Solution:

(i) According to the given question, We know that

\[\angle DCE\text{ }=\text{ }{{90}^{o}}~\angle CDE\]

\[=\text{ }{{90}^{o}}-\text{ }{{40}^{o}}~=\text{ }{{50}^{o}}\]

Hence,

\[\angle DEC\text{ }=\angle OCB\text{ }=\text{ }{{50}^{o}}\]

(ii) In \[\vartriangle BOC\], we have

\[\angle AOC\text{ }=\angle OCB\text{ }+\angle OBC\] [Exterior angle property of a triangle]

\[\angle OBC\text{ }=\text{ }{{80}^{o}}-\text{ }{{50}^{o}}~=\text{ }{{30}^{o}}\]  [Given ∠AOC = 80^o]

Hence, \[\angle ABC\text{ }=\text{ }{{30}^{o}}\]