Question

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.

Answer 1

Since ABCD is a cyclic quadrilateral, therefore, ∠BCD + ∠BAD = 180°  ...(Since opposite angles of a cyclic quadrilateral are supplementary)

`=>` ∠BCD + 70° = 180°

`=>` ∠BCD = 180° − 70° = 110°

In ΔBCD, we have,

∠CBD + ∠BCD + ∠BDC = 180°

`=>` 30° + 110° + ∠BDC = 180°

`=>` ∠BDC = 180° − 140°

`=>` ∠BDC = 40°