3. Let $O$ be the circumcenter of $\triangle A B C$, $I$ be the incenter of $\triangle O B C$, and $\angle B I C=140^{\circ}$. Then $\angle A=$
3. $50^{\circ}$ or $130^{\circ}$.

From the problem, we have
$$
\begin{array}{l}
\angle B I C=90^{\circ}+\frac{1}{2} \angle B O C=140^{\circ} \\
\Rightarrow \angle B O C=100^{\circ} .
\end{array}
$$

As shown in Figure 6, if $\angle A$ is acute or right, then
$$
\angle A=\frac{1}{2} \angle B O C=50^{\circ} \text {. }
$$

As shown in Figure 7, if $\angle A$ is obtuse, then
$$
\begin{array}{l}
180^{\circ}-\angle A=\frac{1}{2} \angle B O C \\
\Rightarrow \angle A=130^{\circ} .
\end{array}
$$

In conclusion, $\angle A$ is $50^{\circ}$ or $130^{\circ}$.