We are told that triangle \(ABC\) is contained within a circle with center \(C\) and points \(A\) and \(B\) lie on the circle.
We are also given that area of the circle is \(25\pi\).
We know that area of triangle \(= \pi \times radius^2\). Thus we can say: \(\pi \times radius^2 = 25 \times \pi\)
\(⇒ \pi \times radius^2 = 25 \times \pi\)
\(⇒ radius^2=25\)
\(⇒ radius=\sqrt{25}\)
\(⇒ radius=5\)
We are given that \(\angle ACB = 60^o\)
After all this information, we can draw:
Since \(AC=CB\), the angles opposite to them i.e., \(\angle A\) and \(\angle B\) will also be same
Thus we can say: \(\angle A+\angle B +\angle C = 180^o\)
\(⇒ \angle A + \angle A + 60 = 180\)
\(⇒ 2\times \angle A = 120\)
\(⇒ \angle A = 60^o\)
Thus \(\angle A = \angle B = \angle C = 60^o\)
So, triangle ABC is an equilateral triangle. So the possible value of \(AB = 5\).
Hence the right answer is
Option C.