(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.