# In figure, $\Delta \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$

Question.

In figure, $\triangle \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$ and $\angle \mathrm{CDO}=70^{\circ}$. Find $\angle \mathrm{DOC}, \angle \mathrm{DCO}$ and $\angle \mathrm{OAB}$.

Solution:

From figure,

$\angle D O C+125^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{DOC}=180^{\circ}-125^{\circ}=55^{\circ}$

$\angle \mathrm{DCO}+\angle \mathrm{CDO}+\angle \mathrm{DOC}=180^{\circ}$

(Sum of three angles of $\Delta \mathrm{ODC}$ )

$\Rightarrow \angle \mathrm{DCO}+70^{\circ}+55^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{DCO}+125^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{DCO}=180^{\circ}-125^{\circ}=55^{\circ}$

Now, we are given that $\Delta \mathrm{ODC} \sim \Delta \mathrm{OBA}$

$\Rightarrow \angle \mathrm{OCD}=\angle \mathrm{OAB}$

$\Rightarrow \angle \mathrm{OAB}=\angle \mathrm{OCD}=\angle \mathrm{DCO}=55^{\circ}$

i.e., $\angle \mathrm{OAB}=55^{\circ}$

Hence, we have

$\angle \mathrm{DOC}=55^{\circ}, \angle \mathrm{DCO}=55^{\circ}, \angle \mathrm{OAB}=55^{\circ}$