Question

In fig 10.36, A,B and C are three points on a circle with centre O such that $<BOC={30}^0$ and $<AOB={60}^0$. If D is a point on the circle other that the are ABC, find <ADC.

Answer 1

$\angle AOB+\angle BOC={60}^o+{30}^o={90}^o$
$\therefore \angle AOC={90}^o$ (angle subtended at the centre)
$\angle ADC=?$
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining of the circle
$\therefore$ Angle subtended at the centre
$\angle AOC=2\times \angle ADC$
${90}^o=2\times \angle ADC=90$
$\therefore \angle ADC=\frac{90}{2}={45}^o$
$\therefore \angle ADC={45}^o$