Question

o is the center of the circle such that angle AOC = 130


In figure O is the centre of the circle such that ∠AOC = 130° then ∠ABC =



(A) 130° (B) 115° (C) 65° (D) 165°

Answer 1

my ans Ans. =65° abc

Answer 2

Join A and C to any point P in major arc AC.

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

$\therefore \angle APC = \frac{1}{2} \angle AOC \\ = > \frac{1}{2} (130 \degree) \\ = > 65 \degree$

Now, ABCP is a cyclic quadrilateral and sum of opposite angles in a cyclic quadrilateral is 180°

$\therefore\angle APC + \angle ABC = 180 \degree$

$= > \angle ABC = 180 \degree - \angle APC$

$= > 180 \degree - 65 \degree$

$= > 115 \degree$

Therefore, $\boxed{\angle ABC = 115\degree}$