Problem about finding pi

I was feeling bored one day and I suddenly wanted to find out how people get the value pi. Before I have done any research, I used my high school math knowledge and tried to find the value of pi.
First I thought, pi can be think of as the ratio of the circumferece of a half circle to the radius, and a circle with the concept of limit can be imagined as a polygon with infinity sides(or in another word, a circle can be formed by infinity number of isoscele triangles with two equal sides as radius)
So by using cosine rule I found the extremely small and identical narrow isoscele triangle's third edge and multiplied it by the number that is enough to form a half circle and divide the total by the radius.
Below is my working
Assume that an isoscele triangle has a side and b side being equal and they are also equal to the radius of a circle. c is the third side of the triangle

Applying Cosine Rule, I obtain the square of third side c^2 as
c^2 = a^2 + b^2 - 2ab Cos(C) note capital C is the corresponding angle of side c

since a = b = r
c^2 = 2(r^2) - 2(r^2) Cos(C)
= 2(r^2)(1-Cos(C))

let C = 180/n n is the number of triangles use to form a half circle
the reason C being 180/n, is so that each identical triangle has 180/n degrees as its third angle so when there are n triangles the total angles would sum up as 180, a half circle i.e. n(180/n) = 180

The sum of the third side of every identical triangle = n x c
c = [2(r^2)(1-Cos(C))]^1/2 square root
= r x [2(1-Cos(C))]^1/2 r^2's square root is r
so sum of sides = nc = nr[2(1-Cos(C))]^1/2

divide sum of sides by the radius r we get

n[2(1-Cos(C))]^1/2 Do this in order to obtain ratio of sum of c sides of the triangles to the radius

lim n[2(1-Cos(C))]^1/2 = pi
n->infinity

The limit concept I said at the beginning