AB is the diameter of a circle with centre O. P be a point on it. If $$\angle POA = 120^\circ$$. Then, $$\angle PBO = ?$$

Given,  $$\angle POA = 120^\circ$$

From the figure,

$$\angle POA$$ + $$\angle POB$$ = $$180^\circ$$

$$=$$>  $$120^\circ$$ + $$\angle POB$$ = $$180^\circ$$

$$=$$>  $$\angle POB$$ = $$60^\circ$$

OP and OB are radius of the circle

$$=$$>  OP = OB

In $$\triangle\ $$OPB, angles opposite to OP and OB are equal

$$\angle PBO$$ = $$\angle BPO$$

$$\angle POB$$ + $$\angle PBO$$ + $$\angle BPO$$ = $$180^\circ$$

$$=$$>  $$60^\circ$$ + $$\angle PBO$$ + $$\angle PBO$$ = $$180^\circ$$

$$=$$>  2$$\angle PBO$$ = $$120^\circ$$

$$=$$>  $$\angle PBO$$ = $$60^\circ$$

Hence, the correct answer is Option D