Question

prove that sum of the measure of the external angles of any polygon is 360​

Answer 1

Answer:

GIVEN: A polygon with ‘n’ number of sides , <1,<2,<3,<4,<5,…… <n are exterior angles & A,B,C,D,E…..are interior angles.

TO PROVE: <1 + <2 + <3 + <4 + <5 + ……<n = 360°

The sum of interior angles of any polygon

=(n-2)*180 ………(formula used)

PROOF: <1 + <A =180° ………(1)

<2 + <B = 180° ……………(2)

<3 + <C = 180° ………..(3)

<4 + < D = 180° ………..(4)

<5 + <E = 180° ……….(5)

And so on up to n times..

By adding all above (1)+(2)+(3)+(4)+(5)+….(n)

<1+<2+<3+<4+<5+…….<n = 180°n - (A+B+C+D+E+…..n)

= 180n -{ ( n-2)*180 }

= 180n - 180n + 2*180

= 2*180

= 360°

=> <1+<2+<3+<4+<5+…..<n = 360°

[ HENCE PROVED] ●

PRACTICAL METHOD: This could also be proved practically in much easier way. In the above figure , observe rays EAp, which is rotated anti clockwise. Doing so, it replaces ABq, again it is rotated. It replaces BCr. Again rotated. It replaces CDs, then replaces DEt, then comes to original position EAp.

This way the initial ray forming exterior angles , takes complete one rotation in every polygon with any number of sides…

In one complete rotation , angle formed = 360°