4. In $\triangle A B C$, $A B=A C=b, B C=a$. If $\frac{a}{b}=\frac{\sqrt{5}-1}{2}$, then $\triangle A B C$ is called a "golden triangle". In a golden $\triangle A B C$, the measure of $\angle A$ is ( ).
(A) $36^{\circ}$
(B) $54^{\circ}$
(C) $72^{\circ}$
(D) $108^{\circ}$
4. A.

As shown in Figure 6, on $AC$, intercept $AD = BC = a$, then $CD = b - a$.
Given $\frac{a}{b} = \frac{\sqrt{5} - 1}{2}$, we have
$$
2a + b = \sqrt{5}b.
$$

Squaring both sides, we get
$$
\begin{array}{l}
4a^2 + 4ab + b^2 = 5b^2 \\
\Rightarrow a^2 + ab = b^2 \Rightarrow \frac{b - a}{a} = \frac{a}{b} \Rightarrow \frac{CD}{BC} = \frac{BC}{AC}.
\end{array}
$$

Since $\angle BCD = \angle ACB$, then $\triangle BCD \sim \triangle ACB$.
Thus, $\angle BDC = \angle ABC$.
Since $AB = AC$, we have $\angle ABC = \angle ACB$, so $\angle BDC = \angle BCD$.
Therefore, $BD = BC = AD = a$.
Hence, $\angle A = \angle ABD$.
Let $\angle A = \alpha$, then
$$
\angle ABD = \alpha, \angle BDC = \angle BCD = \angle ABC = 2\alpha.
$$

In $\triangle ABC$,
$$
\alpha + 2\alpha + 2\alpha = 180^\circ \Rightarrow \alpha = 36^\circ.
$$