Question

If the difference between an interior angle of a regular polygon of $\left( {n + 1} \right)$ sides and an interior angle of a regular polygon of $n$ sides is ${4^ \circ }$; find the value of $n$. Also, state the difference between their exterior angles.
A. $n = 9$ and difference between exterior angles ${4^ \circ }$
B. $n = 5$ and difference between exterior angles ${22^ \circ }$
C. $n = 11$ and difference between exterior angles ${12^ \circ }$
D. none of these

Answer 1

Hint:
We will first write the measure of interior angle with $\left( {n + 1} \right)$ sides and $n$ sides. Then, their difference is given as 4. We will use this condition to form an equation in $n$. We will solve the quadratic equation and find the value of $n$. Next, we will use the formula, $\dfrac{{{{360}^ \circ }}}{N}$, where $N$ is the number of sides of polygon to find the measure of exterior angle of polygon with $n$ and $n + 1$ sides and take their difference.

Complete step by step solution:
We are given that the polygon has $\left( {n + 1} \right)$ sides.
We know that the an interior angle of a polygon with $\left( {n + 1} \right)$ sides is equal to $\dfrac{{{{180}^ \circ }\left( {\left( {n + 1} \right) - 2} \right)}}{{n + 1}} = \dfrac{{{{180}^ \circ }\left( {n - 1} \right)}}{{n + 1}}$
Similarly, a polygon with $n$ sides has each interior angle of measure $\dfrac{{{{180}^ \circ }\left( {n - 2} \right)}}{n}$
We are given that the difference in the measure of interior angle of a regular polygon with $\left( {n + 1} \right)$ sides and $n$ sides is ${4^ \circ }$
Therefore,
$\dfrac{{{{180}^ \circ }\left( {n - 1} \right)}}{{n + 1}} - \dfrac{{{{180}^ \circ }\left( {n - 2} \right)}}{n} = {4^ \circ }$
Take 180 common and divide the equation by 4.
$
  {180^ \circ }\left( {\dfrac{{\left( {n - 1} \right)}}{{n + 1}} - \dfrac{{^ \circ \left( {n - 2} \right)}}{n}} \right) = {4^ \circ } \\
   \Rightarrow 45\left( {\dfrac{{\left( {n - 1} \right)}}{{n + 1}} - \dfrac{{^ \circ \left( {n - 2} \right)}}{n}} \right) = 1
$
Take the LCM and simplify the expression.
$
  45\left( {\dfrac{{n\left( {n - 1} \right) - \left( {n + 1} \right)\left( {n - 2} \right)}}{{n\left( {n + 1} \right)}}} \right) = 1 \\
   \Rightarrow 45\left( {{n^2} - n - {n^2} + 2n - n + 2} \right) = {n^2} + n \\
   \Rightarrow 90 = {n^2} + n \\
   \Rightarrow {n^2} + n - 90 = 0
$
Factorise the above equation.
$
  {n^2} + 10n - 9n - 90 = 0 \\
   \Rightarrow n\left( {n + 10} \right) - 9\left( {n + 10} \right) = 0 \\
   \Rightarrow \left( {n - 9} \right)\left( {n + 10} \right) = 0 \\
   \Rightarrow n = 9, - 10 $
But, $n$ cannot be negative.
Then, $n = 9$
The exterior angle is equal to $\dfrac{{{{360}^ \circ }}}{N}$, where $N$ is the number of sides of a polygon.
Hence, the each exterior angle of a polygon with $n = 9$ sides is
$\dfrac{{{{360}^ \circ }}}{9} = {40^ \circ }$
And each exterior angle of a polygon with $n + 1 = 9 + 1 = 10$ sides is
$\dfrac{{{{360}^ \circ }}}{{10}} = {36^ \circ }$
Thus, the difference in the measure of exterior angle is ${40^ \circ } - {36^ \circ } = {4^ \circ }$

Hence, option A is correct.

Note:
In a regular polygon all the interior angles, exterior angles and all the sides are of equal measure. If a polygon has $n$ sides, then the measure of each of its interior angle is given by $\dfrac{{{{180}^ \circ }\left( {n - 2} \right)}}{n}$ and measure of each exterior angle is given by $\dfrac{{{{360}^ \circ }}}{n}$