To round up my sines and cosines, in an earlier post I forgot to mention that I have to elaborate a bit about the law of Sines. If you draw a circle through the three vertices of the triangle this is called the *circumscribed circle*. If the circumscribed circle has a diameter of the Law of Sines states that

You can prove this as follows (As always: click on the pictures for a larger version):

You see a diagonal line running through the circle with a length (of course) of . This line both splits the big triangle upper angle and the small orange isosceles triangle upper angle below. But there are also two isosceles triangles to the left and the right with angles (left) and (right). Three angles make radians. Thus for the remaining angle of both triangles in the middle ( and ) it means

and

Now look at the remaining angles of those two with the diagonal of the circle. That total angle also adds up to . So the remaining two triangles, which both add up to the top angle of the orange isosceles triangle are respectively

and

This means that for the two triangles that share the same circle sector of the circumscribed circle the top angles we can write:

if the top angle of the large triangle is then the top angle of the small triangle is or .

So we can write those angles as and :

In this picture the orange isosceles triangle below thus has a top angle of . You see the interior bisector of the top angle. From this it follows that both the angles to the right and the left of the bisector ar . The basis of the triangle with length is also divided in two, so each half has length .

Now we can calculate :

Calculating we get:

and thus