Question

In the given figure, O is the centre of a circle. If ∠AOD = 140° and ∠CAB = 50°, calculate
(i) ∠EDB,
(ii) ∠EBD.

Answer 1

O is the centre of the circle where ∠AOD = 140° and ∠CAB = 50°.
(i) ∠BOD = 180° – ∠AOD
= (180° – 140°) = 40°
We have the following:
OB = OD (Radii of a circle)
∠OBD = ∠ODB

In ΔOBD, we have:
∠BOD + ∠OBD + ∠ODB = 180°
⇒ ∠BOD + ∠OBD + ∠OBD = 180° [∵ ∠OBD = ∠ODB]
⇒ 40° +2∠OBD = 180°
⇒ 2∠OBD = (180° – 40°) = 140°
⇒ ∠OBD = 70°
Since ABCD is a cyclic quadrilateral, we have:
∠CAB + ∠BDC = 180°
⇒ ∠CAB + ∠ODB + ∠ODC = 180°
⇒ 50° + 70° + ∠ODC = 180°
⇒ ∠ODC = (180° – 120°) = 60°
∴ ∠ODC = 60°
∠EDB = (180° – (∠ODC + ∠ODB)
= 180° – (60° + 70°)
= 180° – 130° = 50°
∴ ∠EDB = 50°

(ii) ∠EBD = 180° - ∠OBD
= 180° - 70°
= 110°