In $$\triangle ABC, \angle C=30^\circ$$. If the bisectors of the angle B and angle C meet at a point O in the interior of the triangle, then $$\angle BOC$$ is equal to:

Given ,

$$\angle\ c=30^{\circ\ }\ $$

In triangle$$\ \triangle\ $$ ABC, angle $$\ \angle\ c=30^{\circ\ }$$= 60 degrees. and let angle B = 2b and angle A= 2c.

Thus 30 + 2b + 2c = $$180^{\circ\ }$$, or

2b + 2c = 180 - 30 =120 , or

b + c = 60 degrees.

In triangle OBC, angle BOC +angle OBC + angle OCB = $$180^{\circ\ }$$.

But angle OCB +angle OBC = b + c = 30, so

angle BOC +angle OBC + angle OCB = $$180^{\circ\ }$$, or

angle BOC +30 -$$\angle\ c\ =\ \frac{30^{\circ\ \ }}{2}=15$$ = $$180^{\circ\ }$$ , therefore

angle BOC =$$105^{\circ\ }$$