3. As shown in Figure 2, from a point $A$ outside the circle $\odot O$, draw two tangents $A B$ and $A C$, and connect $B C$. Then the relationship between the perimeter $p$ of $\triangle A B C$ and the length $l$ of the minor arc $\overparen{B C}$ is ( ).
(A) $p>2 l$
(B) $p=2 l$
(C) $p<2 l$
(D) Cannot be determined, depends on the position of $A$
3. (A).

Let the radius of $\odot \bigcirc$ be $R$, connect $O A$ intersecting $B C$ at $D$, and intersecting the obvious area relationship of the arc (Fig. 14)
$S_{\triangle A O C}>S_{\text {segment } DE}$,

we get $\frac{1}{2} R^{2} \operatorname{tg} \theta>\frac{1}{2} R^{2} \theta$,
thus $\theta<\operatorname{tg} \theta$.
Considering $\theta$ as $\frac{\theta}{2}$, similarly we get
$$
\begin{array}{l}
\frac{\theta}{2}<\operatorname{tg} \frac{\theta}{2}=\frac{\sin \theta}{1+\cos \theta} \\
=\frac{\sin \theta(1+\cos \theta)}{(1+\cos \theta)^{2}} \\
<\frac{\sin \theta(1+\cos \theta)}{4 \cos \theta}=\frac{1}{4}(\operatorname{tg} \theta+\sin \theta) .
\end{array}
$$

Thus, $2 \theta<\operatorname{tg} \theta+\sin \theta$.
$$
\begin{aligned}
\therefore 2 l & =4 R \theta<2(R \operatorname{tg} \theta+R \sin \theta)=2(A C+C D) \\
& =A C+A B+B C=p .
\end{aligned}
$$