If one of the interior angles of a regular polygon is equal to $\frac{5}{6}$ times of one of the interior angles of a regular pentagon, then the number of sides of the polygon is

One of interior angles of a regular polygon $=\frac{5}{6}$ one of the interior angle of regular pentagon

$\begin{array}{l}\frac{(n-2)180°}{n}=\frac{5}{6}.\frac{(5-2)180°}{5}\\ \frac{(n-2)180°}{n}=\frac{3\times 30°}{1}\\ 180°n-360°=90°n\\ 180°n-90°n=360°\\ 90°n=360°\end{array}$

$\therefore$ Sides of polygon $n=4$