Find the diametre of the well I'm stuck in solving this problem. 

Below, there is a well and $2$ sticks are thrown inside.The distance from the intersection of the rods to the bottom of the well is $1 m$ .What is the diametre of the well?

I think, this is not possible to find the diametre. Am I right?

 A: Introduce $x=AD$, $y=BC$ and $d=AC$. We have some obvious relations coming from similar triangles:
$$\frac{OK}{x}=\frac{KC}{d}\tag{1}$$
$$\frac{OK}{y}=\frac{AK}{d}\tag{2}$$
Add (1) and (2) and you get:
$$OK\left(\frac 1x+\frac 1y\right)=\frac{AK+KC}{d}=\frac dd$$
$$OK\left(\frac 1x+\frac 1y\right)=1\tag{3}$$
Using Pitagora:
$$d^2+x^2={DC}^2\tag{4}$$
$$d^2+y^2={AB}^2\tag{5}$$
Now you have a system of 3 equations ((3), (4), (5)) with three unknowns ($x$, $y$, $d$), with $d$ being the only "interesting" one. From (4) and (5) you get:
$$x=\sqrt{DC^2-d^2}$$
$$y=\sqrt{AB^2-d^2}$$ 
Replace this into (3) and you will obtain a single equation with a single unknown value:
$$\frac {1}{\sqrt{DC^2-d^2}}+\frac {1}{\sqrt{AB^2-d^2}}=\frac1{OK}\tag{6}$$
For $AB=3$, $DC=2$, $OK=1$, the equation (6) becomes:
$$\frac {1}{\sqrt{4-d^2}}+\frac {1}{\sqrt{9-d^2}}=1$$
...or if you introduce substitution $z=d^2$:
$$\frac {1}{\sqrt{4-z}}+\frac {1}{\sqrt{9-z}}=1$$
This looks like a simple equation to solve but it's not. You'll have to square it twice to get rid of the roots and you will end up with an equation of 4-th degree:
$$z^4-22 z^3+163 z^2-454 z+385=0$$
It can be solved analytically and of four solutions only one is a real, positive number that satisfies the original equation. But that number is ugly, I mean VERY UGLY and I am not going to reproduce it here in full. The approximate value is:
$$z=d^2=1.51582 \implies d=1.23119$$
I have also solved the problem in a slightly different way using trigonometry but it still leads to equation (6). 
EDIT
If you use different values (as suggested in some comments): $AB=38$, $DC=28$, $OK=18$, the equation (6) becomes:
$$\frac {1}{\sqrt{28^2-d^2}}+\frac {1}{\sqrt{38^2-d^2}}=\frac1{18}$$
...and Mathematica says that it has no real positive solutions for $d$. 
A: 
I think the relations (i) and (ii) should be sufficient to solve for $\theta$ and $\phi$ and hence $AC$.
(i) would give : $2\cos\theta = 3\cos\phi$
and (ii) : $2\sin\theta + 3\sin\phi = 6\sin\theta \sin\phi$
solving these would give the answer.
