Question

the angles of a polygon are in A.P. with common difference 5 degree if the smallest angle is 120 degree find the number of sides of the polygon

Answer 1

The angles of a polygon are in A.P. with common difference 5 degree if the smallest angle is 120 degree find the number of sides of the polygon

Let there be in n sides in the polygon.

sum of all n interior angles of polygon = (n – 2) * 180°

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

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

⇒ n/2 [5n + 235] = (n – 2 ) * 180

⇒ 5n2 + 235n = 360n – 720

⇒ 5n2 – 125n + 720 = 0 ⇒ n2 – 25n + 144 = 0

⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9

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

∴ not possible.

Hence n = 9.

Answer 2

Answer:

given that the smallest angle = a = 120

common difference = 5

to find the number of sides of polygon that is the formula is 360/ n

the sum of the interior angles of any polygon is ( n-2)*180 for further information of this formula refer to ncert maths class 8 textbook.

Sn=n/2(2a+(n-1)d)

therefore (n-2)*180=n/2(2a+(n-1)d)

by factorization n=16,9

n=16 not applicable. Check by substituting n=16

Therefore n=9

Step-by-step explanation: