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Question

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

(i) BDC

(ii) BEC

(iii) BAC

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Solution

Given BD is diameter

\angle DBC = 58^o

(i) \angle DCB = 90^o (angle in semi circle)

In \triangle BDC

58+\angle BDC+90=180^o

\angle BDC = 32^o

(ii) ABEC is a cyclic quadrilateral

\angle BDC = \angle BAC = 32^o (angle in the same segment)

Therefore \angle BEC + 32 = 180^o

\angle BEC = 148^o

(iii) \angle BAC = 32^o (angle in the same segment)


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