3. The three sides of one triangle are $a$, $a$, and $b$, and the three sides of another triangle are $a$, $b$, and $b$, where $a > b$. If the smallest interior angles of the two triangles are equal, then $\frac{a}{b}$ equals ( ).
(A) $\frac{\sqrt{3}+1}{2}$
(B) $\frac{\sqrt{5}+1}{2}$
(C) $\frac{\sqrt{3}+2}{2}$
(D) $\frac{\sqrt{5}+2}{2}$
3.B.

As shown in Figure 7, in $\triangle ABC$, $AB = AC = a, BC = b$. $D$ is a point on $AB$ such that $AD = b$. Since $a > b$, $\angle A$ is the smallest angle in $\triangle ABC$, denoted as $\angle A = \theta$. Therefore, the smallest angle of the triangle with sides $b, b, a$ is also $\theta$. Thus, this triangle is congruent to $\triangle ADC$. So,
$$
DC = b, \angle ACD = \theta.
$$

Furthermore, $\triangle ABC \sim \triangle CBD$, hence,
$$
\frac{BC}{AB} = \frac{BD}{BC}, \text{ i.e., } \frac{b}{a} = \frac{a - b}{b}.
$$

Let $x = \frac{a}{b}$, we get the equation $x^2 - x - 1 = 0$.
Solving for $x$, we get $x = \frac{a}{b} = \frac{\sqrt{5} + 1}{2}$.