Question 136

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

Solution

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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