In the figure, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4(A B)$. If the area of $\triangle C A B$ is $2 \sqrt{3}-3$ unit $^2$, when $\frac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\triangle \mathrm{CED}$ is equal to $\_\_\_\_$ .
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
