Mathematics - Complex Number Question with Solution | TestHub

$\begin{array}{rl}& OA=\left|z_1\right|=\sqrt{3+8}=\sqrt{11}\\ & \text{ and }OB=\frac{1}{\sqrt{3}}\left|z_1\right|=\sqrt{\frac{11}{3}}\\ & A{B}^2=O{A}^2+O{B}^2-2\cdot OA\cdot OB\cos \frac{\pi }{6}\\ & =11+\frac{11}{3}-2\cdot \frac{11}{\sqrt{3}}\cdot \frac{\sqrt{3}}{2}\\ & \therefore AB=\sqrt{\frac{11}{3}}\\ & \therefore \text{ Area of }△ABD=\frac{1}{2}\cdot OA\cdot OB\cdot \sin \frac{\pi }{6}\\ & =\frac{11}{4\sqrt{3}}\text{ sq. units }\end{array}$
Here $OB=AB$ and $\angle A=\frac{2\pi }{3}$
$\therefore △ABD$ is an obtuse angled isosceles triangle.