A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130°. Find ∠BAC.

Given: AB is the diameter.

ABCD is the cyclic quadrilateral.

∠ADC = 130°

Since ABCD is the cyclic quadrilateral, therefore sum of opposite angles is 180°.

⇒ ∠ADC + ∠ABC = 180°

⇒ 130° + ∠ABC = 180°

⇒ ∠ABC = 180° - 130°

⇒ ∠ABC = 50°

Now, AB is the diameter and the angle subtended to the circle by the diameter is a right angle.

∴ ∠ACB = 90°

In triangle ACB, sum of all angles of triangle is 180°.

∴ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 90° + 50° + ∠BAC = 180°

⇒ ∠BAC = 180° - 90° - 50°

⇒ ∠BAC = 40°