In the given figure, , =115^ and =70^ . Find 1. 2. 3. 4. - Vidyadip

Question

In the given figure, $\triangle\text{ODC}\sim\triangle\text{OBA},\angle\text{BOC}=115^\circ$ and $\angle\text{CDO}=70^\circ.$
Find

$\angle\text{DOC}$
$\angle\text{DCO}$
$\angle\text{OAB}$
$\angle\text{OBA}$

✓

Answer

$\triangle\text{ODC}\sim\triangle\text{OBC}$
$\angle\text{BOC}=115^\circ$
$\angle\text{CDO}=70^\circ$

$\angle\text{DOC}=(180^\circ-\angle\text{BOC})$

$=(180^\circ-115^\circ)$

$=65^\circ$

$\angle\text{OCD}=180^\circ-\angle\text{CDO}-\angle\text{DOC}$

$\angle\text{OCD}=180^\circ-(70^\circ+65^\circ)$

$=45^\circ$

Now, $\triangle\text{ABO}\sim\triangle\text{ODC}$

$\angle\text{AOB}=\angle\text{OCD }(\text{vert.Opp }\angle\text{s})=45^\circ$

$\angle\text{OAB}=\angle\text{OCD}=45^\circ$

$\angle\text{OBA}=\angle\text{ODC}$ (alternate angles) = 70º

So, $\angle\text{OAB}=45^\circ$ and $\angle\text{OBA}=70^\circ$

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