Question

In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :
(i) ∠BDC (ii) ∠BEC (iii) ∠BAC​

Answer 1

Step-by-step explanation:

In ∆ BCD; ∠DBC = 58°

(i) ∠BCD = 90° (Angle in the semicircle as BD is diameter)

∴ ∠DBC + ∠BCD + ∠BDC = 180°

⇒ 58°+ 90° +∠BDC = 180°

⇒ ∠BDC = 180° – (90° + 58°)

= 180° - 148°

= 32° Ans.

(ii) ∠BEC + ∠BDC = 180° ( ∵BECD is a cyclic quadrilateral)

∠BEC = 180°- ∠BDC = 180°-32°

∠BEC = 148°

(iii) ∠BAC = ∠BDC (Angle of same segment are equal)

∠BAC = 32°

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