In the figure, find $x$, further find $\angle BOC, \angle COD$ and $\angle AOD$.
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To do:
We have to find $x$ and $\angle BOC, \angle COD$ and $\angle AOD$.
Solution:
We know that,
Linear pairs of angles add up to 180 degrees.
In the given figure, $\angle AOD$ and $\angle BOD$ form a linear pair.
Therefore,
$(x+10)^o+[x^o+(x+20^o)]=180^o$ [$\angle BOD=\angle BOC+\angle COD$]
$3x=180^o-10^o-20^o$
$3x=150^o$
$x=50^o$
$\angle BOC=x+20^o=50^o+20^o=70^o$
$\angle COD=x=50^o$
$\angle AOD=x+10^o=50^o+10^o=60^o$
The value of $x$ is $50^o$.
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