In $$\triangle$$ABC, $$\angle$$A = 52$$^\circ$$. Its sides AB and AC are produced to the points D and E respectively. If the bisectors of the $$\angle$$CBD and $$\angle$$BCE meet at point O, then $$\angle$$BOC is equal to:

Given,

In $$\triangle$$ABC,  $$\angle$$A = 52$$^\circ$$

OB is the angular bisector of $$\angle$$CBD

$$=$$>  $$\angle$$OBD = $$\angle$$OBC

Let $$\angle$$OBD = $$\angle$$OBC = $$x$$

OC is the angular bisector of $$\angle$$BCE

$$=$$>  $$\angle$$OCE = $$\angle$$OCB

Let $$\angle$$OCE = $$\angle$$OCB = $$y$$

From the figure,

$$\angle$$ABC + $$\angle$$CBD = 180$$^\circ$$

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

$$=$$> $$\angle$$ABC  = 180$$^\circ$$- 2$$x$$

$$\angle$$ACB + $$\angle$$BCE = 180$$^\circ$$

$$=$$> $$\angle$$ACB + $$y$$ + $$y$$ = 180$$^\circ$$

$$=$$> $$\angle$$ACB = 180$$^\circ$$- 2$$y$$

In $$\triangle$$ABC,

$$\angle$$ABC + $$\angle$$ACB + $$\angle$$BAC = 180$$^\circ$$

$$=$$>  180$$^\circ$$- 2$$x$$ + 180$$^\circ$$- 2$$y$$ + 52$$^\circ$$ = 180$$^\circ$$

$$=$$>  2$$x$$ + 2$$y$$ = 180$$^\circ$$ + 52$$^\circ$$

$$=$$>  2$$(x+y)$$ = 232$$^\circ$$

$$=$$>  $$x+y$$ = 116$$^\circ$$ ....................(1)

In $$\triangle$$OBC,

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

$$=$$>  $$x$$ + $$y$$ + $$\angle$$BOC = 180$$^\circ$$

$$=$$>  116$$^\circ$$ + $$\angle$$BOC = 180$$^\circ$$

$$=$$>  $$\angle$$BOC = 180$$^\circ$$- 116$$^\circ$$

$$=$$>  $$\angle$$BOC = 64$$^\circ$$

Hence, the correct answer is Option A