Exterior angles of polygon - class-VII

Let the number of sides in the polygon is $n$

$We\quad know\quad External\quad angles\quad of\quad a\quad regular\quad polygon\quad are\quad equal.$

Given, regular polygon whose each exterior angle is 40% of a right angle = $ 90^o \times \dfrac{40}{100} = 36^o $
Sum of all exterior angle of any polygon is $ 360^o $
Now,
$ 36^o \times$  number    of   angles  = $ 360^o $
The number  of  angles =$ 10 $
Any polygon have equal number of angles and sides.
Therefore the number of side of the polygon is 10.
Since the number of sides is an integer, therefore their exist a polygon whose each exterior angle is 40% of a right angle. 

Given, a regular polygon whose each exterior angle is $32^o $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = 32^o $
$=> n = \dfrac{45}{4} $
Since, n should be an integer, so it is not possible a regular polygon whose each exterior angle is $32^o $

Let the number of sides of the polygon is n. (which must be an integer)
If each exterior angle is $ 80^o $, then sum of all exterior angle is $ n \times 80^o $.
And Sum of all exterior angles = $ 180^o $
$=>  n \times 80^o = 180^o $
$=> n = 1.25 $
So, it is not possible to have a regular polygon whose each exterior angle is $ 80^o $

Given, a regular polygon whose each exterior angle is $20^o $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = 20^o $
$=> n = 18 $
Since, n should be an integer, so their exist a regular polygon whose each exterior angle is $20^o $

The sum of exterior angles of any convex polygon is 360 degrees. For a regular polygon, each exterior angle is 360/n. Here, 360/n = 24, so n = 360/24 = 15.

The sum of the exterior angles of a regular polygon is $360^o$

Each exterior angle of a regular polygon of 9 sides
$=\dfrac {360^o}{n}$, where $n=9=\left (\dfrac {360^o}{9}\right )^o=40^o$

Given the exterior angle of regular polygon is 12

$\Rightarrow$ Sum of exterior angles of a regular hexagon $=360^o$

Since the number of sides of a regular polygon
$=\dfrac {360}{\text {Exterior angle}}$
$\therefore$ The number of sides of a regular polygon
$=\dfrac {360}{22}[\because$ Exterior angle $=22^o$, given]
$=\dfrac {180}{11}$
Which is not a whole number.
$\therefore$ A regular polygon with measure of each exterior angle as $22^o$ is not possible.

$\Rightarrow$  The sum of the exterior angles of regular octagon is $360^o$.

A regular hexagon has 6 sides. Each exterior angle is 360/6 = 60 degrees.

Let no of sides of the polygon is $n$ 

The exterior angle of a regular polygon is $\dfrac{360}{n}$.

Given, a regular polygon whose each exterior angle is $ \dfrac{1}{8}$ of a right angle = $ \dfrac {1}{8} \times 90^o = \dfrac {45^o}{4} $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = \dfrac {45^o}{4}  $
$=> n = 8 $
Since, n should be an integer, so their exist a regular polygon whose each exterior angle is $ \frac{1}{8}$ of a right angle.

Three of the exterior angles of a hexagon are $ 40^o, 51^o$  and  $86^o $. Each of the remaining exterior angles is $ x^o $.
Sum of all exterior angle of any polygon is $ 360^o $
$ 40^o + 51^o + 86^o + 3 \times x^o = 360^o $
$ => 3 \times x^o = 183^o $
$ => x^o = 61^o $

The sum of the exterior angles of any polygon is always equal to 360.
Exterior angles are $\displaystyle (6x-1)^{\circ}, (10x+2)^{\circ}, (8x+2)^{\circ}, (9x-3)^{\circ}, (5x+4)^{\circ}  and  (12x+6)^{\circ} $
Now, 
 $\displaystyle (6x-1)^{\circ}+ (10x+2)^{\circ}+ (8x+2)^{\circ} + (9x-3)^{\circ} + (5x+4)^{\circ} +  (12x+6)^{\circ} = 360^o $
$ => (50x + 10)^o = 360^o $
$ => x = 7 $
Each Exterior angle 
$ => (6x -1)^o = 6 \times 7 -1 =41^o $
$ => (10x +2)^o = 10 \times 7 +2 =72^o $
$ => (8x +2)^o = 8 \times 7 +2 =58^o $
$ => (9x -3)^o = 9 \times 7 -3 =60^o $
$ => (5x +4)^o = 5 \times 7 +4  =39^o $
$ => (12x +6)^o = 12 \times 7 + 6 =90^o $

In a regular polygon all the exterior angles have the same measure. 

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
The sum of the exterior angles of a polygon is always equal to $360^o$.
The sum of the interior angles of polygon = $180 (n-2)$
=> $180 (n-2) = 4 \times 360$
=> $n -2 = 8$ 
=> $n =10$ 
Number of sides in the polygon = $10$

No matter what type of polygon, the sum of the exterior angles is always equal to $360^o$.
It does not depends upon number of sides of polygon.  

Sum of measure of Interior angles of a regular polygon is calculated as,
$(n-2)180$ where,
n: Number of sides of a regular polygon.
Sum of exterior angles of a regular polygon always add up to $360^{o}$
$\therefore$ ,$(n-2)180=5(360)$
$\therefore n=12$

Two times the interior angle of a regular polygon is equal to seven times is exterior angle.
Each Interior angle of a polygon = $ \dfrac {180^o (n-2)}{n} $
Each Exterior angle of a polygon = $ \dfrac{360^o}{n} $
Now,
$ 2 \times \dfrac {180^o (n-2)}{n} = 7 \times  \dfrac{360^o}{n}  $
$=> n -2 = 7 $
$=> n = 9 $
Number of sides of polygon is 9.
Each Interior angle of a polygon = $ \dfrac {180^o (n-2)}{n} = \dfrac {180^o (9-2)}{9} = 140^o  $

Let $7a$ be the interior angle

For a polygon of 'n' sides, the angle subtended at the centre is $ \dfrac {{360}^{o}}{n} $

Given, angle at the centre $ = {30}^{o} $
$ => \dfrac {{360}^{o}}{n}= {30}^{o} $
$ => n = 12 $

Hence, the polygon has $ 12 $ sides.

Let the number of sides in the regular polygon be $n$

Sum of exterior angles for any polynomial is always $360$. 

A) Exterior angle of m-gon$=\cfrac { (m180)-(m-2)180 }{ m } $

Number of sides in hexagon $=6$
Sum of the interior angles of a polygon$=(n-2)\pi$
$n(Interior\ Angle)=(n-2)\pi$
$\Rightarrow $ Interior Angle $= \dfrac{4}{6}\pi$

Here, let the number of sides of the polygon be $n$