What is the formula for finding the exterior and the interior angles of a polygon?

Hint: We will use the formula for finding the sum of the interior angles of a polygon to find each interior angle. We can use the fact that the sum of the interior angle and the exterior angle is $180{}^\circ .$ Also, we know that the sum of the exterior angles is equal to $360{}^\circ .$

Complete step by step solution:
Suppose that we are given a polygon of $n$ sides. Then we can find the sum of interior angles of the polynomial by the formula $S=180\left( n-2 \right).$
We usually use division to find the value of a single object when the value of a number of objects is given.
So, to find each interior angle, we need to divide the sum of the interior angles by the number of sides.
So, the interior angle of a polygon can be found by $\theta =\dfrac{180\left( n-2 \right)}{n}.$
Let us suppose that we are given a polygon with an interior angle is $\theta .$
We have already learnt that the sum of the exterior angle and the interior angle is equal to $180{}^\circ .$
So, if we represent the exterior angle by $\alpha ,$ we will get $\alpha +\theta =180{}^\circ .$
So, from this we will get the value of exterior angle by $\alpha =180-\theta .$
We can also find the interior angle using another method. Because, we know that the sum of exterior angles is equal to $360{}^\circ .$
So, we can find each exterior angle by $\alpha =\dfrac{360}{n}.$
Hence the interior angle of a polygon is $\dfrac{180\left( n-2 \right)}{n}$ and the exterior angle is $\dfrac{360}{n}.$

Note: We know that the sum of interior angles of a triangle is equal to $180{}^\circ .$ Suppose that we are given a polygon of $n$ sides. If we draw diagonals from each side of the polygon, we will get $n-2$ triangles. So, the sum of interior angles of a polygon is equal to $\left( n-2 \right)180.$