How can I determine the sides in a squished hexagon?

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How can I determine the sides, and the angle with horizontal, in a squished hexagon ?

If I start with a regular hexagon
(coordinates (75,0); (25,0); (0, 50*sin60);(25,100*sin60); (75,100*sin60); (100,50*sin60)
to fit in a smallest rectangle size (w0=100, h0=100*sin60)) - see first picture,
then I resize it to a box of new size (w,h)

How can I determine the new sides of the hexagon, and the angle theta, based on the new sizes w and h ?

I started trying some type of equation, based on right triangle trigonometry, but it got very complicated - even with my initial assumption that the oblique sides will be equal to the horizontal ones, which I see in my picture that they don't seem to be equal...

Is the problem possible, and do I have enough information ?

I really need the size of the oblique sides and/or the angle they make with the horizontal (knowing one would give me the other).

I hope someone can give me a solution...

I tried to figure it out but it seems incorrect.

enter image description here

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I am starting to figure it out. Understanding what the size changes mean, it is easy to find the solution.

The horizontal size change (width change) will not affect height proportions, so of course y = h/2.

Since the vertical size change (change in height) does not affect horizontal proportions, and the side of the initial hexagon is half the total width, the side x in image below may also be w/2 (not sure about that though - will width change affect horizontal proportions ?).

If that is true, from there I can easily find the angle and the oblique side.

enter image description here