好簡單的求多邊形邊數問題！！！！！！

let x be the degree of interior angle
then exterior angle is 180-x
x-(180-x)=120
2x-180=120
x=150

also
x=180*(n-2)/n
150=(180n-360)/n
150n=180n-360
360=30n
n=12

so, n=12

[(n-2)180°]/n - 360°/n=120° (∠ sum of polygon)(ext.∠ of polygon)
180°n-360°-360°=120°n
60°n=720°
n=12

或者可以咁計:
Let an interior angle be y
then an exterior angle wil be y-120°
y+y-120°=180°(∠s on st. line)
2y=300°
y=150°

[(n-2)180°]/n = y(∠ sum of polygon)
180°n-360°=150°n
30°n=360°
n=12

ps.
(n-2)180° = sum of interior angles in a n-gon
[(n-2)180°]/n = each interior angles in a n-gon
360° = sum of exterior angles in a n-gon
360°/n = each exterior angles in a n-gon