In the given figure

Solution:

(i)
It is given that DB is a straight line.
Therefore,

$\angle D O C+\angle C O B=180^{\circ}$

 

$\angle D O C=180^{\circ}-115^{\circ}=65^{\circ}$

(ii)

In $\triangle D O C$, we have:

$\angle O D C+\angle D C O+\angle D O C=180^{\circ}$

Therefore,

$70^{\circ}+\angle D C O+65^{\circ}=180^{\circ}$

$\Rightarrow \angle D C O=180-70-65=45^{\circ}$

(iii)

It is given that $\triangle O D C \sim \triangle O B A$

Therefore,

$\angle O A B=\angle O C D=45^{\circ}$

(iv)

Again, $\triangle O D C \sim \triangle O B A$

Therefore,

$\angle O B A=\angle O D C=70^{\circ}$