Question

Let $O$ be the origin. $P$ is any point on a circle with $OP$ as diameter. Two points $Q$ and $R$ are on same side of $OP$ such that $\angle POQ=\angle QOR=\theta .$ Let $P,Q,R$ be $z_1,z_2,z_3$ respectively such that $2\sqrt{3}{z}_2^2=(2+\sqrt{3})z_1{z}_3.$ Then the value of $\theta$ in degree measure is

Answer 1

$\frac{z_2}{z_1}=\frac{OQ}{OP}{e}^{i\theta }\Rightarrow e^{i\theta }=\frac{z_2}{z_1}\cdot \frac{OP}{OQ}$
$\frac{z_3}{z_1}=\frac{OR}{OP}{e}^{i2\theta }\Rightarrow e^{i2\theta }=\frac{z_3}{z_1}\cdot \frac{OP}{OR}$

In $△POQ,$
$\cos \theta =\frac{OQ}{OP}\Rightarrow \frac{OP}{OQ}=\frac{1}{\cos \theta }$
In $△POR,$
$\cos 2\theta =\frac{OR}{OP}\Rightarrow \frac{OP}{OR}=\frac{1}{\cos 2\theta }$
Therefore, $e^{i\theta }=\frac{z_2}{z_1}\cdot \frac{1}{\cos \theta }\cdots \left(1\right)$
and $e^{i2\theta }=\frac{z_3}{z_1}\cdot \frac{1}{\cos 2\theta }\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$\frac{{\cos }^2\theta }{\cos 2\theta }=\frac{z_2^2}{z_1{z}_3}=\frac{2+\sqrt{3}}{2\sqrt{3}}$ (Given)
$\Rightarrow {\cos }^2\theta =\frac{2+\sqrt{3}}{4}$
$\Rightarrow 2{\cos }^2\theta =\frac{2+\sqrt{3}}{2}$
$\Rightarrow 2{\cos }^2\theta =1+\cos {30}^{\circ }$
$\therefore \theta ={15}^{\circ }$