Question

In the given figure, O is centre of a circle in which $\angle AOC={100}^{\circ }$. Side AB of quad. OABC has been produced to D. Then, $\angle$CBD = ?

(a) ${50}^{\circ }$

(b) ${40}^{\circ }$

(c) ${25}^{\circ }$

(d) ${80}^{\circ }$

Answer 1

ANSWER:
(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE.
Then $\angle$AEC=1/2∠AOC=1/2×100°=50°
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior $\angle$CBD = $\angle$AEC = 50°