In the figure (ii) given below, O is the center of the circle. If ∠BAD = 75° and BC = CD, find:

(i) ∠BOD

(ii) ∠BCD

(iii) ∠BOC

(iv) ∠OBD

Join OC and BD as shown in the figure below:

(i) ∠BOD = 2 × ∠BAD (∵ angle subtended by an arc at center is double the angle subtended at any point on the remaining part of the circle.)

∠BOD = 2 × 75° = 150°.

Hence, the value of ∠BOD = 150°.

(ii) 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°.

(iii) Join OC.

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

∠BOC = $\dfrac{1}{2}$∠BOD = $\dfrac{1}{2} \times 150°$ = 75°.

Hence, the value of ∠BOC = 75°.

(iv) 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°.