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