Higher Paper 2

\begin{aligned} &\text{In } \triangle ABD:\; \angle A=110^\circ,\; \angle D=40^\circ \Rightarrow \angle ABD=30^\circ.\\ &AD\parallel BC \Rightarrow \angle ABC=110^\circ.\\ &\angle CBD=110^\circ-30^\circ=80^\circ.\\[2mm] &\text{Sine Rule: } \dfrac{BD}{\sin 110^\circ}=\dfrac{8.4}{\sin 40^\circ}\Rightarrow BD\approx 12.28\text{ cm}.\\[2mm] &\text{Area}(BCD)=\tfrac12\times 14.2\times 12.28\times\sin 80^\circ\approx 85.9\text{ cm}^2. \end{aligned}