Calculate the value of $$\angle OBC+\angle BAC$$, if O is the circum-centre of the triangle ABC inscribed in the circle.

O is the circumcentre of the triangle, => $$\angle$$ BOC = $$2\angle$$ BAC -------------(i)

OB = OC = radius of circle => $$\angle$$ OBC = $$\angle$$ OCB

In $$\triangle$$ OBC, => $$\angle$$ OBC + $$\angle$$ OCB + $$\angle$$ BOC = $$180^\circ$$

=> $$2\angle$$ OBC + $$2\angle$$ BAC = $$180^\circ$$     [Using equation (i)]

=> $$2$$($$\angle$$ OBC + $$\angle$$ BAC) = $$180^\circ$$

=> $$\angle$$ OBC + $$\angle$$ BAC = $$\frac{180}{2}=90^\circ$$

=> Ans - (B)