In the figure, θ _ 1 + θ _ 2 = (π)/(2) and 3 (BE) = 4(AB) . If the...

Given,

And ${\theta}_{1} + {\theta}_{2} = \dfrac{\pi}{2}$

Now, Let $AB = x, BD = y$

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

$⇒ \sqrt{3}\left(y + DE\right) = 4x$

$⇒ DE = \dfrac{4x}{\sqrt{3}} - y$

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

$⇒ \dfrac{1}{2}xy = 2\sqrt{3} - 3$

$⇒ y = \dfrac{4\sqrt{3} - 6}{x}$

Now finding,

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

${tan \theta}_{1} = \dfrac{y}{x} = \dfrac{4\sqrt{3} - 6}{{x}^{2}}$

Now taking $tan$ both of ${\theta}_{1} + {\theta}_{2} = \dfrac{\pi}{2}$ we get,

${tan \theta}_{1} \cdot {tan \theta}_{2} = 1$

$⇒ \left(\dfrac{4}{\sqrt{3}} - \dfrac{4\sqrt{3} - 6}{{x}^{2}}\right) \cdot \left(\dfrac{4\sqrt{3} - 6}{{x}^{2}}\right) = 1$

$⇒ \dfrac{4\sqrt{3} - 6}{{x}^{2}} = \sqrt{3}or\dfrac{1}{\sqrt{3}}$

So, $\dfrac{{\theta}_{2}}{{\theta}_{1}}$ is maximum when $\dfrac{4\sqrt{3} - 6}{{x}^{2}} = \dfrac{1}{\sqrt{3}}$

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

$⇒ {x}^{2} = 12 - 6\sqrt{3}$

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

$⇒ x = 3 - \sqrt{3}$ and ${\theta}_{2} = 60 °$

$⇒ \tan 60 ° = \dfrac{DE}{CD} ⇒ DE = x \cdot \sqrt{3} = 3\sqrt{3} - 3$

And $\cos 60 ° = \dfrac{CD}{CE} ⇒ CE = 2x = 6 - 2\sqrt{3}$

So, the Perimeter of $∆ CED = CD + DE + CE = 6$