Example 2 Draw three rays $O A, O B, O C$ from point $O$, with the angles between each pair being $60^{\circ}, 90^{\circ}, 120^{\circ}$, respectively. What is the minimum value of the angles between the angle bisectors of these three angles?
See Figure $15-2$. Without loss of generality, let the angle between $O A$ and $O B$ be $60^{\circ}$, the angle between $O B$ and $O C$ be $120^{\circ}$, and the angle between $O C$ and $O A$ be $90^{\circ}$. Take points $A^{\prime}, B^{\prime}, C^{\prime}$ on $O A, O B, O C$ respectively, such that $O A^{\prime}=O B^{\prime}=O C^{\prime}=1$.

Construct the angle bisectors of $\angle A O B, \angle B O C, \angle C O A$ to intersect $A^{\prime} B^{\prime}, B^{\prime} C^{\prime}, C^{\prime} A^{\prime}$ at $D, E, F$ respectively, then $O D \perp A^{\prime} B^{\prime}, O E \perp B^{\prime} C^{\prime}, O F \perp C^{\prime} A^{\prime}$, and $D, E, F$ are the midpoints of $A^{\prime} B^{\prime}, B^{\prime} C^{\prime}, C^{\prime} A^{\prime}$ respectively.
In the equilateral $\triangle O A^{\prime} B^{\prime}$, $A^{\prime} B^{\prime}=1$
In the isosceles right $\triangle O A^{\prime} C^{\prime}$, $A^{\prime} C^{\prime}=\sqrt{2}$
In $\triangle O B^{\prime} C^{\prime}$, by the cosine rule we get
$$
\begin{array}{c}
B^{\prime} C^{\prime}=\sqrt{1^{2}+1^{2}-2 \times 1^{2} \times \cos 120^{\circ}}=\sqrt{3} \\
\text { Also, } O D=1 \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{2} \cdot O E=1 \cdot \sin 30^{\circ}=\frac{1}{2}, O F=\frac{A^{\prime} C^{\prime}}{2}=\frac{\sqrt{2}}{2}
\end{array}
$$

And $D E=\frac{A^{\prime} C^{\prime}}{2}=\frac{\sqrt{2}}{2}, E F=\frac{A^{\prime} B^{\prime}}{2}=\frac{1}{2}, D F=\frac{B^{\prime} C^{\prime}}{2}=\frac{\sqrt{3}}{2}$
Therefore, $\cos \angle D O E=\frac{O D^{2}+O E^{2}-D E^{2}}{2 O D \cdot O E}=\frac{\sqrt{3}}{3}, \cos \angle E O F=\frac{O E^{2}+O F^{2}-E F^{2}}{2 O E \cdot O F}=\frac{\sqrt{2}}{2}$
$$
\cos \angle F O D=\frac{O F^{2}+O D^{2}-D F^{2}}{2 O F \cdot O D}=\frac{\sqrt{6}}{6}
$$

Since $\frac{\sqrt{2}}{2}=\max \left\{\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{3}, \frac{\sqrt{6}}{6}\right\}$
Therefore, $\angle E O F=\min \{\angle D O E, \angle E O F, \angle F O D\}$
Therefore, the minimum angle between the angle bisectors of these three angles is $\angle E O F=45^{\circ}$.