In the figure, θ 1 + θ 2 = π 2 and 3 BE = 4 AB . If the area of ∆ CAB is 2 3 - 3  …

Given,

And ${\mathrm{\theta }}_1+{\mathrm{\theta }}_2=\frac{\mathrm{\pi }}{2}$

Now, Let $\mathrm{AB} =x, \mathrm{BD}=y$

Also given, $\sqrt{3}\mathrm{BE}=4\mathrm{AB}$

$\Rightarrow \sqrt{3}\left(y+\mathrm{DE}\right)=4x$

$\Rightarrow \mathrm{DE}=\frac{4x}{\sqrt{3}}-y$

Now given area of triangle $∆\mathrm{CAB}=2\sqrt{3}-3$

$\Rightarrow \frac{1}{2}xy=2\sqrt{3}-3$

$\Rightarrow y=\frac{4\sqrt{3}-6}{x}$

Now finding,

${\mathrm{tan\theta }}_2=\frac{\left(\frac{4x}{\sqrt{3}}-\mathrm{y}\right)}{x}=\frac{4}{\sqrt{3}}-\frac{(4\sqrt{3}-6)}{x^2}$

${\mathrm{tan\theta }}_1=\frac{y}{x}=\frac{4\sqrt{3}-6}{x^2}$

Now taking $\tan$ both of ${\mathrm{\theta }}_1+{\mathrm{\theta }}_2=\frac{\mathrm{\pi }}{2}$ we get,

${\mathrm{tan\theta }}_1·{\mathrm{tan\theta }}_2=1$

$\Rightarrow \left(\frac{4}{\sqrt{3}}-\frac{4\sqrt{3}-6}{x^2}\right)·\left(\frac{4\sqrt{3}-6}{x^2}\right)=1$

$\Rightarrow \frac{4\sqrt{3}-6}{x^2}=\sqrt{3} \mathrm{or} \frac{1}{\sqrt{3}}$

So, $\frac{{\mathrm{\theta }}_2}{{\mathrm{\theta }}_1}$ is maximum when $\frac{4\sqrt{3}-6}{x^2}=\frac{1}{\sqrt{3}}$

$\Rightarrow x^2=\sqrt{3}\left(4\sqrt{3}-6\right)$

$\Rightarrow x^2=12-6\sqrt{3}$

$\Rightarrow x^2={\left(3-\sqrt{3}\right)}^2$

$\Rightarrow x=3-\sqrt{3}$ and ${\mathrm{\theta }}_2=60°$

$\Rightarrow \tan 60°=\frac{\mathrm{DE}}{\mathrm{CD}}\Rightarrow \mathrm{DE}=x·\sqrt{3}=3\sqrt{3}-3$

And $\cos 60°=\frac{\mathrm{CD}}{\mathrm{CE}}\Rightarrow \mathrm{CE}=2x=6-2\sqrt{3}$

So, the Perimeter of $∆\mathrm{CED}=\mathrm{CD}+\mathrm{DE}+\mathrm{CE}=6$