14. In the given figure, O is the centre
of a circle. If AOD = 140° and
ZCAB = 50°, calculate
(i) ZEDB, (ii) ZEBD.
Answers
Answer:
(i) ∠EBD = 50°
∠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)
∠EBD = 110°
∠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°
Answer:
∠EDB = 50°
Step-by-step explanation:
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°