In $\triangle A B C$, we have
$\begin{aligned}
& \frac{A B}{\sin C}=\frac{B C}{\sin A}=\frac{A C}{\sin B} \\
\Rightarrow \quad & \frac{A B}{\sin \left(180^{\circ}-4 \theta\right)}=\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta} \\
\Rightarrow \quad & \frac{A B}{\sin 4 \theta}=\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta}
\end{aligned}$
$\begin{aligned} & \text { Now, } \frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta} \\ & \Rightarrow \quad 16 \sin \theta=9 \sin 3 \theta \\ & \Rightarrow \quad 16 \sin \theta=9\left(3 \sin \theta-4 \sin ^3 \theta\right) \\ & \Rightarrow \quad 16=27-36 \sin ^2 \theta \Rightarrow 36 \sin ^2 \theta=11 \\ & \Rightarrow \quad \sin \theta=\frac{\sqrt{11}}{6} \Rightarrow \cos \theta=\frac{5}{6} \\ & \end{aligned}$
$\text { Now, } \begin{aligned}
\sin 4 \theta & =4 \sin \theta \cos \theta\left(1-2 \sin ^2 \theta\right) \\
& =4 \times \frac{\sqrt{11}}{6} \times \frac{5}{6}\left(1-2 \times \frac{11}{36}\right)=\frac{35 \sqrt{11}}{162}
\end{aligned}$
Again, we have
$\begin{aligned}
\frac{A B}{\sin 4 \theta} & =\frac{9}{\sin \theta} \\
\Rightarrow \quad A B & =\frac{9 \sin 4 \theta}{\sin \theta}=9 \times \frac{35 \sqrt{11}}{162} \times \frac{6}{\sqrt{11}}=\frac{35}{3}
\end{aligned}$