Three. (20 points) Given a positive number $t$, let two non-zero vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in the same plane satisfy
$$
|\boldsymbol{a}+\boldsymbol{b}|=t|\boldsymbol{a}-\boldsymbol{b}| .
$$

Try to find the range of the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$.
Three, let $|\boldsymbol{a}|=k|\boldsymbol{b}|, k>0, \boldsymbol{a}$ and $\boldsymbol{b}$ have an angle $\theta$ between them. Then from $|\boldsymbol{a}+\boldsymbol{b}|=t|\boldsymbol{a}-\boldsymbol{b}|$ we get
$$
\begin{array}{l}
(\boldsymbol{a}+\boldsymbol{b}) \cdot(\boldsymbol{a}+\boldsymbol{b})=t^{2}(\boldsymbol{a}-\boldsymbol{b}) \cdot(\boldsymbol{a}-\boldsymbol{b}) \\
\Rightarrow|\boldsymbol{a}|^{2}+|\boldsymbol{b}|^{2}+2|\boldsymbol{a}||\boldsymbol{b}| \cos \theta \\
\quad=t^{2}\left(|\boldsymbol{a}|^{2}+|\boldsymbol{b}|^{2}-2|\boldsymbol{a}||\boldsymbol{b}| \cos \theta\right) \\
\Rightarrow k^{2}+1+2 k \cos \theta=t^{2}\left(k^{2}+1-2 k \cos \theta\right) \\
\Rightarrow \cos \theta=\frac{t^{2}-1}{t^{2}+1} \cdot \frac{k^{2}+1}{2 k} .
\end{array}
$$

If $t=1$, then from equation (1) we get $\cos \theta=0, \theta=90^{\circ}$. If $t>1$, then from equation (1) we get
$$
\cos \theta \geqslant \frac{t^{2}-1}{t^{2}+1} \cdot \frac{2 k}{2 k}=\frac{t^{2}-1}{t^{2}+1},
$$

with equality holding if and only if $k=1$.
Thus $0 \leqslant \theta \leqslant \arccos \frac{t^{2}-1}{t^{2}+1}$.
If $01$, then $\theta \in\left[0, \arccos \frac{t^{2}-1}{t^{2}+1}\right]$;
If $0<t<1$, then $\theta \in\left[\arccos \frac{t^{2}-1}{t^{2}+1}, \pi\right]$.