② The angles ofa polygon are in A. P, whose common difference is 5°

1.	② The angles ofa polygon are in A. P, whose common difference is 5°. If the least angle2beradian, then find the number of sides of the polygon.
Answer» Let there be in n sides in the polygon. Then by geometry, sum of all n interior angles of polygon = (n – 2) * 180° Also the angles are in A. P. with the smallest angle = 120° , common difference = 5° ∴ Sum of all interior angles of polygon = n/2[2 * 120 + ( n – 1) * 5 Thus we should have n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180 ⇒ n/2 [5n + 235] = (n – 2 ) * 180 ⇒ 5n^2+ 235n = 360n – 720 ⇒ 5n^2– 125n + 720 = 0 ⇒ n^2– 25n + 144 = 0 ⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9 Also if n = 16 then 16thangle = 120 + 15 * 5 = 195° > 180° ∴ not possible. Hence n = 9.

② The angles ofa polygon are in A. P, whose common difference is 5°. If the least angle2beradian, then find the number of sides of the polygon.

Answer»

Let there be in n sides in the polygon.

Then by geometry, sum of all n interior angles of polygon = (n – 2) * 180°

Also the angles are in A. P. with the smallest angle = 120° , common difference = 5°

∴ Sum of all interior angles of polygon

= n/2[2 * 120 + ( n – 1) * 5

Thus we should have

n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180

⇒ n/2 [5n + 235] = (n – 2 ) * 180

⇒ 5n^2+ 235n = 360n – 720

⇒ 5n^2– 125n + 720 = 0 ⇒ n^2– 25n + 144 = 0

⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9

Also if n = 16 then 16thangle = 120 + 15 * 5 = 195° > 180°

∴ not possible.

Hence n = 9.