In figure A, B and C are three points on a circle with centre O such that \(\angle{BOC}\) = \(30^\circ\) and \(\angle{AOB}\) = \(60^\circ\). If D is a point on the circle other than the arc ABC, find \(\angle{ADC}\).
Let PQ = QR = PR = X
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Answer :

Here, \(\angle{AOC}\) = \(\angle{AOB}\) + \(\angle{BOC}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\)

\(\therefore \) Arc ABC makes \(90^\circ\) at the centre of the circle.

Thus, \(\angle{ADC}\) = (\(\frac{1}{2} \)) \(\angle{AOC}\)
(Since, The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle)
\(\Rightarrow\) = (\(\frac{1}{2} \) ) \(90^\circ\) = \(45^\circ\)

Hence, \(\angle{ADC}\) = \(45^\circ\).

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