How do you find exterior/interior angles in polygons?

For a regular polygon.

- Sum of exterior angles = 360

Formula for an individual exterior angle ---> 360 / n

- Sum of Interior Angles = 180(n - 2)

Individual Interior Angles = 180(n - 2) / n

1) n = 29

Individual Interior Angles = 180(n - 2) / n

= 180(29 - 2) / 29 = 167.6°

2) Individual exterior angle ---> 360 / n

36 = 360 / n ---> n = 360 / 36 = 10 sides.

3) Sum of Interior Angles = 180(n - 2)

4500 = 180n - 360 ---> 180n = 4860

n = 27 sides

The sum of the interior angles of a polygon is S-2 (180) where S is the number of sides.

1.) (29 - 2) (180) = 27 * 180 = 4860 degrees

http://www.mathopenref.com/polygoninteriorangles.h...

2) The sum of the exterior angles of a polygon is always 360 degrees

since the polygon is regular all angles are the same

360/S = 36 ; S = 360/36 = 10 sides

http://www.mathopenref.com/polygonexteriorangles.h...

3 as in question 1 number of sides = N

(n-2)(180) = 4500 ; n-2 = 4500/180 = 25 ; n = 25 +2 = 27 sides

Extrrior angles add up to 360, which is fairly handy to prove. Imagive going for walks along such exterior angles. You can at one in all them, and then you possibly can walk around, finally you have got ended up where you could have began. As a result, you will have accomplished a 360 rotation. Additional facet notes is that each exterior perspective is a hundred and eighty - z which z is the perspective meavsure of a normal polygon. We all know that z in comparative with the number of aspects yiels (n-2)180/n. So, we have n(180-one hundred eighty(n-2)/n) => 180n(1-(n-2)/n) = 180n([n-(n-2)]/n) = 180(n-(n-2)) = one hundred eighty(2) = 360. Oh an ingeneral, sum of interrior angles is a hundred and eighty(n-2) the place n is the number of facets ;)