In figure A, B and C are three points on a circle with centre O such that $$\angle{BOC}$$ = $$30^\circ$$ and $$\angle{AOB}$$ = $$60^\circ$$. If D is a point on the circle other than the arc ABC, find $$\angle{ADC}$$.Let PQ = QR = PR = X

Here, $$\angle{AOC}$$ = $$\angle{AOB}$$ + $$\angle{BOC}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$

$$\therefore$$ Arc ABC makes $$90^\circ$$ at the centre of the circle.

Thus, $$\angle{ADC}$$ = ($$\frac{1}{2}$$) $$\angle{AOC}$$
(Since, The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle)
$$\Rightarrow$$ = ($$\frac{1}{2}$$ ) $$90^\circ$$ = $$45^\circ$$

Hence, $$\angle{ADC}$$ = $$45^\circ$$.