4. Given that the cosine value of the dihedral angle formed by the side face and the base of a regular tetrahedron is $\frac{1}{6}$, then the ratio of the height $h$ to the radius $r$ of the inscribed sphere of this tetrahedron is
A. 5
B. 6
C. 7
D. 8
Answer
As shown in the figure, let the side length of the base $\triangle ABC$ be $2\sqrt{3}$, then $AE=3 \Rightarrow DE=1, PE=6 \Rightarrow PD=\sqrt{35}$.
Let the center of the inscribed sphere be $O, \triangle POF \sim \triangle PED$ $\Rightarrow \frac{PO}{PE}=\frac{OF}{ED} \Rightarrow \frac{\sqrt{35}-r}{6}=r \Rightarrow 7r=\sqrt{35}$ $\Rightarrow r=\frac{\sqrt{35}}{7}=\frac{1}{7}h$, so the answer is $C$.