In $$\triangle$$ABC, $$\angle$$A = $$50^\circ$$. Its sides AB and AC are produced to the point D and E. If the bisectors of the $$\angle$$CBD and $$\angle$$BCE meet at the point O, then $$\angle$$BOC will be equal to:

As per the question,

$$\angle ACB=180-\angle ECO -------------(i)$$

$$\angle ABC=180-\angle BDC -------------(ii)$$

Now, in $$\triangle ABC$$,

$$\Rightarrow \angle BAC+\angle ABC+\angle BCA=180^\circ --------------(iii)$$

From the equation (i), (ii) and (iii)

$$\Rightarrow 50+180-\angle ECO+180-\angle BDC=180^\circ$$

$$\Rightarrow \angle ECO +\angle BDC=230^\circ --------(iv)$$

$$\Rightarrow \angle ECO=2\angle OBC$$

$$\Rightarrow \angle ECB=2\angle OCB$$

$$\Rightarrow 2\angle OCB +2\angle OBC=230^\circ$$

$$\Rightarrow \angle OCB +\angle OBC=115^\circ$$

Now, In $$\triangle OCB$$

$$\Rightarrow \angle OBC+\angle OCB+\angle BOC=180$$

$$\Rightarrow 115+\angle BOC=180^\circ$$

$$\Rightarrow \angle BOC=180^\circ-115^\circ=65^\circ$$