rectangular

Hello, perash!

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

Quote:

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