Solution:

According to the given question,

Given, \[\angle AOB\text{ }=\text{ }{{140}^{o}}~\]and \[\angle OAC\text{ }=\text{ }{{50}^{o}}\]

(i) So,

\[\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }Reflex\text{ }(\angle AOB)\]

\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }({{360}^{o}}-\text{ }{{140}^{o}})\text{ }=\text{ }{{110}^{o}}\]

[Angle at the center is double the angle at the circumference subtend by the same chord]

(ii) In quadrilateral \[OBCA,\]

\[\angle OBC\text{ }+\angle ACB\text{ }+\angle OCA\text{ }+\angle AOB\text{ }=\text{ }{{360}^{o}}~\] [Angle sum property of a quadrilateral]

\[\angle OBC\text{ }+\text{ }{{110}^{o}}~+\text{ }{{50}^{o}}~+\text{ }{{140}^{o}}~=\text{ }{{360}^{o}}\]

Hence, \[\angle OBC\text{ }=\text{ }{{360}^{o}}-\text{ }{{300}^{o}}~=\text{ }{{60}^{o}}\]