I'll try and explain the details of my suggestion. Made two sketches in paint, they dont look very good but hopfully you'll understand what Im trying to draw.

Give the time difference between point A (0,0) and point B (10,0) it's possible to draw arc line and the position of the sender is located somewhere along this arc.

Sketch of arc

(This is probably not a very accurate drawing)

If you then find the same arc between point C (0,10) and point A (0,0) then the two arc will intersect at one point.

So I need to calculate the arc between point A and point B. If you take a look at this sketch you'll see a snapshoot of one possible solution.

From the above I made this simple formula :

y1 + delay = y2

using pythagorean theorem you can re-write it like this :

sqrt(a^2 + h^2) + delay = sqrt(b^2 + h^2)

You can also write 'b' as 'a-10' - so the you can update the above formula :

sqrt(a^2 + h^2) + delay = sqrt( (10-a)^2 + h^2)

The I remove the square root on the right hand side by raising both sides to the power of 2

( sqrt(a^2 + h^2) + delay)^2 = (10-a)^2 + h^2

After a bit of tweaking I get this forumla

sqrt(a^2 + h^2) = (100 - 20a - delay^2) / (2 * delay)

What I finally end up with is :

(400a^2 / (4 * delay^2) - a^2 - ((1000a-10*a*delay^2) / delay^2) + (( 100 - delay^2) / (2*delay))^2 - h^2)

So what I bascilly do is use a simple for loop where I set h=0 and increase it by 0.05 for each iteration, when I have both the delay and h I can insert them into the above forumla and I get an answer (actually you get two answers from a second degree equation, so thats why I have a method call isAnswerCorrect() which checks which answer is correct. Once I've found a I insert it into an array along with h. This process is done form h=0 till h=10 with baby steps of 0,05.

Not sure if this even makes sense. Please let me know if something was unclear or plain wrong.

Btw, how do you insert a picture into the post ?