好簡單的求多邊形邊數問題！！！！！！

Let n be the number of side of the polygon
exterior angle = 360/n
interior angle = ((n-2)*180)/n
So we get:
(360/n)+120=((n-2)*180)/n
(360/n)+120=(180n-360)/n
(360/n)+120=180-360/n
(360/n)+(360/n)=60
720/n=60
n=12

n = 12.
The polygon has 12 sides.
It is a dodecagon

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

或者可以咁計:
[(n-2)180°]/n - 360°/n=120° (∠ sum of polygon)(ext.∠ of polygon)
180°n-360°-360°=120°n
60°n=720°
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

exterior angle=360/n
interior angle=(n-2)x180/n
360/n+120=(n-2)x180/n
360/n+120=(180n-360)/n
360+120n=180n-360
60n=720
n=12