In the given figure, O is the centre of the circle, ∠BAD = 75° and chord BC = chord CD. Find : (i) ∠BOC (ii) ∠OBD (iii) ∠BCD

(i) As equal chords of a circle subtend equal angles at the center and chord BC = chord CD, so ∠BOC = ∠COD.

∠BOD = 2 × ∠BAD

∠BOD = 2 × 75°

∠BOD = 150°

∠BOC = $\dfrac{1}{2}$ ∠BOD

= $\dfrac{1}{2}$ × 150° = 75°.

Hence, the value of ∠BOC = 75°.

(ii) Join BD.

Since, OB = OD

∴ ∠OBD = ∠ODB = x

Since sum of angles of triangle = 180°

In △OBD

⇒ ∠BOD + ∠OBD + ∠ODB = 180°

⇒ 150° + x + x = 180°

⇒ 150° + 2x = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = 15°.

Hence, the value of ∠OBD = 15°.

(iii) ABCD is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 75° = 180°

⇒ ∠BCD = 180° - 75°

⇒ ∠BCD = 105°.

Hence, the value of ∠BCD = 105°.