AB is diameter of a circle having centre at O. P is a point on the circumference of the circle. If $$\angle$$POA = 120$$^\circ$$, then the measure of $$\angle$$PBO is

Given,  $$\angle$$POA = 120$$^\circ$$

In $$\triangle\ $$OPB,

OP = OB

$$=$$>  $$\angle$$PBO = $$\angle$$BPO

Let $$\angle$$PBO = $$\angle$$BPO = x

From the figure,

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

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

$$=$$>   $$\angle$$POB = 60$$^\circ$$

In $$\triangle\ $$OPB,

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

$$=$$>  60$$^\circ$$ + x + x = 180$$^\circ$$

$$=$$>  2x = 120$$^\circ$$

$$=$$>   x =  60$$^\circ$$

$$=$$>  $$\angle$$PBO = x = 60$$^\circ$$

Hence, the correct answer is Option D