Question

In the given figure, O is the centre of a circle and $\angle AOC={130}^{\circ }.Then\angle ABC=?$

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

(b) ${65}^{\circ }$

(c) ${115}^{\circ }$

(d) ${130}^{\circ }$

Answer 1

ANSWER:
( c ) 115°
Take a point D on the remaining part of the circumference.
Join AD and CD.
Then $\angle ADC$=1/2$\angle AOC$=1/2×130°=65°
In cyclic quadrilateral ABCD, we have:
$\angle ABC$ + $\angle ADC$ = 180° (Opposite angles of a cyclic quadrilateral)
⇒ $\angle ABC$ + 65° = 180°
⇒ $\angle ABC$ = (180° - 65°) = 115°