Let A, B, and C be the centers of the smaller circles, let E be the point of tangency of circles B and C, and let D be the center of the large circle, as indicated in the figure below.

Then triangle ABC is an equilateral triangle with side lengths 2r, and triangle CED is a right triangle with length of CE equal to r.  Further, because triangle ABC is equilateral, angle Θ is half of angle ACB, or π/6 .  If R is the radius of the large circle, then clearly

Now,

so that

and therefore,

Hence,