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Why the sum of exterior angles of any n-sided polygon is 360°?

 


Why the sum of exterior angles of any n-sided polygon is 360°?


1. Firstly, let’s take any n-sided polygon and construct their exterior angles taken either clockwise or anti-clockwise as shown in the given figure.


 

2.     In the given figure, a, b, c, d, e,… are the interior angles and a’, b’, c’, d’, e’,… are the exterior angles of n-sided polygon ABCDE.

3.     We know, the no. of exterior angles formed will be equal to the no. Of sides of polygon (n).

4.     By the rule of straight angles, we have:

   a + a’ = 180° ,  b + b’ = 180°

  c + c’ = 180°,  d + d’ = 180°,  e + e’ = 180° and so on..

Here, adding them all, we get:

   (a + a’) + (b + b’) + (c + c’) + (d + d’) + (e + e’) +… = n*180° 

               [ there are n straight angles when the sides are extended as in the above figure.]

or, (a + b + c + d + e +…) + (a’ + b’ + c’ + d’ + e’ +..) = 180°n ………………… (I)

 

5.     We also know that sum of interior angles of any n-sided polygon is

   (n-2)*180° i.e. (a + b + c + d + e +…) = (n-2)*180°……………(II)

 

6.     Combining equations (I) and (II), we have:

   (a’ + b’ + c’ + d’ + e’ +..) + (n-2)*180° = 180°n

  or,  (a’ + b’ + c’ + d’ + e’ +..) = 180°n - [(n-2)*180° ]

  or,  (a’ + b’ + c’ + d’ + e’ +..) = 180°n - [180°n - 360°]

  or,  (a’ + b’ + c’ + d’ + e’ +..) = 180°n - 180°n + 360°

      or,  (a’ + b’ + c’ + d’ + e’ +..) = 360°,,


Thus,

  The sum of exterior angles of any n-sided polygon is 360°.


-Nischal karki

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