In the given figure, is the center of a circle. If and calculate (i) (ii)

Question

Philoid · Accepted 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)