What's the most elementary way to solve this trigonometric problem? 
A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.
This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.
Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).
Thanks in advance.

We know the side $c=2$ and the altitude $h_c=1$, hence the area of the triangle is $S=\frac12ch_c=1$.
We also have Heron's formula for the area:
$$S=\sqrt{p(p-a)(p-b)(p-c)}$$
With $p=\dfrac{a+b+c}2=3$.
Hence
$$(p-a)(p-b)=\frac{S^2}{p(p-c)}=\frac13$$
$$p^2-(a+b)p+ab=p(p-a-b)+ab=\frac13$$
$$ab=\frac13+3=\frac{10}{3}$$
Now, we know $a+b$ and $ab$, hence:
$$(a-b)^2=(a+b)^2-4ab=16-\frac{40}{3}=\frac83$$
And since the larger side is $b$,
$$a-b=-2\sqrt{\frac23}$$
$$a=\frac12(a+b+a-b)=\frac12(4-2\sqrt{\frac23})=2-\sqrt{\frac23}$$
Now, with Pythagoras' theorem in the triangle BDC:
$$y=\sqrt{a^2-1}=\frac13\sqrt{33-12\sqrt6}=\frac13\sqrt{24-2\times3\times2\sqrt6+9}=\frac{2\sqrt6-3}3=\frac{2\sqrt6}3-1$$
