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${\text{(a)}}$ Is it possible to have a regular polygon with measure of each exterior angle as ${\text{2}}{{\text{2}}^0}$?
${\text{(b)}}$ Can it be an interior angle of a regular polygon? Why?

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Hint- Here, the property of a regular polygon is used (i.e., the sum of all the exterior angles which are equal to each other of a regular polygon is always ${360^0}$).


${\text{(a)}}$ No, it is not possible to have a regular polygon with measure of each exterior angle of ${\text{2}}{{\text{2}}^0}$ because ${\text{2}}{{\text{2}}^0}$ is not a multiple of ${360^0}$. Since, the sum of all the exterior angles of a regular polygon is always ${360^0}$ and all the exterior angles of a regular polygon are equal in measure.


${\text{(b)}}$ If the interior angle of a regular polygon is ${\text{2}}{{\text{2}}^0}$, then the measure of exterior angle of that regular polygon will be \[\left( {{{180}^0} - {\text{2}}{{\text{2}}^0}} \right) = {158^0}\]. Clearly, \[{158^0}\] is not a multiple of ${360^0}$. So, it is not possible to have a regular polygon with a measure of each interior angle of ${\text{2}}{{\text{2}}^0}$.

Note- In these types of problems, if the interior angle of the regular polygon is given then it is converted into the exterior angle of the regular polygon. Then using properties of a regular polygon like the sum of all the exterior angles is always equal to ${360^0}$ and each exterior angle is equal, we have to check whether these properties hold true or false. If they hold then that regular polygon is possible else, it is not.