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).