Question

Find the measure of each interior and exterior angle of a regular $40-\mathrm{gon}$.

Answer 1

Step 1: Find the measure of each interior angle.

Measure of each interior angle of regular polygon of $n$ sides is given by the formula, $\frac{\left(n-2\right)·180°}{n}$.

We know that, $40-\mathrm{gon}$ is a polygon with $40$ sides.

Substitute $n=40$ in $\frac{\left(n-2\right)·180°}{n}$.

$\begin{array}{rcl}\frac{\left(n-2\right)·180°}{n}& =& \frac{\left(40-2\right)·180°}{40}\\ & =& \frac{38·180°}{40}\\ & =& 171°\end{array}$

Therefore, measure of each interior angle of $40-\mathrm{gon}$ is $171°$.

Step 2: Find the measure of each exterior angle.

We know that, sum of interior angle and exterior angle is $180°$.

So, exterior angle is given by difference between $180°$ and interior angle.

Then, exterior angle is $180°-171°=9°$.

Hence, measure of each interior and exterior angle of a regular $40-\mathrm{gon}$ is $171°$ and $9°$.