(i)     Find the size of angle ${\rm{ACB}}$.

(ii)     Show that the size of angle ${\rm{DCE}}$ is $85.5^\circ$, correct to one decimal place.

The surveyor measures the size of angle ${\text{CDE}}$ to be twice that of angle ${\text{DEC}}$.

(i)     Using angle ${\text{DCE}} = 85.5^\circ$, find the size of angle ${\text{DEC}}$.

(ii)     Find the length of ${\text{DE}}$.

Calculate the area of triangle ${\text{DEC}}$.

(i)     $\cos {\rm{A\hat CB}} = \frac{{{{10}^2} + {{12}^2} - {{15}^2}}}{{2 \times 10 \times 12}}$     (M1)(A1)

Note: Award (M1) for substituted cosine rule,

(A1) for correct substitution.

${\rm{A\hat CB}} = 85.5^\circ \;\;\;({\text{85.4593}} \ldots {\text{)}}$     (A1)(G2)

(ii)     ${\rm{D\hat CE}} = {\rm{A\hat CB}}\;\;\;{\text{and}}\;\;\;{\rm{A\hat CB}} = 85.5^\circ \;\;\;({\text{85.4593}} \ldots ^\circ {\text{)}}$     (A1)

OR

${\rm{B\hat CE}} = 180^\circ  - 85.5^\circ  = 94.5^\circ \;\;\;{\text{and}}\;\;\;{\rm{D\hat CE}} = 180^\circ  - 94.5^\circ  = 85.5^\circ$     (A1)

Notes: Both reasons must be seen for the (A1) to be awarded.

${\rm{D\hat CE}} = 85.5^\circ$     (AG)

(i)     ${\rm{D\hat EC}} = \frac{{180^\circ  - 85.5^\circ }}{3}$     (M1)

${\rm{D\hat EC}} = 31.5^\circ$     (A1)(G2)

(ii)     $\frac{{\sin (31.5^\circ )}}{9} = \frac{{\sin (85.5^\circ )}}{{{\text{DE}}}}$     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule, (A1) for correct substitution.

${\text{DE}} = 17.2{\text{ (km)}}(17.1718 \ldots )$.     (A1)(ft)(G2)

$0.5 \times 17.1718 \ldots  \times 9 \times \sin (63^\circ )$     (A1)(ft)(M1)(A1)(ft)

Note: Award (A1)(ft) for $63$ seen, (M1) for substituted triangle area formula, (A1)(ft) for $0.5 \times 17.1718 \ldots  \times 9 \times \sin ({\text{their angle CDE}})$.

OR

${\text{(triangle height}} = ){\text{ }}9 \times \sin (63^\circ )$     (A1)(ft)(A1)(ft)

${\text{0.5}} \times {\text{17.1718}} \ldots  \times {\text{9}} \times {\text{sin(their angle CDE)}}$     (M1)

Note: Award (A1)(ft) for $63$ seen, (A1)(ft) for correct triangle height with their angle ${\text{CDE}}$, (M1) for ${\text{0.5}} \times {\text{17.1718}} \ldots  \times {\text{9}} \times {\text{sin(their angle CDE)}}$.

$= 68.9{\text{ k}}{{\text{m}}^2}\;\;\;(68.8509 \ldots )$     (A1)(ft)(G3)

Notes: Units are required for the last (A1)(ft) mark to be awarded.

Follow through from parts (b)(i) and (b)(ii).

Follow through from their angle ${\text{CDE}}$ within this part of the question.