In $$\triangle$$ABC, $$\angle$$ A= 72$$^\circ$$. 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 = 72$$^\circ$$

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$$y$$ + 72$$^\circ$$ = 180$$^\circ$$

$$=$$> 2$$x$$ + 2$$y$$ = 180$$^\circ$$ + 72$$^\circ$$

$$=$$> 2$$(x+y)$$ = 252$$^\circ$$

$$=$$> $$x+y$$ = 126$$^\circ$$ ....................(1)

In $$\triangle$$OBC,

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

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

$$=$$> 126$$^\circ$$ + $$\angle$$BOC = 180$$^\circ$$

$$=$$> $$\angle$$BOC = 180$$^\circ$$- 126$$^\circ$$

$$=$$> $$\angle$$BOC = 54$$^\circ$$

Hence, the correct answer is Option B