Q. 153.5( 6 Votes )

# <span lang="EN-US

(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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