(i) From the figure we know that ∠ AOB and ∠ COD are vertically opposite angles

So we get

∠ AOB = ∠ COD = 105^o

Consider △ AOB

By sum property of a triangle

∠ OAB + ∠ AOB + ∠ ABO = 180^o

By substituting the values in above equation

35^o + 105^o + ∠ ABO = 180^o

On further calculation

∠ ABO = 180^o – 35^o – 105^o

By subtraction

∠ ABO = 180^o – 140^o

∠ ABO = 40^o

(ii) We know that AB || DC and BD is a transversal

From the figure we know that ∠ ABD and ∠ CDB are alternate angles

It can be written as

∠ CDO = ∠ CDB = ∠ ABD = ∠ ABO = 40^o

So we get

∠ ODC = 40^o

(iii) We know that AB || CD and AC is a transversal

From the figure we know that ∠ ACB and ∠ DAC are alternate opposite angles

So we get

∠ ACB = ∠ DAC = 40^o

(iv) We know that ∠ B can be written as

∠ B = ∠ CBD + ∠ ABO

So we get

∠ CBD = ∠ B – ∠ ABO

In a parallelogram we know that the sum of all the angles is 360^o

So we get

∠ A + ∠ B + ∠ C + ∠ D = 360^o

It can be written as

2 ∠ A + 2 ∠ B = 360^o

By substituting values in the above equation

2 (40^o + 35^o) + 2 ∠ B = 360^o

On further calculation

2 (75^o) + 2 ∠ B = 360^o

So we get

150^o + 2 ∠ B = 360^o

2 ∠ B = 360^o – 150^o

By subtraction

2 ∠ B = 210^o

By division

∠ B = 105^o

So we get

∠ CBD = ∠ B – ∠ ABO

By substituting values

∠ CBD = 105^o – 40^o

By subtraction

∠ CBD = 65^o