A quadrilateral ABCD is inscribed in a circle with centre O. If  $$\angle$$BOC=92$$^\circ$$ and $$\angle$$ADC=112$$^\circ$$, then $$\angle$$ABO  is equal to:

In $$\triangle\ $$OBC,

OC = OB

$$=$$>  $$\angle$$OBC = $$\angle$$OCB

Let $$\angle$$OBC = $$\angle$$OCB = x

$$\angle$$OBC + $$\angle$$OCB + $$\angle$$BOC = 180$$^\circ$$

$$=$$>  x + x + 92$$^\circ$$ = 180$$^\circ$$

$$=$$>  2x = 88$$^\circ$$

$$=$$>   x = 44$$^\circ$$

$$=$$>  $$\angle$$OBC = $$\angle$$OCB = 44$$^\circ$$

In cyclic quadrilateral ABCD, sum of opposite angles = 180$$^\circ$$

$$=$$>  $$\angle$$ABC + $$\angle$$ADC = 180$$^\circ$$

$$=$$>  $$\angle$$ABO + $$\angle$$OBC + $$\angle$$ADC = 180$$^\circ$$

$$=$$>  $$\angle$$ABO + 44$$^\circ$$ + 112$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$ABO + 156$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$ABO = 24$$^\circ$$

Hence, the correct answer is Option A