In the figure,theta_1+theta_2=frac{pi}{2} and | vedclass

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(C) Let $AB = x$. From the figure,$AC = x 	an 	heta_1$ and $CD = x$. Also,$BD = AC = x 	an 	heta_1$ and $DE = CD 	an 	heta_2 = x 	an 	heta_2$.

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$6$

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(C) Let $AB = x$. From the figure,$AC = x 	an 	heta_1$ and $CD = x$. Also,$BD = AC = x 	an 	heta_1$ and $DE = CD 	an 	heta_2 = x 	an 	heta_2$.Given $\sqrt{3}(BE) = 4(AB)$,so $\sqrt{3}(BD + DE) = 4x$.$\sqrt{3}(x 	an 	heta_1 + x 	an 	heta_2) = 4x \implies \sqrt{3}(	an 	heta_1 + \cot 	heta_1) = 4$ (since $	heta_1 + 	heta_2 = \frac{\pi}{2}, 	an 	heta_2 = \cot 	heta_1$).$\sqrt{3}(	an 	heta_1 + \frac{1}{	an 	heta_1}) = 4 \implies 3 	an^2 	heta_1 - 4\sqrt{3} 	an 	heta_1 + 3 = 0$.Solving for $	an 	heta_1$,we get $	an 	heta_1 = \sqrt{3}$ or $	an 	heta_1 = \frac{1}{\sqrt{3}}$.If $	an 	heta_1 = \sqrt{3}$,then $	heta_1 = \frac{\pi}{3}$ and $	heta_2 = \frac{\pi}{6}$. Then $\frac{	heta_2}{	heta_1} = \frac{1}{2}$.If $	an 	heta_1 = \frac{1}{\sqrt{3}}$,then $	heta_1 = \frac{\pi}{6}$ and $	heta_2 = \frac{\pi}{3}$. Then $\frac{	heta_2}{	heta_1} = 2$.Since $\frac{	heta_2}{	heta_1}$ is largest,we take $	heta_1 = \frac{\pi}{6}$.Area of $	riangle CAB = \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes x 	imes (x 	an \frac{\pi}{6}) = \frac{1}{2} x^2 \frac{1}{\sqrt{3}} = 2\sqrt{3}-3$.$x^2 = 2\sqrt{3}(2\sqrt{3}-3) = 12 - 6\sqrt{3} = (3-\sqrt{3})^2 \implies x = 3-\sqrt{3}$.Perimeter of $	riangle CED = CD + DE + CE = x + x 	an 	heta_2 + \sqrt{x^2 + (x 	an 	heta_2)^2} = x(1 + 	an \frac{\pi}{3} + \sec \frac{\pi}{3}) = x(1 + \sqrt{3} + 2) = x(3+\sqrt{3})$.Perimeter $= (3-\sqrt{3})(3+\sqrt{3}) = 9 - 3 = 6$.

In the figure,$\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(BE)=4(AB)$. If the area of $\triangle CAB$ is $2\sqrt{3}-3 \text{ unit}^2$,when $\frac{\theta_2}{\theta_1}$ is the largest,then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.

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