Inside a square ABCD, $$\ \triangle BEC\ $$is an equilateral triangle. If CE and BD interesect at O, then $$\ \angle BOC\ $$ is equal to

In square ABCD, $$\triangle$$ BEC is an equilateral triangle

Each angle of an equilateral triangle is 60$$^\circ$$

$$\Rightarrow$$ $$\angle$$ OCB $$= 60^\circ$$

$$\angle$$ DBC $$= \frac{90^\circ}{2} = 45^\circ$$ ($$\because$$ BD is diagonal of ABCD)

In $$\triangle$$ OBC,

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

$$60^\circ+45^\circ+\angle BOC = 180^\circ$$

$$\therefore \angle BOC = 75^\circ$$