In the figure, ∠DBC=58∘. BD is a diameter of the circle. Calculate :
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Given is diameter
(i) (angle in semi circle)
In
(ii) is a cyclic quadrilateral
(angle in the same segment)
Therefore
(iii) (angle in the same segment)