Question

Find $\angle ABC$ = ?
$\left[3Marks\right]$

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.
$\left[1Marks\right]$

Arc ABC subtends $\angle AOC$ at centre and $\angle APC$ on the circumference of the circle (at the point P).
Since the angle subtended by an arc at the centre is double the angle subtended by it on the circumference of the circle, we have
$\angle APC=\frac{1}{2}\angle AOC.$
$⟹\angle APC=\frac{1}{2}\times {130}^{\circ }={65}^{\circ }$
$\left[1Marks\right]$

Since APCB is a cylic quadrilateral, we have its opposite angles to be supplementary.
$⟹\angle ABC+\angle APC={180}^{\circ }$
$⟹\angle ABC+{65}^{\circ }={180}^{\circ }$
$⟹\angle ABC={180}^{\circ }-{65}^{\circ }={115}^{\circ }$
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