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

(i) We know that

∠BOD + ∠AOD = 180^o

By substituting the values

∠BOD + 140^o = 180^o

On further calculation

∠BOD = 180^o – 140^o

By subtraction

∠BOD = 40^o

We know that OB = OD

So we get ∠OBD = ∠ODB

Consider △ OBD

By using the angle sum property

∠BOD + ∠OBD + ∠ODB = 180^o

We know that ∠OBD = ∠ODB

So we get

40^o + 2 ∠OBD = 180^o

On further calculation

2 ∠OBD = 180^o – 40^o

By subtraction

2 ∠OBD = 140^o

By division

∠OBD = 70^o

We know that ABCD is a cyclic quadrilateral

∠CAB + ∠BDC = 180^o

∠CAB + ∠ODB + ∠ODC = 180^o

By substituting the values

50^o + 70^o + ∠ODC = 180^o

On further calculation

∠ODC = 180^o – 50^o – 70^o

By subtraction

∠ODC = 180^o – 120^o

So we get

∠ODC = 60^o

Using the angle sum property

∠EDB + ∠ODC + ∠ODB = 180^o

By substituting the values

∠EDB + 60^o + 70^o = 180^o

On further calculation

∠EDB = 180^o – 60^o – 70^o

By subtraction

∠EDB = 180^o – 130^o

So we get

∠EDB = 50^o

(ii) We know that

∠EDB + ∠OBD = 180^o

By substituting the values

∠EDB + 70^o = 180^o

On further calculation

∠EDB = 180^o – 70^o

By subtraction

∠EDB = 110^o