28. [16] Let $A B C$ be a triangle, and let points $P$ and $Q$ lie on $B C$ such that $P$ is closer to $B$ than $Q$ is. Suppose that the radii of the incircles of triangles $A B P, A P Q$, and $A Q C$ are all equal to 1 , and that the radii of the corresponding excircles opposite $A$ are 3,6 , and 5 , respectively. If the radius of the incircle of triangle $A B C$ is $\frac{3}{2}$, find the radius of the excircle of triangle $A B C$ opposite $A$.
Answer: 135
Let $t$ denote the radius of the excircle of triangle $\triangle A B C$.
Lemma: Let $\triangle A B C$ be a triangle, and let $r$ and $r_{A}$ be the inradius and exradius opposite $A$. Then
$$
\frac{r}{r_{A}}=\tan \frac{B}{2} \tan \frac{C}{2} .
$$

Proof. Let $I$ and $J$ denote the incenter and the excenter with respect to $A$. Let $D$ and $E$ be the foot of the perpendicular from $I$ and $J$ to $B C$, respectively. Then
$$
\begin{aligned}
r=I D=B I \sin \frac{B}{2} \\
r_{A}=J E=B J \sin \frac{180^{\circ}-B}{2}=B J \cos \frac{B}{2} \\
B I=B J \tan \angle A J B=B Y \tan \frac{C}{2} .
\end{aligned}
$$

The last equation followed from
$$
\angle A J B=180^{\circ}-\angle A B J-\angle J A B=\frac{180^{\circ}-B}{2}-\frac{A}{2}=\frac{C}{2} .
$$

Hence
$$
\frac{r}{r_{A}}=\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} \cdot \frac{B I}{B J}=\tan \frac{B}{2} \cdot \tan \frac{C}{2} .
$$

Noting $\tan \frac{\angle A P B}{2} \tan \frac{\angle A P Q}{2}=\tan \frac{\angle A Q P}{2} \tan \frac{\angle A Q C}{2}=1$ and applying the lemma to $\triangle A B C, \triangle A B P$, $\triangle A P Q$, and $\triangle A Q C$ give
$$
\begin{aligned}
\frac{3 / 2}{t} & =\tan \frac{\angle A B C}{2} \cdot \tan \frac{\angle A C B}{2} \\
& =\left(\tan \frac{\angle A B C}{2} \cdot \tan \frac{\angle A P B}{2}\right) \cdot\left(\tan \frac{\angle A P Q}{2} \cdot \tan \frac{\angle A Q P}{2}\right) \cdot\left(\tan \frac{\angle A Q C}{2} \cdot \tan \frac{\angle A C B}{2}\right) \\
& =\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{1}{5}
\end{aligned}
$$

Therefore, $t=135$.