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

(i) 170°

(ii) 138°

(i) 170°

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

Given that each interior angle is 170°,

⇒ $\dfrac{(2n - 4) \times 90°}{n}$ = 170°

⇒ (2n - 4) x 90° = 170°n

⇒ 180°n - 360° = 170°n

⇒ 180°n - 170°n = 360°

⇒ 10°n = 360°

⇒ n = $\dfrac{360°}{10°}$

⇒ n = 36

Hence, a regular polygon is possible when each interior angle is 170°.

(ii) 138°

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

Given that each interior angle is 138°,

⇒ $\dfrac{(2n - 4) \times 90°}{n}$ = 138°

⇒ (2n - 4) x 90° = 138°n

⇒ 180°n - 360° = 138°n

⇒ 180°n - 138°n = 360°

⇒ 42°n = 360°

⇒ n = $\dfrac{360°}{42°}$

⇒ n = $\dfrac{60°}{7°}$

⇒ n = $8\dfrac{4}{7}$

Hence, a regular polygon is not possible when each interior angle is 138°.