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

As,

$\cos B=\frac{3}{5}\Rightarrow \sin B=\frac{4}{5}$

Given:

$R=5$

By sine rule, we have

$\frac{c}{\sin C}=2R$

$\Rightarrow \frac{5}{10}=\sin C$

$\Rightarrow C=30°$
Now,

$\frac{b}{\sin B}=2R$

$\Rightarrow b=\frac{4}{5}\times 10=8$

Now, by cosine formula

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\Rightarrow \frac{3}{5}=\frac{a^2+25-64}{2\left(5\right)a}$

$\Rightarrow a^2-6a-39=0$

$\Rightarrow a=\frac{6\pm \sqrt{192}}{2}=\frac{6\pm 8\sqrt{3}}{2}$

$\Rightarrow a=\left(3+4\sqrt{3}\right)$ 

(Reject $a=3-4\sqrt{3}<0$)

Now, $\Delta =\frac{abc}{4R}=\frac{\left(3+4\sqrt{3}\right)·8·5}{4·5}=2\left(3+4\sqrt{3}\right)$

$\Rightarrow \Delta =\left(6+8\sqrt{3}\right)$ sq. units.