Question

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

Answer 1

Take any point P on the circumference of the circle as shown.
Join AP and CP.
∵ABC subtends ∠AOC at centre O and ∠APC at any point P on the circumference of the circle.
∴∠AOC=2∠APC [∵ Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference]
⟹∠APC= $\frac{1}{2}$∠AOC [∵AOC $={130}^{\circ }$ ]
⟹ $\frac{1}{2}\times {130}^{\circ }={65}^{\circ }$ .
∵ ABCP is a cycle quadrilateral,
⟹∠APC+∠ABC$={180}^{\circ }$ ...[∵ sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$]
⟹${65}^{\circ }$ +∠ABC=${180}^{\circ }$
∴∠ABC$={180}^{\circ }-{65}^{\circ }$
$={115}^{\circ }$ .