given an isosceles triangle , whose one angle is 2/3 and the radius of its incircle = √3. then find the area of the triangle.

$\frac{2}{3} \mathrm{radian}={60}^{\mathrm{o}}$

That is, one angle of the isosceles triangle is ${60}^{\mathrm{o}}$
$\Rightarrow \mathrm{All} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{isosceles} \mathrm{triangle} \mathrm{are} {60}^{\mathrm{o}}$
$\Rightarrow$The triangle is equilateral.



Now, inradius is given as $\sqrt{3}$
$\Rightarrow \mathrm{OQ}=\sqrt{3}$

$\Rightarrow \mathrm{OA}=2\sqrt{3}$   [since O divides AQ in the ratio 2:1]

So, AQ=OQ+OA=$\sqrt{3}+2\sqrt{3}=3\sqrt{3}$

Now, in right angled $△AQB, A{B}^2=A{Q}^2+B{Q}^2$
$$\begin{gathered} \Rightarrow A{B}^2={\left(\sqrt{3}+2\sqrt{3}\right)}^2 + {\left(\frac{AQ}{2}\right)}^2 \\ \Rightarrow \frac{3}{4}A{B}^2=27 \\ \Rightarrow AB=\frac{3}{2} \end{gathered}$$

Now, $$\begin{gathered} BC=AB \\ \Rightarrow BC=\frac{3}{2} \end{gathered}$$

Therefore, area of the triangle = $$\begin{gathered} =\frac{1}{2}\times AQ\times BC \\ =\frac{1}{2}\times 3\sqrt{3}\times \frac{3}{2} \\ =\frac{9\sqrt{3}}{4} \end{gathered}$$