In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?
Given: ∠AOC = 100°
Here,
(Exterior ∠AOC) = 360° – (interior ∠AOC)
(Exterior ∠AOC) = 360° – 100°
(Exterior ∠AOC) = 260°
We know that,
(Exterior ∠AOC) = 2 × ∠ADC
260° = 2 × ∠ABC
∠ABC = = 130°
∴ ∠ABC = 130°
Here,
∠ABD = ∠ABC + ∠CBD
180° = 130° + ∠CBD
∠CBD = 180° – 130° = 50°
∴ ∠CBD = 50°