Question

In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find:

1. ∠DAB
2. ∠DBA

Answer 1

i. ∠DAB + ∠BCD = 180°   ...(Opposite angle of a cyclic quadrilateral)

∠DAB + 130° = 180°

∠DAB = 180° – 130°

∠DAB = 50°

ii. ∠ADB = 90°   ...(Angle in semicircle)

In ΔADB,

∠DAB + ∠ADB + ∠DBA = 180°   ...(Angle sum property)

50° + 90° + ∠DBA = 180°

∠DBA = 180° – 140°

∠DBA = 40°