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Answer :
We have, AB || DE,
Therefore, by alternate angles theorem, we get,
\(\angle{AED}\) = \(\angle{BAE}\)
Also, given that, \(\angle{BAC}\) = \(35^\circ\) and \(\angle{BAC}\) = \(\angle{BAE}\)
\(\therefore \) , \(\angle{AED}\) = \(35^\circ\)
Now, in \(\triangle{DCE}\),
\(\angle{DCE}\) + \(\angle{CED}\) + \(\angle{CDE}\) = \(180^\circ\)
because, Sum of all angles of triangle is equal to 180.
Thus, by putting given values, we get,
\(\Rightarrow \) \(\angle{DCE}\) + \(35^\circ\) + \(53^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{DCE}\) = \(180^\circ\) - \(88^\circ\)
\(\Rightarrow \) \(\angle{DCE}\) = \(92^\circ\)