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°