1. A, B and C are three points on a circle such that the angles subtended by the chords AB and AC at the centre 0 are 90° and 110° respectively. Further suppose that the centre ‘0’ lies in the interior L BAC. The L BAC is
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By: anil on 05 May 2019 01.53 am
As per the given condition,
$$angle$$AOB=90, $$angle$$AOC= 110
Thus $$angle$$BOC= 360-90-110= 160
$$angle$$BAC= $$frac{ angle BOC}{2}$$= 80 (B)
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$$angle$$AOB=90, $$angle$$AOC= 110
Thus $$angle$$BOC= 360-90-110= 160
$$angle$$BAC= $$frac{ angle BOC}{2}$$= 80 (B)