First, a correction, \theta 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 H= L cos(\theta/2), the altitude of the triangle, and the third side L sin(\theta/2). 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
\frac{r}{H-r}= \frac{L sin(\theta/2)}{L}
That will allow you to write the formula in terms of the variable \theta and r, which is a constant.