In the given figure, $\triangle\text{ODC}\sim\triangle\text{OBA},\angle\text{BOC}=115^\circ$ and $\angle\text{CDO}=70^\circ.$
Find
Find
- $\angle\text{DOC}$
- $\angle\text{DCO}$
- $\angle\text{OAB}$
- $\angle\text{OBA}$
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$\triangle\text{ODC}\sim\triangle\text{OBC}$
$\angle\text{BOC}=115^\circ$
$\angle\text{CDO}=70^\circ$
- $\angle\text{DOC}=(180^\circ-\angle\text{BOC})$
$=65^\circ$
- $\angle\text{OCD}=180^\circ-\angle\text{CDO}-\angle\text{DOC}$
$=45^\circ$
- Now, $\triangle\text{ABO}\sim\triangle\text{ODC}$
$\angle\text{OAB}=\angle\text{OCD}=45^\circ$
- $\angle\text{OBA}=\angle\text{ODC}$ (alternate angles) = 70º
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