In the given figure
Question:

In the given figure $\triangle O D C \sim \triangle O B A, \angle B O C=115^{\circ}$ and $\angle C D O=70^{\circ}$. Find

(i) ∠DOC
(ii) ∠DCO
(iii) ∠OAB
(iv) ∠OBA.

 

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,

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

 

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