In the given figure, O is the centre of the circle. Then, find ∠CBD.

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Then, ∠AOC=2∠APC

=>∠APC = 1/2 ∠AOC = 1/2 x 100° = 50°

Now, ∠APC + ∠ABC = 180°

[∵ sum of opposite angles of a cyclic quadrilateral is 180°]

=> 50° +∠ABC = 180°

=∠ABC = 180°- 50° = 130°

Also,∠ABC + ∠CBD = 180° [linear pair axiom]

=> 130° +∠CBD =180°

=>∠CBD = 180° - 130° = 50°