In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that \(\angle{BEC}\) = \(130^\circ\) and \(\angle{ECD}\) = \(20^\circ\). Find \(\angle{BAC}\).
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Answer :

In \(\triangle{ABC}\),

\(\angle{AED}\) = \(180^\circ\) -\(130^\circ\) = \(50^\circ\)
(linear pair of angles)

\(\therefore \) \(\angle{CED}\) = \(\angle{AED}\) = \(50^\circ\)
(Vertically opposite angles)

Also, \(\angle{ABD}\) = \(\angle{ACD}\)
(Since, the angles in the same segment are equal)
\(\Rightarrow \) \(\angle{ABE}\) = \(\angle{ECD}\)
\(\therefore \) \(\angle{ABE}\) = \(180^\circ\) ...(ii)

Now, in \(\triangle{CDE}\),
\(\angle{BAC}\) + \(20^\circ\) + \(50^\circ\) = \(180^\circ\)
(from eq.(i) and (ii))

\(\Rightarrow \) \(\angle{BAC}\) + \(70^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{BAC}\) = \(180^\circ\) - \(70^\circ\)
\(\therefore \) \(\angle{BAC}\) = \(110^\circ\)

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