Question

If $O$ is the centre of the circle and $A,B$ and $C$ are points on its circumference and $\angle AOC={130}^{\circ }$, find $\angle ABC$.

• A. ${105}^{\circ }$
• B. ${115}^{\circ }$
• C. ${110}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is B ${115}^{\circ }$

Take any point $P$ on the circumference of the circle as shown.

Join $AP$ and $CP$.

$\because ABC$ subtends $\angle AOC$ at centre $O$ and $\angle APC$ at any point $P$ on the circumference of the circle.
$\therefore \angle AOC=2\angle APC$ [$\because$ Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference]
$⟹\angle APC=\frac{1}{2}\angle AOC[\because AOC={130}^{\circ }]$
$⟹$ $\frac{1}{2}\times {130}^{\circ }={65}^{\circ }$.

$\because$ $ABCP$ is a cycle quadrilateral,
$⟹\angle APC+\angle ABC={180}^{\circ }$ ...[$\because$ sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$]
$⟹{65}^{\circ }+\angle ABC={180}^{\circ }$
$\therefore \angle ABC={180}^{\circ }-{65}^{\circ }={115}^{\circ }$.

Hence, option $B$ is correct.