In the given figure, C and D are points on the semi-circle described on AB as diameter.

Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.

As ABCD is a cyclic quadrilateral, we have

⇒ ∠BCD + ∠BAD = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

⇒ ∠BCD + 70° = 180°

⇒ ∠BCD = 180° - 70° = 110°

In ∆BCD,

⇒ ∠CBD + ∠BCD + ∠BDC = 180° [By angle sum property of triangle]

⇒ 30° + 110° + ∠BDC = 180°

⇒ ∠BDC = 180° - 140° = 40°.

Hence, ∠BDC = 40°.