In the given figure, is a diameter of a circle with center If and are straight lines, meeting at such that and find

(i) (ii) (iii)


(i)


BDA = 90°= EDB [Semi circle angle]


In triangle EBD,


DBE + EDB + BED = 180°


DBE + 90° + 25° = 180°


DBE + 115° = 180°


DBE = 65°


Now,


DBC + DBE = 180°[CBE is a straight line]


DBC + 65° = 180°


DBC = 115°


(ii)


DCB = BAD [Angle in the same segment]


DCB = 35°


(iii) BDC = 30°


In triangle BCD,


BDC + DCB + DBC = 180°


BDC + 35° + 115° = 180°


BDC + 150° = 180°


BDC = 30°


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