The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.
Solution
We know that sum of exterior angles of a polynomial is 360°
(i) If sides of a regular polygon = n – 1
Then each angle = `360^circ/("n" - 1)`
and if sides are n + 1, then
each angle = `360^circ/("n" + 1)`
According to the condition,
`360^circ/("n" - 1) - 360^circ/("n" + 1)=9`
`=> 360 [1/("x" - 1) - 1/("x" + 1)] = 9`
`=> 360 [("n" + 1 - "n" + 1)/("n" - 1)("n" + 1)] = 9`
`=> (2 xx 360)/("n"^2 - 1) = 9`
`=> "n"^2 - 1 = (2 xx 360)/9 = 80`
`=> n^2 - 1 = 80`
`=> n^2 = 1 - 80 = 0`
⇒ n2 - 81 = 0
⇒ (n)2 - (9)2 = 0
⇒ (n + 9)(n - 9) = 0
Either n + 9 = 0. then n = -9 which is not possible being negative,
or n - 9 = 0, then n = 9
∴ n = 9
∴ No. of. sides of a regular polygon = 9
