Mathematics - Properties of Triangles Question with Solution | TestHub

In $△ABC$, by sine rule, $$\begin{array}{rl}& \frac{a}{\sin A}=\frac{2\sqrt{2}}{\sin 30^{\circ }}=\frac{4}{\sin C}\\ \Rightarrow & C=45^{\circ },C^{\prime }=135^{\circ }\\ \Rightarrow & A=180^{\circ }-\left(45^{\circ }+30^{\circ }\right)=105^{\circ }\end{array}$$
When $C^{\prime }=135^{\circ }$, then $$\begin{array}{rl}& A=180^{\circ }-\left(135^{\circ }+30^{\circ }\right)=15^{\circ }\\ & \text{ Area of }△ABC=\frac{1}{2}AB\times AC\sin A\\ & =\frac{1}{2}\times 4\times 2\sqrt{2}\sin \left(105^{\circ }\right)\\ & =4\sqrt{2}\times \frac{\sqrt{3}+1}{2\sqrt{2}}=2(\sqrt{3}+1)\\ & \text{ Area of }△AB{C}^{\prime }=\frac{1}{2}AB\times AC\sin A\\ & =\frac{1}{2}\times 4\times 2\sqrt{2}\sin \left(15^{\circ }\right)=2(\sqrt{3}-1)\end{array}$$ Difference of areas of triangles $$=|2(\sqrt{3}+1)-2(\sqrt{3}-1)|=4$$
$$AD=2,DC=2$$ Difference of areas of $△ABC$ and $△AB{C}^{\prime }$ $$\begin{array}{rl}& =\text{ Area of }△AC{C}^{\prime }\\ & =\frac{1}{2}AD\times C{C}^{\prime }=\frac{1}{2}\times 2\times 4=4\end{array}$$