The difference between any two consecutive interior angles of a polygon is `5o`. If the smallest angle is `120o`, find the number of the sides of the polygon.

Answer

N/a

Explanation

Let there are 'n' sides in the polygen. Sum of interior angles of polygen `=(2n-4)xx90^(@).` and sum of 'n' angles of A.P.
`=(n)/(2)[2xx120^(@)+(n-1).5^(@)]=(n)/(2)(5n+235)^(@)`
`"Now "(n)/(2)(5n+235)^(@)=(2n-4)xx90^(@)`
`rArr" "n(n+47)=2(2n-4)xx18`
`rArr" "n^(2)+47n=72n-144`
`rArr" "n^(2)-25n+144=0`
`rArr" "(n-9)(n-16)=0`
`rArr" "n-9=0 or n-16=0`
`rArr" "n=9 or n=16`
`T_(9)=a+8d=8d=120^(@)+8xx5^(@)=160^(@)`
`"and "T_(16)=a+15d=120^(@)+15xx5^(@)=195^(@)`
But no angle of the polygon be `195^(@)`.
Therefore, number of sides in polygon are 9.