1. ## rectangular

$ABCD$ rectangular , $AXY$ Equilateral Triangle
X∈BC , Y∈CD and $AB=22 , BC=14\sqrt 3$

find $AX$

2. In the triangle $\Delta ADY$ we have $\sin (60) = \frac{{AD}}{{AY}}$.
Now solve for AY.
Why does AY = AX?

3. Hello, perash!

I hope I interpreted this correctly.
Even then, I have an intricate solution.
. . Maybe someone can simplify it.

Rectangle $ABCD$, equilateral triangle $AXY.$
$X$ is on $BC,\;Y$ is on $CD.\;\;AB = 22,\;BC = 14\sqrt{3}$

Find $AX.$
Code:
            X
B * - -*- - - - - - - * C
|        *          |
|   * 60°   * a     |
|              *    |
|  *a             * |
22 |               60° * Y
| *             *   |
|           * a     |
|* 60°  *           |
|   * θ             |
A * - - - - - - - - - * D
14√3

Let $a = AX$ = side of the triangle.

Let $\theta \,=\,\angle YAD$
Then $30^o-\theta \,=\,\angle XAB$

In right triangle $YBA\!:\;\;\cos\theta \:=\:\frac{14\sqrt{3}}{a}\quad\Rightarrow\quad a \:=\:\frac{14\sqrt{3}}{\cos\theta}$ .[1]

In right triangle $XBA\!:\;\;\cos(30-\theta) \:=\:\frac{22}{a}\quad\Rightarrow\quad a \:=\:\frac{22}{\cos(30-\theta)}$ .[2]

Equate [1] and [2]: . $\frac{14\sqrt{3}}{\cos\theta} \:=\:\frac{22}{\cos(30-\theta)}\quad\Rightarrow\quad14\sqrt{3}\cos(30-\theta) \:=\:22\cos\theta$

. . $14\sqrt{3}(\cos30\cos\theta + \sin30\sin\theta) \:=\:22\cos\theta$ . $\Rightarrow\quad 14\sqrt{3}\left(\frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta\right) \:=\:22\cos\theta$

. . $21\cos\theta + 7\sqrt{3}\sin\theta \:=\:232\cos\theta \quad\Rightarrow\quad 7\sqrt{3}\sin\theta \:=\:\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta}\:=\:\frac{1}{7\sqrt{ 3}}$

Hence: . $\tan\theta \:=\:\frac{1}{7\sqrt{3}}\quad\Rightarrow\quad\thet a \:=\:\tan^{\text{-}1}\!\left(\frac{1}{7\sqrt{3}}\right) \:\approx\:4.715^o$

Substitute into [1]: . $a \:=\:\frac{14\sqrt{3}}{\cos4.715^o} \:\approx\:24.331$