5. In a right-angled triangle, the sines of the three interior angles form a geometric progression. Then the smallest angle of the triangle is equal to ( ).
(A) $\arcsin \frac{\sqrt{5}-1}{2}$
(B) $\arccos \frac{\sqrt{5}-1}{2}$
(C) $\arcsin \frac{\sqrt{5}+1}{4}$
(D) $\arccos \frac{\sqrt{5}+1}{4}$
5.A.

Let the three sides be $1$, $x$, and $x^{2}$ $(0<x<1)$. Then the sine of the smallest angle of the triangle is $x^{2}$.
From $1^{2}=x^{2}+\left(x^{2}\right)^{2}$, we solve to get $x^{2}=\frac{\sqrt{5}-1}{2}$.