Question

In the given figure, O is the centre of a circle. If $\angle AOD={140}^{\circ }and\angle CAB={50}^{\circ }$ , calculate

(i) $\angle$ EDB, (ii) $\angle$EBD.

Answer 1

ANSWER:
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 Δ OBD, 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 ° - ∠O BD
= 180 ° - 70°
= 110°