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