Knowing the hypotenuse and the direction of the adjacent, how would I get the length of the adjacent

Question

I have a good understanding of the basic SohCahToa trig functions, but this kinda stumped me, since I don't have two parts of the information that is needed, here is a example image: https://i.imgur.com/ljztgBm.png

I have though of several methods, such as using an angle function between H and X, and a few others, but all seem too hacky, if anyone has a better idea, I'd really appreciate it :-) least number of steps the better

If I had a guess, the answer would involve X and H being used to find the length of Y

known variables: A,X,H,P1-4, A is always vertical, the pink is always a rectangle

desired information: the length of Y, which is the width of the pink rectangle

Answer 1

Let the left top corner of the rectangle be P5. Call a=|P5-P3| (distance between P5 and P3) and b = |P5-P1|. Then

$$H^2 = b^2 + (X+a)^2$$

$$A^2 = b^2 + a^2 $$

Solve the second one for b, i.e $b^2 = A^2 - a^2$. Then put this into the first equation:

$$H^2 = A^2 -a^2 + X^2 + 2aX + a^2$$

Then

$$a = \frac{H^2 - A^2 - X^2}{2X}$$

Then $Y = X + a = X + \frac{H^2 - A^2 - X^2}{2X} = \frac{X^2 + H^2 - A^2 }{2X}$

Answer 2

In the triangle $P_1P_2P_4$, we have $$\cos(\widehat{P_2P_1P_4})=\frac{x^2+H^2-A^2}{2xH}$$

In the pink right triangle, $$\cos(\widehat{P_2P_1P_4})=\frac{Y}{H}$$

Can you deduce $Y$?