We are given that two tangents $\(\mathrm{OA} \& \mathrm{OB}\)$, of length 3 inches each, intersect at point O and form an angle of 60 degrees (Two tangents meeting at a common point outside the circle are equal in length).


As triangle $\(\mathrm{AOC} \&\)$ triangle CBO are congruent triangles, we get angle $\(\mathrm{AOC}=\)$ angle $\(\mathrm{COB}=30\)$ degrees each. Also as radius is always perpendicular on tangent, we get angle $\mathrm{CAO}=90$ degree .

Now, using the measure of angle CAO ( $\(=90\)$ degree) \& angle AOC ( $\(=30\)$ degrees), in triangle ACO , we get angle $\(\mathrm{ACO}=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)=180^{\circ}-120^{\circ}=60^{\circ}\)$


As shown in the figure to the right, in a triangle having angles $\(90,60 \& 30\)$, the corresponding opposite sides are $\(2 \mathrm{a}, \mathrm{a} \sqrt{3} \& \mathrm{a}\)$ respectively.



Using the $\(90,60,30\)$ triangle rule, in the triangle ACO , the side opposite to 60 degrees i.e. CO is given as 3 , so we get the side opposite to 30 degrees as $\(\frac{3}{\sqrt{3}}=\sqrt{3}=\mathrm{AO}=\)$ radius of the circle.

So, we get the area of the circle as $\(\pi r^2=\pi(\sqrt{3})^2=3 \pi\)$

Hence the answer is (C).