In the figure (i) given below, if ∠DBC = 58° and BD is a diameter of the circle, calculate

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

(i) Given,

∠DBC = 58°

From figure,

∠BCD = 90° (∵ angle in semicircle is equal to 90°.)

Since sum of angles in triangle is 180°.

∴ In △BCD

⇒ ∠DBC + ∠BCD + ∠BDC = 180°
⇒ 58° + 90° + ∠BDC = 180°
⇒ 148° + ∠BDC = 180°
⇒ ∠BDC = 180° - 148°
⇒ ∠BDC = 32°.

Hence, the value of ∠BDC = 32°.

(ii) Considering quadrilateral BDCE.

From figure,

BDCE is a cyclic quadrilateral.

Since opposite angles sum is 180° in cyclic quadrilateral

⇒ ∠BDC + ∠BEC = 180°
⇒ 32° + ∠BEC = 180°
⇒ ∠BEC = 180° - 32°
⇒ ∠BEC = 148°.

Hence, the value of ∠BEC = 148°.

(iii) From figure,

∠BAC = ∠BDC (∵ angles in same segment are equal.)

∴ ∠BAC = 32°.

Hence, the value of ∠BAC = 32°.