Let $O$ be the origin, the point $A$ be $z_1=\sqrt{3}+2 \sqrt{2} i$, the point $B\left(z_2\right)$ be such that $\sqrt{3}\left|z_2\right|=\left|z_1\right|$ and $\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}$. Then

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Aicharya · Accepted Answer

$\begin{aligned} & O A=\left|z_1ight|=\sqrt{3+8}=\sqrt{11} \ & 	ext { and } O B=\frac{1}{\sqrt{3}}\left|z_1ight|=\sqrt{\frac{11}{3}} \ & A B^2=O A^2+O B^2-2 \cdot O A \cdot O B \cos \frac{\pi}{6} \ & \quad=11+\frac{11}{3}-2 \cdot \frac{11}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \ & 	herefore \quad A B=\sqrt{\frac{11}{3}} \ & 	herefore \quad 	ext { Area of } 	riangle A B D=\frac{1}{2} \cdot O A \cdot O B \cdot \sin \frac{\pi}{6} \ & \quad=\frac{11}{4 \sqrt{3}} 	ext { sq. units } \end{aligned}$ Here $O B=A B$ and $\angle A=\frac{2 \pi}{3}$ $	herefore \quad 	riangle A B D$ is an obtuse angled isosceles triangle.

Let $O$ be the origin, the point $A$ be $z_1=\sqrt{3}+2 \sqrt{2} i$, the point $B\left(z_2\right)$ be such…

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