In Fig. 16.193, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If APB=70°, find ACB.

O is the centre of the smaller circle.

APB = 70°


By degree measure theorem,


AOB = 2 APB


AOB = 2 × 70°


= 140°


Therefore,


AOBC is a cyclic quadrilateral


ACB + AOB = 180°


ACB + 140° = 180°


ACB = 40°


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