The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9°.

Solution:

Let the number of sides in the first polygon be 5x and the number of sides in the second polygon be 4x.

We know:

Angle of an $n$-sided regular polygon $=\left(\frac{n-2}{n}\right) 180^{\circ}$

Thus, we have:

Angle of the first polygon $=\left(\frac{5 x-2}{5 x}\right) 180^{\circ}$

Angle of the second polygon $=\left(\frac{4 x-2}{4 x}\right) 180^{\circ}$

Now,

 $\left(\frac{5 x-2}{5 x}\right) 180-\left(\frac{4 x-2}{4 x}\right) 180=9$

$\Rightarrow 180\left(\frac{4(5 x-2)-5(4 x-2)}{20 x}\right)=9$

$\Rightarrow \frac{20 x-8-20 x+10}{20 x}=\frac{9}{180}$

$\Rightarrow \frac{2}{20 x}=\frac{1}{20}$

$\Rightarrow \frac{2}{x}=1$

$\Rightarrow x=2$

Thus, we have:

Number of sides in the first polygon = 5x = 10

Number of sides in the second polygon = 4x = 8