Question
In Fig. $OB$ is perpendicular to $OA$ and $ \text{BOC} = 49^\circ $. Find $\angle\text{AOD}$. 
✓
Answer
From the given figure, we have $\angle\text{DOB}+\angle\text{BOC}=180^\circ$ [Liner pair]
$\Rightarrow\angle\text{DOB}+49^\circ=180^\circ$
$[\because\angle\text{BOC=49}^\circ]$
$\Rightarrow\angle\text{DOB}=180^\circ-49^\circ=131^\circ$ Now, $\angle\text{DOB}+\angle\text{BOA}+\angle\text{AOB}=360^\circ$
$[\because $ sum of all the angles around apoint is $360^\circ ]$
$\Rightarrow131^\circ+90^\circ+\angle\text{AOD}=360^\circ$
$[\because\ \angle\text{DOB}=131^\circ,\angle\text{BOA}=90^\circ]$
$\Rightarrow221^\circ+\angle\text{AOD}=360^\circ$
$\Rightarrow\angle \text{AOD}=360^\circ-221^\circ=139^\circ$
$\Rightarrow\angle\text{DOB}+49^\circ=180^\circ$
$[\because\angle\text{BOC=49}^\circ]$
$\Rightarrow\angle\text{DOB}=180^\circ-49^\circ=131^\circ$ Now, $\angle\text{DOB}+\angle\text{BOA}+\angle\text{AOB}=360^\circ$
$[\because $ sum of all the angles around apoint is $360^\circ ]$
$\Rightarrow131^\circ+90^\circ+\angle\text{AOD}=360^\circ$
$[\because\ \angle\text{DOB}=131^\circ,\angle\text{BOA}=90^\circ]$
$\Rightarrow221^\circ+\angle\text{AOD}=360^\circ$
$\Rightarrow\angle \text{AOD}=360^\circ-221^\circ=139^\circ$
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