The interior angle of a regular polygon is 144°. Find the interior angle of a polygon, which has double the number of sides as the first polygon. 

Let the number of sides of the first polygon = n

$n=\frac{360°}{180°- Interior angle}=\frac{360°}{180°-144°}=10$

The number of sides of second polygon =20

Interior angle of second polygon = $\frac{(2n-4)90°}{n}$

$=\frac{(2\times 20-4)\times 90°}{20}$

$=162°$