Animated Solution for Mathematics - Trigonometry: In the figure, ${\theta }_1+{\theta }_2=\frac{\pi }{2}$ and $\sqrt{3}(BE)=4(AB)$. If the area of $\Delta CAB$ is $2\sqrt{3}-3{\text{ unit}}^2$, when $\frac{{\theta }_2}{{\theta }_1}$ is the largest, then the perimeter (in unit) of $\Delta CED$ is equal to

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering a hidden symmetry. When you look at this diagram, do not see just lines and angles; see a structure, a rectangle $ACDB$ that acts as the foundation for two distinct worlds: the lower triangle $\Delta CAB$ and the upper triangle $\Delta CED$.

Our journey begins by defining our variables. Let us set the base $AB=x$ and the height $BD=y$. Because $ACDB$ is a rectangle, we immediately know that $AC=y$ and $CD=x$. This is our anchor.

We are given the area of $\Delta CAB$ as $2\sqrt{3}-3$. Since the area of a right-angled triangle is $\frac{1}{2}\cdot \text{base}\cdot \text{height}$, we have:

This gives us a crucial relationship:

Keep this in your pocket; we will need it later.

Now, look at the vertical segment $BE$. The problem gives us a beautiful constraint: $\sqrt{3}(BE)=4(AB)$. Since $BE=BD+DE=y+DE$, we can write:

Rearranging this, we find $DE=\frac{4x}{\sqrt{3}}-y$. This is the key to the upper triangle.

In $\Delta CED$, $\tan {\theta }_2=\frac{DE}{CD}=\frac{DE}{x}$. Substituting our expression for $DE$, we get:

Since $\tan {\theta }_1=\frac{y}{x}$, we have arrived at the elegant relation:

Here is where the magic happens. We are told ${\theta }_1+{\theta }_2=\frac{\pi }{2}$. This means $\tan {\theta }_2=\cot {\theta }_1=\frac{1}{\tan {\theta }_1}$.

Substituting this into our previous equation, we get:

Multiplying by $\tan {\theta }_1$ and rearranging, we obtain the quadratic equation:

Solving this using the quadratic formula, we find $\tan {\theta }_1=\sqrt{3}$ or $\tan {\theta }_1=\frac{1}{\sqrt{3}}$.

To maximize the ratio $\frac{{\theta }_2}{{\theta }_1}$, we need ${\theta }_1$ to be as small as possible. Thus, we choose $\tan {\theta }_1=\frac{1}{\sqrt{3}}$, which means ${\theta }_1=30^{\circ }$ and ${\theta }_2=60^{\circ }$.

Substituting $\tan 30^{\circ }=\frac{y}{x}=\frac{1}{\sqrt{3}}$ back into our area equation, we find $x=3-\sqrt{3}$. Now, calculating the sides of $\Delta CED$:

Adding these together, the perimeter is:

The terms cancel with such satisfying precision. You have mastered the geometry, and the final answer is 6.