Question

Look at the polygon alongside. Observe that, on extending a side of a polygon, we obtain a linear pair of angles, given by an interior angle of the polygon and an exterior angle of the polygon, such that,

Interior angle + exterior angle = 180 degrees​

Answer 1

Step-by-step explanation:

Solution

consider a quadrilateral ABCD, with all the four sides extended in order to obtain four exterior angles

$\tt \angle1, \angle2. \angle3 , and \angle4$

$\tt \: We \: have, \: \angle \: a+ \angle1=180° \\ \tt\angle b+ \angle2=180° \\ \tt \angle c+ \angle3=180° \\ \tt \angle d+ \angle4=180°$

Add the four relations.

$\bf \angle a+ \angle1+ \angle b+ \angle2+ \angle c+ \angle3+ \angle d+ \ angle4 \\ \bf =180° +180° +180° +180°$

Rearrange the terms to get,

$\bf( \angle1+ \angle2+ \angle3+ \angle4)+( \angle a+ \angle b+ \angle c+ \angle d) \\ \bf=720°$

We know that, angle a+ angle b+ angle c+ angle d=360°

(Sum of the interior angles of a quadrilateral)

$\tt \Rightarrow( \angle1+ \angle2+ \angle3+ \angle4)+360° =720° \\ \tt \Rightarrow( \angle1+ \angle2+ \angle3+ \angle4)=720° -360° \\ \tt \Rightarrow( \angle1+ \angle2+ \angle3+ \angle4)=360°$

Thus, sum of the exterior angles of a quadrilateral is 360 degrees

We can see that,

Sum of exterior angles of a quadrilateral

=4×180°- (Sum of interior angles of a quadrilateral)

Continuing in the similar manner, for an n-sided polygon, we will have,

Sum of exterior angles of an n-sided polygon

=n×180° - (Sum of interior angles of an n-sided polygon)

Sum of exterior angles of an n-sided polygon =(n-n+2)×180° =360°

Sum of exterior angles of an n-sided polygon =n×180° - (n-2)×180°

Therefore,

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