# angles of polygon

• Sep 12th 2006, 05:32 AM
helkp
angles of polygon
Find the number of sides of a regular polygon in which the interior angle is four times the exterior angle
• Sep 12th 2006, 05:39 AM
ThePerfectHacker
Quote:

Originally Posted by helkp
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 is
180(n-2)
And there are n angles as well as sides.
Thus, each angle has,
Code:

```180(n-2) ----------     n```
Degree.

The sum of all exterior angles is always 360 degree.
Divided by the number of angles n thus each one has,
Code:

```  360   ----     n```
By the conditions of the problem you are trying to solve,
Code:

```180(n-2)      4*360 ------------  = ----       n              n```
• Sep 12th 2006, 06:42 AM
Soroban
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?