In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :
(i) ∠BDC (ii) ∠BEC (iii) ∠BAC
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Step-by-step explanation:
Question :
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :
(i) ∠BDC (ii) ∠BEC (iii) ∠BAC
Solution :
In ∆ BCD; ∠DBC = 58°
(i) ∠BCD = 90° (Angle in the semicircle as BD is diameter)
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
(ii) ∠BEC + ∠BDC = 180° ( ∵ BECD is a cyclic quadrilateral)
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
(iii) ∠BAC = ∠BDC (Angle of same segment are equal)
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