Animated Solution for Mathematics - Trigonometry: If in a triangle $ABC$, $AB=5$ units, $\angle B={\cos }^{-1}\left(\frac{3}{5}\right)$ and radius of circumcircle of $\Delta ABC$ is 5 units, then the area (in sq. units) of $\Delta ABC$ is:

Unlocking the Triangle

Welcome, future engineer! Today, we are not just solving a geometry problem; we are unraveling the elegant dance of a triangle inscribed within a circle.

In the world of JEE Advanced, geometry is not about memorizing formulas; it is about seeing the hidden connections between lengths, angles, and the circumcircle that binds them. Let us embark on this journey together.

Decoding the Given Information

Imagine you are standing before a circle with a radius $R=5$. Inside, a triangle $\Delta ABC$ is perfectly nestled.

We are given the side $AB=c=5$ and the angle $\angle B={\cos }^{-1}(\frac{3}{5})$. The moment you see a circumradius $R$ and a side length $c$, your mind should immediately race to the Sine Rule.

It is the golden key that unlocks the relationship between the sides and the circumcircle:

But before we can use it, we need to understand our angles. We are given $\cos B=\frac{3}{5}$. Using the fundamental identity ${\sin }^{2}B+{\cos }^{2}B=1$, we find:

We have our first piece of the puzzle!

The Hunt for the Missing Pieces

Now, let us apply the Sine Rule to find the missing side $b$ (which is $AC$). The rule states $\frac{b}{\sin B}=2R$.

Substituting our known values, we get:

This simplifies beautifully to $b=10\times \frac{4}{5}=8$. We now have side $b=8$ and side $c=5$. We are closing in on the solution!

The Cosine Rule Challenge

We have two sides and an angle, but we are missing the third side, $a$. This is where the Cosine Rule becomes our most trusted ally.

The formula $\cos B=\frac{a^2+c^2-b^2}{2ac}$ allows us to relate all three sides to the angle $B$. Let us substitute our values carefully:

This simplifies to $\frac{3}{5}=\frac{a^2+25-64}{10a}$, which further reduces to $\frac{3}{5}=\frac{a^2-39}{10a}$.

Cross-multiplying gives us $6a=a^2-39$, or the quadratic equation $a^2-6a-39=0$. Do not let this quadratic equation intimidate you; it is simply a gatekeeper to the final answer.

The Grand Finale

Using the quadratic formula $a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we find:

Since a side length must be positive, we reject the negative root and accept $a=3+4\sqrt{3}$.

Finally, we calculate the area using the elegant formula $\text{Area}=\frac{abc}{4R}$. Substituting our values:

The $5$ cancels out, the $8$ divided by $4$ leaves $2$, and we are left with $2(3+4\sqrt{3})=6+8\sqrt{3}$.

And there it is! The area of our triangle is $6+8\sqrt{3}$ square units. You have successfully navigated the constraints, applied the laws of trigonometry, and emerged victorious.