First, a correction, [itex]\theta[/itex] in your formula is the vertex angle, not half the vertex angle.
Half of the isosceles triangle is a right triangle with hypotenuse L and one side [itex]H= L cos(\theta/2)[/itex], the altitude of the triangle, and the third side [itex]L sin(\theta/2)[/itex]. If you draw a line from the center of the circle to point at which the line L is tangent to the circle, that also gives a right triangle (a radius of a circle is always perpendicular to a tangent) similar to the first right triangle. The hypotenuse of this smaller right triangle is H-r and the "opposite side" has length r. That is
[tex]\frac{r}{H-r}= \frac{L sin(\theta/2)}{L}[/tex]
That will allow you to write the formula in terms of the variable [itex]\theta[/itex] and r, which is a constant.