In $$\triangle$$ABC, $$\angle$$A = 50$$^\circ$$. If the bisectors of the angle B and angle C, meet at a point O, then $$\angle$$BOC is equal to:

BO is the bisector of angle B.

Let $$\angle$$OBC = $$\angle$$OBA = y

CO is the bisector of angle C.

Let $$\angle$$OCA = $$\angle$$OCB = x

From $$\triangle$$ABC,

$$\angle$$A + $$\angle$$B + $$\angle$$C = 180$$^\circ$$

50$$^\circ$$ + 2y + 2x = 180$$^\circ$$

2(x+y) = 130$$^\circ$$

x + y = 65$$^\circ$$......(1)

From $$\triangle$$OBC,

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

y + x + $$\angle$$BOC = 180$$^\circ$$

65$$^\circ$$ + $$\angle$$BOC = 180$$^\circ$$ [From (1)]

$$\angle$$BOC  = 115$$^\circ$$

Hence, the correct answer is Option D