10. In the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, let the left focus be $F$, the right vertex be $A$, and the upper endpoint of the minor axis be $B$. If the eccentricity of the ellipse is $\frac{\sqrt{5}-1}{2}$, then $\angle A B F=$ $\qquad$ .
$$
\begin{array}{l}
e=\frac{\sqrt{5}-1}{2} \Rightarrow(2 e+1)^{2}=5 \Rightarrow e^{2}+e-1=0 \\
\Rightarrow c^{2}+a c-a^{2}=0 . \text { In } \triangle A B F \text {, } \\
|A B|^{2}+|B F|^{2}=\left(a^{2}+b^{2}\right)+a^{2}=3 a^{2}-c^{2} \\
=a^{2}+2\left(c^{2}+a c\right)-c^{2}=a^{2}+2 a c+c^{2}=|A F|^{2} \\
\Rightarrow A B \perp B F \Rightarrow \angle A B F=90^{\circ} .
\end{array}
$$