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In the given figure, $∆ODC \sim ∆OBA, \angle BOC = 125^o$ and $\angle CDO = 70^o$. Find $\angle DOC, \angle DCO$ and $\angle OAB$.
Given:
$∆ODC \sim ∆OBA, \angle BOC = 125^o$ and $\angle CDO = 70^o$.
To do:
We have to find $\angle DOC, \angle DCO$ and $\angle OAB$.
Solution:
 From the given figure,
$\angle DOC=180^o-125^o$
$=55^o$ (Linear pair)
In $\triangle DOC,$
$\angle DCO+\angle ODC+\angle DOC=180^o$
$\angle DCO+70^o+55^o=180^o$
$\angle DCO=180^o-125^o$
$=55^o$
$\triangle ODC \sim \triangle OBA$
This implies,
$\angle OAB=\angle OCD$
$=55^o$
Therefore,
$\angle DOC=55^o, \angle DCO=55^o$ and $\angle OAB=55^o$.
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