Find the number of sides of a regular polygon in which the interior angle is four times the exterior angle
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Find the number of sides of a regular polygon in which the interior angle is four times the exterior angle
The sum of the angles a regular polygon isQuote:
Originally Posted by helkp
180(n-2)
And there are n angles as well as sides.
Thus, each angle has,
Degree.Code:180(n-2)
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n
The sum of all exterior angles is always 360 degree.
Divided by the number of angles n thus each one has,
By the conditions of the problem you are trying to solve,Code:360
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n
Code:180(n-2) 4*360
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n n
Hello, helkp!
It would help if you knew some formulas.
If you don't, you could wait for years for Divine Inspiration . . .
Quote:
Find the number of sides of a regular polygon in which the interior angle
is four times the exterior angle
A regular polygon of n sides has a total of 180(n - 2) degrees in its interior.
So each interior angle has: 180(n - 2)/n degrees.
An exterior angle is the supplement of an interior angle.
Hence, an exterior angle has: . 180 - 180(n - 2)/2 = 360/n degrees.
We are told that: .[Interior angle] .= .4 × [Exterior angle]
. . . .So we have: . 180(n - 2)/n . = .4 × . (360/n)
Can you finish it now?