Find $AC$.

Find the area of triangle $ABC$.

The area of triangle $ACD$ is half the area of triangle $ABC$.

Find the possible values of $\theta$.

Given that $\theta$ is obtuse, find $CD$.

evidence of choosing sine rule     (M1)

eg$\;\;\;\frac{{{\text{AC}}}}{{\sin {\rm{C\hat BA}}}} = \frac{{{\text{AB}}}}{{\sin {\rm{A\hat CB}}}}$

correct substitution     (A1)

eg$\;\;\;\frac{{{\text{AC}}}}{{\sin 44^\circ }} = \frac{{15}}{{\sin 83^\circ }}$

$10.4981$

${\text{AC}} = 10.5{\text{ }}{\text{ (cm)}}$     A1     N2

[3 marks]

finding ${\rm{C\hat AB}}$ (seen anywhere)     (A1)

eg$\;\;\;180^\circ  - 44^\circ  - 83^\circ ,{\rm{ C\hat AB}} = 53^\circ$

correct substitution for area of triangle $ABC$     A1

eg$\;\;\;\frac{1}{2} \times 15 \times 10.4981 \times \sin 53^\circ$

62.8813

${\text{area}} = 62.9{\text{ }}{\text{ (c}}{{\text{m}}^2}{\text{)}}$     A1     N2

[3 marks]

correct substitution for area of triangle $DAC$    (A1)

eg$\;\;\;\frac{1}{2} \times 6 \times 10.4981 \times \sin \theta$

attempt to equate area of triangle $ACD$ to half the area of triangle $ABC$     (M1)

eg$\;\;\;{\text{area ACD}} = \frac{1}{2} \times {\text{ area ABC; 2ACD}} = {\text{ABC}}$

correct equation     A1

eg$\;\;\;\frac{1}{2} \times 6 \times 10.4981 \times \sin \theta  = \frac{1}{2}(62.9),{\text{ }}62.9887\sin \theta  = 62.8813,{\text{ }}\sin \theta  = 0.998294$

$86.6531$, $93.3468$

$\theta  = 86.7^\circ {\text{ }},{\text{ }}\theta  = 93.3^\circ {\text{ }}$     A1A1     N2

[5 marks]

Note:     Note: If candidates use an acute angle from part (c) in the cosine rule, award M1A0A0 in part (d).

evidence of choosing cosine rule     (M1)

eg$\;\;\;{\text{C}}{{\text{D}}^2} = {\text{A}}{{\text{D}}^2} + {\text{A}}{{\text{C}}^2} - 2 \times {\text{AD}} \times {\text{AC}} \times \cos \theta$

correct substitution into rhs     (A1)

eg$\;\;\;{\text{C}}{{\text{D}}^2} = {6^2} + {10.498^2} - 2(6)(10.498)\cos 93.336^\circ$

$12.3921$

$12.4{\text{ }}{\text{ (cm)}}$     A1     N2

[3 marks]

Total [14 marks]