Question 91, Vector $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ have an angle $\theta$ between them, $|\overrightarrow{\mathrm{OA}}|=2,|\overrightarrow{\mathrm{OB}}|=1, \overrightarrow{\mathrm{OP}}=\mathrm{t} \overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OQ}}=(1-\mathrm{t}) \overrightarrow{\mathrm{OB}}$, $|\overrightarrow{P Q}|=f(t)$ reaches its minimum value at $t=t_{0}$, if $0<t_{0}<\frac{1}{5}$, find the range of $\theta$.

Starting from the 91st, the solution is as follows:
$$
\begin{array}{l}
\mathrm{f}(\mathrm{t})=|\overrightarrow{\mathrm{PQ}}|=\sqrt{(\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}) \cdot(\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}})}=\sqrt{[(1-\mathrm{t}) \overline{\mathrm{OB}}-\mathrm{t} \overrightarrow{\mathrm{OA}}] \cdot[(1-\mathrm{t}) \overrightarrow{\mathrm{OB}}-\mathrm{t} \overrightarrow{\mathrm{OA}}]} \\
=\sqrt{(1-\mathrm{t})^{2}|\overline{\mathrm{OB}}|^{2}+\mathrm{t}^{2}|\overrightarrow{\mathrm{OA}}|^{2}-2 \mathrm{t}(1-\mathrm{t}) \overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}} \\
=\sqrt{(1-\mathrm{t})^{2}+4 \mathrm{t}^{2}-4 \mathrm{t}(1-\mathrm{t}) \cos \theta} \\
=\sqrt{(5+4 \cos \theta) \mathrm{t}^{2}-2(1+2 \cos \theta) \mathrm{t}+1}
\end{array}
$$

Notice that $5+4 \cos \theta>0$, so $f(t)=\sqrt{(5+4 \cos \theta) t^{2}-2(1+2 \cos \theta) \mathrm{t}+1}$, achieves its minimum value at $t=$ $\frac{1+2 \cos \theta}{5+4 \cos \theta}$. Therefore, $0<\frac{1+2 \cos \theta}{5+4 \cos \theta}<\frac{1}{5}$. Since $-1 \leq \cos \theta \leq 1$, it follows that $-\frac{1}{2}<\cos \theta<0$, so $\theta \in\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right)$.