What is the area of an isosceles triangle with one angle 120 and inscr

Isosceles triangle with one angle 120 => That means other two angles are 30 and 30.
If the triangle is inscribed in a circle, then the center O will be equidistance from all the vertices=> distance will be equal to the radius.
Let triangle is ABC and AB= AC = 2x
Drop a perpendicular from point A to side BC, It will intersect at the midpoint. Let the midpoint is D.
In triangle ADB, we can apply the 30-60-90 rule:
AB = 2x, AD = x, BD = \(\sqrt{3}\)x
In triangle ODB=> radius=OB = 2, then OD = 2-x, BD = 3x and OB = 2
Using pythagorus theorum:
\((2-x)^2\) + \((\sqrt{3}x)^2\) = \(2^2\)
upon solving this:
4x(x-1) = 0
=> x= 1 => means the triangle is above the diameter.

Then the area = \(\frac{1}{2}\) * BC * AD
=> \(\frac{1}{2}\) * 2\(\sqrt{3}\) * 1
=> \(\sqrt{3}\)\(\)

IMO the answer is A.

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