AB is a chord of a circle with centre O. C is a point on the circle i…

OA = OB = radius of the circle

So, \(\angle\)ABO = \(\angle\)BAO = \({50}^\circ\)

So, \(\angle\)AOB = \({180}^\circ\) - (\({50}^\circ\) - \({50}^\circ\))

⇒ \(\angle\)AOB = \({180}^\circ\) - \({100}^\circ\)

⇒ \(\angle\)AOB = \({80}^\circ\)

Let A and B meet at point D in the major sector

Now,

\(\angle\)ADB = \({80}^\circ\)/2 = \({40}^\circ\)

ACBD is a cyclic quadrilateral

So, \(\angle\)ACB = \({180}^\circ\) - \({40}^\circ\)

⇒ \(\angle\)ACB = \({140}^\circ\)

Therefore, \(\angle\)ACB is \({140}^\circ\).