9. (20 points) In $\triangle A B C$, the measures of the three interior angles satisfy $\frac{\angle A}{\angle B}=\frac{\angle B}{\angle C}=\frac{1}{3}$.
Find the value of $T=\cos A+\cos B+\cos C$.
Given $\angle A=\theta, \angle B=3 \theta, \angle C=9 \theta$.
From $\theta+3 \theta+9 \theta=\pi \Rightarrow \theta=\frac{\pi}{13}$.
Then $T=\cos \theta+\cos 3 \theta+\cos 9 \theta$
$$
\begin{aligned}
= & \cos \theta+\cos 3 \theta-\cos 4 \theta \\
= & 2 \cos \theta \cdot \cos 2 \theta-2 \cos ^{2} 2 \theta+1>1, \\
T^{2}= & (\cos \theta+\cos 3 \theta+\cos 9 \theta)^{2} \\
= & \cos ^{2} \theta+\cos ^{2} 3 \theta+\cos ^{2} 9 \theta+2 \cos \theta \cdot \cos 3 \theta+ \\
& 2 \cos \theta \cdot \cos 9 \theta+2 \cos 3 \theta \cdot \cos 9 \theta \\
= & \frac{1+\cos 2 \theta}{2}+\frac{1+\cos 6 \theta}{2}+\frac{1+\cos 8 \theta}{2}+ \\
& (\cos 2 \theta+\cos 4 \theta)+(\cos 8 \theta+\cos 10 \theta)+ \\
& (\cos 6 \theta+\cos 12 \theta) .
\end{aligned}
$$

And $T=\cos \theta+\cos 3 \theta+\cos 9 \theta$
$$
=-\cos 12 \theta-\cos 10 \theta-\cos 4 \theta \text {, }
$$

Thus, $2 T^{2}-T=3+3 \sum_{k=1}^{6} \cos 2 k \theta$.
$$
\text { Let } P=\sum_{k=1}^{6} \cos 2 k \theta \text {. }
$$

Then $P \times 2 \sin \theta$
$$
\begin{array}{l}
=\sum_{k=1}^{6}[\cos (2 k+1) \theta-\cos (2 k-1) \theta] \\
=-\sin \theta .
\end{array}
$$

Therefore, $P=-\frac{1}{2}$.
Hence, $2 T^{2}-T=3-\frac{3}{2}=\frac{3}{2}$, which means
$$
4 T^{2}-2 T-3=0 \text {. }
$$

Since $T>1$, solving this equation gives $T=\frac{1+\sqrt{13}}{4}$.