In Fig. 16.181, O is the centre of the circle. Find ∠CBD.
Given that,
∠BOC = 100o
By degree measure theorem,
∠AOC = 2 ∠APC
100o = 2 ∠APC
∠APC = 50o
Therefore,
∠APC + ∠ABC = 180o (Opposite angles of a cyclic quadrilateral)
50o + ∠ABC = 180o
∠ABC = 130o
Therefore,
∠ABC + ∠CBD = 180o (Linear pair)
130o + ∠CBD = 180o
∠CBD = 50o