In the given figure, $∆ODC \sim ∆OBA, \angle BOC = 125^o$ and $\angle CDO = 70^o$. Find $\angle DOC, \angle DCO$ and $\angle OAB$.
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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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Updated on: 10-Oct-2022

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