# <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°

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°

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Arc of Circle And Related IMP Qs40 mins  IMP Qs Related with Circles For Boards44 mins  Chords Of Circles42 mins  IMP Theorems And Their Application43 mins  Circles- All kinds of Questions36 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

In Fig. 16.195, <RD Sharma - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

PQRS is a RD Sharma - Mathematics

ABCD is a RD Sharma - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

ABCD is a RD Sharma - Mathematics