A regular polygon has $n$ sides . When the number of sides is doubled, each interior angle increases by $20^{\circ}$. Find $n$.
My workings till I got stuck
$1$ int. angle of $n$ sides $=180^{\circ}n-360^{\circ}/n$
$1$ int. angle of $2n$ sides $=200^{\circ}n-360^{\circ}/n$
$1$ ext. angle of $2n$ sides $=-20^{\circ}n+360^{\circ}/n$
$n=360$ divide $1$ ext. angle …
I've done it till $-20n+360=360$
Then I got stuck. Can I get help? Thanks in advance!
Best Answer
The total of the angles in an $n$-gon is $180(n-2)$, so each angle is
$$a(n) = \frac{180(n-2)}{n}$$
with the angle measured in degrees.
So you want to find $n$ where $a(2n) = a(n) + 20$. This is the same as solving
$$\frac{180(2n-2)}{2n} = \frac{180(n-2)}{n} + 20.$$
Multiplying through by $n$ gives
$$180(n-1) = 180(n-2) + 20n,$$
so $n=9$.