Question
The difference between the exterior angles of two regular polygons, having the sides equal to $(n – 1)$ and $(n + 1)$ is $9^\circ$. Find the value of $n.$
✓
Answer
We know that sum of exterior angles of a polynomial is $360^{\circ}$
$(i)$ If sides of a regular polygon $=n-1$
Then each angle $=\frac{360^{\circ}}{\mathrm{n}-1}$
and if sides are $n+1$, then
each angle $=\frac{360^{\circ}}{\mathrm{n}+1}$
According to the condition,
$\frac{360^{\circ}}{\mathrm{n}-1}-\frac{360^{\circ}}{\mathrm{n}+1}=9$
$ \Rightarrow 360\left[\frac{1}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+1}\right]=9$
$ \Rightarrow 360\left[\frac{\mathrm{n}+1-\mathrm{n}+1}{\mathrm{n}-1}(\mathrm{n}+1)\right]=9$
$ \Rightarrow \frac{2 \times 360}{\mathrm{n}^2-1}=9$
$ \Rightarrow \mathrm{n}^2-1=\frac{2 \times 360}{9}=80$
$ \Rightarrow n^2-1=80$
$ \Rightarrow n^2=1-80=0$
$ \Rightarrow \mathrm{n}^2-81=0$
$\Rightarrow(n)^2-(9)^2=0$
$ \Rightarrow(n+9)(n-9)=0$
Either $n+9=0$. then $n=-9$ which is not possible being negative, or $\mathrm{n}-9=0$, then $\mathrm{n}=9$
$\therefore \mathrm{n}=9$
$\therefore$ No. of. sides of a regular polygon $=9$
$(i)$ If sides of a regular polygon $=n-1$
Then each angle $=\frac{360^{\circ}}{\mathrm{n}-1}$
and if sides are $n+1$, then
each angle $=\frac{360^{\circ}}{\mathrm{n}+1}$
According to the condition,
$\frac{360^{\circ}}{\mathrm{n}-1}-\frac{360^{\circ}}{\mathrm{n}+1}=9$
$ \Rightarrow 360\left[\frac{1}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+1}\right]=9$
$ \Rightarrow 360\left[\frac{\mathrm{n}+1-\mathrm{n}+1}{\mathrm{n}-1}(\mathrm{n}+1)\right]=9$
$ \Rightarrow \frac{2 \times 360}{\mathrm{n}^2-1}=9$
$ \Rightarrow \mathrm{n}^2-1=\frac{2 \times 360}{9}=80$
$ \Rightarrow n^2-1=80$
$ \Rightarrow n^2=1-80=0$
$ \Rightarrow \mathrm{n}^2-81=0$
$\Rightarrow(n)^2-(9)^2=0$
$ \Rightarrow(n+9)(n-9)=0$
Either $n+9=0$. then $n=-9$ which is not possible being negative, or $\mathrm{n}-9=0$, then $\mathrm{n}=9$
$\therefore \mathrm{n}=9$
$\therefore$ No. of. sides of a regular polygon $=9$
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