In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

(i) Given that BD is a diameter of the circle.

We know that,

Angle in a semicircle is a right angle.

So, ∠BCD = 90°

Also given that,

∠DBC = 58°

In ∆BDC,

⇒ ∠DBC + ∠BCD + ∠BDC = 180° [Angle sum property of triangle]

⇒ 58° + 90° + ∠BDC = 180°

⇒ 148° + ∠BDC = 180°

⇒ ∠BDC = 180° - 148° = 32°.

Hence, ∠BDC = 32°.

(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.

So, in cyclic quadrilateral BECD

⇒ ∠BEC + ∠BDC = 180°

⇒ ∠BEC + 32° = 180°

⇒ ∠BEC = 180° - 32° = 148°

Hence, ∠BEC = 148°.

(iii) In cyclic quadrilateral ABEC,

⇒ ∠BAC + ∠BEC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ ∠BAC + 148° = 180°

⇒ ∠BAC = 180° - 148°= 32°.

Hence, ∠BAC = 32°.