3. In $\triangle A B C$, it is known that $\angle A=120^{\circ}$. Let the vectors be
$$
\begin{aligned}
\alpha & =\frac{\overrightarrow{B A}}{|\overrightarrow{B A}| \cos A}+\frac{\overrightarrow{B C}}{|\overrightarrow{B C}| \cos C}, \\
\beta & =\frac{\overrightarrow{C A}}{|\overrightarrow{C A}| \cos A}+\frac{\overrightarrow{C B}}{|\overrightarrow{C B}| \cos B} .
\end{aligned}
$$

Then the angle between $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$ is equal to
3. $60^{\circ}$.

Notice that $\boldsymbol{\alpha} \cdot \overrightarrow{C A}=\boldsymbol{\beta} \cdot \overrightarrow{B A}=0$, i.e., $\boldsymbol{\alpha} \perp \overrightarrow{C A}, \boldsymbol{\beta} \perp \overrightarrow{B A}$.
Thus, the angle between $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$ is equal to or supplementary to $\angle A$.
$$
\begin{aligned}
\text { Also, } \boldsymbol{\alpha} \cdot \boldsymbol{\beta}= & \frac{\overrightarrow{B A} \cdot \overrightarrow{C B}}{|\overrightarrow{B A}| \cdot|\overrightarrow{C B}| \cos A \cdot \cos B}
\end{aligned}+
$$

Clearly, $\cos B \cdot \cos C>-\cos A>0$.
Then $\boldsymbol{\alpha} \boldsymbol{\beta}>0$.
Therefore, the angle between $\alpha$ and $\beta$ is equal to $60^{\circ}$.