4. Two spheres with radius $r$ are tangent to each other, and both are tangent to the two faces of a dihedral angle. A third sphere is also tangent to the two faces of the dihedral angle and is tangent to the two spheres with radius $r$. Given that the plane angle of the dihedral angle is $60^{\circ}$, and the radius of the third sphere is greater than $r$. Then the radius of the third sphere is ( ).
(A) $\frac{15-\sqrt{11}}{3} r$
(B) $\frac{8+\sqrt{7}}{6} r$
(C) $\frac{5+\sqrt{13}}{3} r$
(D) $\frac{6+\sqrt{11}}{4} r$
4.C.

Let the radius of the large sphere be $R$. Denote the centers of the two small spheres as $O_{1}$ and $O_{2}$, their tangency point as $\mathrm{A}$, the center of the large sphere as $O_{3}$, and one of the half-planes of the dihedral angle as $\alpha$. The tangency point of the large sphere with the plane $\alpha$ is $B$. Construct a plane $O_{1} O_{2} C \parallel \alpha$ through $O_{1} O_{2}$, intersecting $O_{3} B$ at point $C$. It is easy to see that the dihedral angle $O_{3}-O_{1} O_{2}-C$ is $30^{\circ}$, and $O_{3} \mathrm{~A}, O_{3} \mathrm{C}$ are both perpendicular to $O_{1} O_{2}$, so $\angle O_{3} A C=30^{\circ}$.
At the same time, $O_{3} A=\sqrt{(R+r)^{2}-r^{2}}, O_{3} C=R-r$. In the right triangle $\triangle O_{3} A C$,
$$
O_{3} C=O_{3} A \sin \angle O_{3} A C=\frac{1}{2} O_{3} A,
$$

which means $R-r=\frac{1}{2} \sqrt{(R+r)^{2}-r^{2}}$.
Solving this equation for $R$, taking the solution where $R>r$, we get
$$
R=\frac{5+\sqrt{13}}{3} r .
$$