A, B and C are three points on a circle such that the angles subtended by the chord AB and AC at the centre O are $$110^\circ$$ and $$130^\circ$$, respectively. Then the value of $$\angle BAC$$ is:

In $$\triangle$$ OBA,

$$\angle O = 110 \degree$$

$$\angle OBA = \angle OAB$$ (OB = OA = radius)

$$\angle O + \angle OBA + \angle OAB = 180\degree$$

$$110 + \angle OAB + \angle OAB = 180\degree$$

$$2 \angle OAB = 70\degree$$

$$\angle OAB = 35\degree$$

In $$\triangle$$ OAC,

$$\angle O = 130 \degree$$

$$\angle OAC = \angle OCA$$ (OA = OC = radius)

$$\angle O + \angle OAC + \angle OCA = 180\degree$$

$$130 + \angle OAC + \angle OAC = 180\degree$$

$$2 \angle OAC = 50\degree$$

$$ \angle OAC = 25\degree$$

Now,

$$\angle BAC = \angle OAB + \angle OAC = 35 \degree + 25\degree = 60\degree$$