Answer :

(i)


BOD + AOD = 180°[AOB is a straight line]


BOD + 140° = 180°


BOD = 40°


OB = OD


OBD = ODB


In triangle AOD,


BOD + OBD + ODB = 180°[Sum of angles of triangle]


40°+ 2 OBD = 180°


2 OBD = 140°


OBD = 70°


OBD = ODB = 70°


ABDC is a cyclic quadrilateral.


CAB + BDC = 180°


CAB + ODB + ODC = 180°


50°+ 70°+ ODC = 180°


ODC = 60°


Now,


EDB = 180° – BDC [Because CDE is a straight line]


EDB = 180° – (ODB + ODC)


EDB = 180° – (70°+ 60°)


EDB = 180° – 130°


EDB = 50°


(ii)


BOD + AOD = 180°[AOB is a straight line]


BOD + 140° = 180°


BOD = 40°


OB = OD


OBD = ODB


In triangle AOD,


BOD + OBD + ODB = 180°[Sum of angles of triangle]


40°+ 2 OBD = 180°


2 OBD = 140°


OBD = 70°


OBD = ODB = 70°


Now,


EBD + OBD = 180°[Because OBE is a straight line]


EBD + 70° = 180°


EBD = 110°


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