Question

In given figure o is centre of the circle if angle Aoc = 130 °then angle Abc

Answer 1

Answer:

answer is 65 degree.

Step-by-step explanation:

i dont have the figure in question but if i am correct in what i assumed then :-

there is a triangle fitted inside the circle touching every vertex to its circumference and o is center

angle aoc = 130 degree

hence angle abc is half of angle aoc

i.e. 65 degree.

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