1). If the sum of the interior angles of a n-gon is eight times of the sum of its exterior angles.find n
2) The size of each interior angle of an n-sided regular ploygan is two-thirds of that of a 20-sided regular polygan.Find the value of n.
兩題form2數學
2010-02-18 9:29 pm
回答 (4)
2010-02-19 8:09 pm
✔ 最佳答案
1) sum of interior angles = 8 x sum of exterior angles=>(n-2)*180 = 8 *360
=>n-2 = 16
=>n=18//
2) (n-2)*)180/n=(2/3) (20-2)*180/20
=>(n-2)/n=18/30
=>(n-2)/n=3/5
=>5n-10=3n
=>2n=10 =>n=5//
2010-02-19 3:01 am
x deg = x degrees
------------------------------
sum of the int. angles of an n-sided polygon = (n-2) 180 deg
int. angles of an n-sided regular polygon = (n-2) 180 deg / n
sum of the int. angles of an n-sided polygon = 360 deg
------------------------------
1.
(n-2) 180 deg = 8 (360 deg)
n-2 = 8 (360)/180
n = 16+2
n=18
2.
(n-2) 180 deg / n = [(20-2) 180 deg / 20] (2/3)
(n-2) / n = 18/30
30n-60=18n
12n=60
n=5
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sum of the int. angles of an n-sided polygon = (n-2) 180 deg
int. angles of an n-sided regular polygon = (n-2) 180 deg / n
sum of the int. angles of an n-sided polygon = 360 deg
------------------------------
1.
(n-2) 180 deg = 8 (360 deg)
n-2 = 8 (360)/180
n = 16+2
n=18
2.
(n-2) 180 deg / n = [(20-2) 180 deg / 20] (2/3)
(n-2) / n = 18/30
30n-60=18n
12n=60
n=5
2010-02-18 10:03 pm
1) sum of interior angles = 8 x sum of exterior angles
=>(n-2)*180 = 8 *360
=>n-2 = 16
=>n=18//
2) (n-2)*)180/n=(2/3) (20-2)*180/20
=>(n-2)/n=18/30
=>(n-2)/n=3/5
=>5n-10=3n
=>2n=10 =>n=5//
=>(n-2)*180 = 8 *360
=>n-2 = 16
=>n=18//
2) (n-2)*)180/n=(2/3) (20-2)*180/20
=>(n-2)/n=18/30
=>(n-2)/n=3/5
=>5n-10=3n
=>2n=10 =>n=5//
2010-02-18 10:02 pm
1)
(n-2)(180)=8(360)
n=18
2)
(n-2)(180)/n=(2/3)(20-2)(180)/20
(n-2)/n=3/5
n=5
(n-2)(180)=8(360)
n=18
2)
(n-2)(180)/n=(2/3)(20-2)(180)/20
(n-2)/n=3/5
n=5
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