In the given figure, O is the centre of the circle. If ∠ADC = 140°, find ∠BAC.

From figure,

∠ACB = 90° [Angle in a semicircle is a right angle]

We know that,

In a cyclic quadrilateral, the sum of opposite angles is 180°.

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° - 140°

⇒ ∠ABC = 40°.

By angle sum property of a triangle we get,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ ∠BAC + 40° + 90° = 180°

⇒ ∠BAC + 130° = 180°

⇒ ∠BAC = 180° - 130°

⇒ ∠BAC = 50°.

Hence, ∠BAC = 50°.