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

It is given that each interior angle of a regular polygon is 144°.

According to the properties of polygons, if a regular polygon has n sides, each of its interior angles $\dfrac{(2n - 4) \times 90°}{n}$.

⇒ $\dfrac{(2n - 4) \times 90°}{n}$ = 144°

⇒ (2n - 4) x 90° = 144°n

⇒ 180°n - 360° = 144°n

⇒ 180°n - 144°n = 360°

⇒ 36°n = 360°

⇒ n = $\dfrac{360°}{36°}$

⇒ n = 10

So, the first polygon has 10 sides.

The number of sides of the second polygon is double that of the first polygon:

n = 2 x 10 = 20

For this second polygon, each of its interior angles is $\dfrac{(2n - 4) \times 90°}{n}$.

⇒ Angle = $\dfrac{(2 \times 20 - 4) \times 90°}{20}$

= $\dfrac{(40 - 4) \times 90°}{20}$

= $\dfrac{36 \times 90°}{20}$

= $\dfrac{3240°}{20}$

= 162°

Hence, the interior angle of the second polygon is 162°.