How can you use trigonometry to help you find the area of a regular polygon?

If we have a regular polygon, we can divide it on congruent triangles, and than find the area of polygon n*Area of triangle.

1)    If you given the radius of the polygon, then side of the triangle =radius of polygon.
n - number of sides of the polygon.
To find the angle ( angle between 2 radii)  in this triangle between 2 equals side =360/n.
Area of the triangle then A=(1/2)ab sin C= (1/2)r*r *sin(360/n)
A=(1/2)r²*sin(360/n)
A(polygon)=n*(1/2)r²*sin(360/n)
2) If you are given the side of the polygon.
You still need to  divide it into triangles and find an area of the triangle.
Find one of the angles of the triangle.
Angle of the triangle = 360/n  
Area triangle =1/2*base*height
base=side of the polygon (s)
To find the height we need to look at a small right triangle,
in this triangle angle near the center of polygon =(360/2n).
Opposite leg to this angle =s/2
tan(360/2n)= opposite/adjacent =(s/2)/height
height=(s/2)/tan(360/2n)=s/(2tan(360/2n)
The area of the triangle = (1/2) *s*s/(2tan(360/2n)=s²/(4tan(360/2n))
The area of polygon =n*s²/(4tan(360/2n))

But I think it is better to solve problem when you know what are given at least.