Find cone vertex angle formed by roll-bending a rectangular sheet of paper ( $ 3 \times \sqrt 5) $ to a circular cone shape in 3D so that $P^{\prime}$ falls on P, top edge line $P^{\prime}A $ falls along $PA$ and side edge $P^{\prime} B^{\prime} $ falls on $PB$. ( Points $P,B$ are the same in space after roll). Contact of surfaces takes place on back side of paper,the bent paper cone now points up towards the number 3 shown, at some places roll rotation can be more than a turn when viewed along cone axis.
EDIT 1& 2
The conclusion is that as long as PB falls on P'B' the cone angle will be same.
If I understand the situation correctly, the cone is folded thus:
(source: nominal-animal.net)
and the shaded area below is what actually is folded into a truncated cone:
(source: nominal-animal.net)
If we extend the lines $\overline{BP}$ and $\overline{B'P'}$, they eventually intersect, with angle $\alpha$. The original diagram gives us that angle, because we have a right triangle ($(\sqrt{5})^2=5=1^2+2^2$), and left side of the sheet is parallel to $\overline{B'P'}$: $$\sin{\alpha} = \frac{1}{\sqrt{5}}\tag{1}\label{1}$$
Whenever we fold a circular sector of paper (sector angle $\alpha$, radius $s$) into a right circular cone, the length of the perimeter is $p = 2 \pi s \alpha / 360°$. For the resulting right circular cone, the radius is $r = p / (2 \pi) = s \alpha / 360°$.
If we look at a cross section of the right circular cone, we see the axis, radius $r$, and side $s$ form a right triangle, with the angle $\theta$ at the apex being half the cone vertex angle $\phi$, $\phi = 2 \theta$: $$\sin(\theta) = r/s = \frac{\alpha}{360°}$$ and therefore $$\phi = 2 \arcsin \left ( \frac{ \alpha }{ 360° } \right )\tag{2}\label{2}$$
Note that equation $\eqref{2}$ applies in general, not just in this special case. Might be useful when making party hats.
If we plug in $\eqref{1}$ into $\eqref{2}$, we get $$\phi = 2 \arcsin \left ( \frac{ \arcsin \left ( \frac{1}{\sqrt{5}} \right ) }{ 360° } \right ) \approx 8.5°$$