Correct option is (D) 35°
Angle subtended by an arc at centre of the circle is double than the angle made by that arc at any point on the circumference of the circle.
\(\therefore\) \(\angle AOB=2\angle ACB\)
\(\Rightarrow\) \(\angle ACB=\frac{\angle AOB}2\)
\(=\frac{140^\circ}2=70^\circ\) ___________(1)
Now, in \(\triangle ACO\;\&\;\triangle BCO,\)
AC = BC (Given)
AO = BO (radii)
OC = OC (Common)
\(\therefore\) \(\triangle ACO\cong\triangle BCO\) (By SSS congruence criteria)
\(\therefore\) \(\angle ACO=\angle BCO\) (By CPCT)
\(\Rightarrow\) \(\angle ACO+\angle BCO\) \(=\angle ACB=70^\circ\) (From (1))
\(\Rightarrow\) \(2\angle ACO=70^\circ\)
\(\Rightarrow\) \(\angle ACO=\frac{70^\circ}2=35^\circ\)
\(\Rightarrow\) \(x=35^\circ\) \((\because\) \(\angle ACO=x)\)
