Question

In the given figure,O is the centre of a circle in which angle AOC=100°.Side AB of quadrilateral OABC has been produced to D.Then,angle CBD is?

Answer 1

Answer:

option a) 50° is the correct answer

Step-by-step explanation:

It is given that AOC=100° and Side AB of quadrilateral OABC has been produced to D.

To find <ABC

<ABC = 180 - (<AOC/2)

<ABC = 180 - 100/2 = 180 - 50 = 130°

To find <CBD

<CBD and <ABC are linear pairs.

Therefore <CBD + <ABC = 180

<CBD= 180 - <ABC = 180 - 130 = 50°

Therefore option a) 50° is the correct answer.

Answer 2

ANSWER :

∠AOC=100°

ABCE is a cyclic quadrilateral.

∠AEC=  50 D egree

∠AOC=50°      [ Since as angle made by the chord at the surface of the circle is half of the angle made by the same chord at the center]

In quadrilateral ABCE;

∠AEC+∠ABC=180°⟹∠ABC=130°     [ Since in a cyclic quadrilateral sum of diagonal angle is 180°]

∠ABC+∠CBD=180°

⟹∠CBD=50°

Hence;

∠CBD=50° and

∠ABC=130°