10. In the ellipse $\frac{x^{2}}{a^{2}} \div \frac{y^{2}}{b^{2}}=1(a>b>0)$, 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 $\angle A B F=$ $\qquad$ .
$10.90^{\circ}$.
Let $c$ be the semi-focal distance of the ellipse, then from $\frac{c}{a}=\frac{\sqrt{5}-1}{2}$ we get
$$
\begin{array}{l}
c^{2}+a c-a^{2}=0 . \\
\text { Also }|A B|^{2}=a^{2}+b^{2},|B F|^{2}=a^{2}, \\
\text { so }|A B|^{2}+|B F|^{2}=2 a^{2}+b^{2}=3 a^{2}-c^{2} . \\
\text { And }|A F|^{2}=(a+c)^{2}=a^{2}+2 a c+c^{2} \\
\quad=3 a^{2}-c^{2}=|A B|^{2}+|B F|^{2},
\end{array}
$$

Therefore, $\angle A B F=90^{\circ}$.