Is it possible to have a regular polygon whose each exterior angle is:

(i) 80°

(ii) 40% of a right angle

(i) 80°

According to the properties of a polygon, if there are n sides, then each of its exterior angles is $\dfrac{360°}{n}$.

Given that each exterior angle is 80°,

⇒ $\dfrac{360°}{n}$ = 80°

⇒ n = $\dfrac{360°}{80°}$

⇒ n = 4.5

Hence, a regular polygon is not possible when each exterior angle is 80°.

(ii) 40% of a right angle is:

= 40% x 90°

= $\dfrac{40}{100} \times 90°$

= $\dfrac{3600°}{100}$

= 36°

According to the properties of a polygon, if there are n sides, then each of its exterior angles is $\dfrac{360°}{n}$.

Given that each exterior angle is 36°,

⇒ $\dfrac{360°}{n}$ = 36°

⇒ n = $\dfrac{360°}{36°}$

⇒ n = 10

Hence, a regular polygon is possible when each exterior angle is 40% of a right angle.