\text{ (m)}}$ award, at most, (M1)(A1)(A0).