Question

In the adjoining figure, AB is the diameter of the circle with centre O. If ∠BCD = 120°, calculate:
(i) ∠BAD (ii) ∠DBA

Answer 1

(i) Since AOB is a diameter
∴ ∠ADB = 90°        ...( C is a semi-circle)
Also, ABCD is a cyclic quadrilateral.
∴ ∠BCD + ∠BAD = 180°
∠BAD = 180° - 120°
⇒ ∠BAD = 60°

(ii) Now, In Δ BAD,
∠BAD + ∠BDA + ∠DBA = 180°
60° + 90° + ∠DBA = 180°
∠DBA = 180° - 150°
∠DBA = 30°