Question

In the given figure, if $\angle AOC={130}^{\circ },$ find the value of $\angle ABC$ . [3 Marks]

Answer 1

Given $\angle AOC={130}^{\circ }.$

Consider the arc ABC. Consider a point P not on this arc, but on the circumference of the circle. Join AP and CP.

Arc ABC subtends $\angle AOC$ at centre and $\angle APC$ on the remaining part of the circle (at the point P). [1 Mark]
Since the angle subtended by an arc at the centre is double the angle subtended by it on any remaining point of the circle, we have
$\angle APC=\frac{1}{2}\angle AOC.$ [1 Mark]
$⟹\angle APC=\frac{1}{2}\times {130}^{\circ }={65}^{\circ }$
Since APCB is a cylic quadrilateral, we have its opposite angles to be supplementary.
$⟹\angle ABC+\angle APC={180}^{\circ }$
$⟹\angle ABC+{65}^{\circ }={180}^{\circ }$
$⟹\angle ABC={180}^{\circ }-{65}^{\circ }={115}^{\circ }$ [1 Mark]