<s>[INST] Finding the length of the side of the equilateral triangle 
Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.
 [/INST] 
Let $|XY|=|YZ|=|ZX|=a$,
$\angle EYZ=\theta$, $|FY|=1$.
Then $\angle XYF=120^\circ-\theta$,
\begin{align}
\triangle EYZ:\quad
a\cos\theta&=2
\tag{1}\label{1}
,\\
\triangle FXY:\quad
a\cos(120^\circ-\theta)&=1
\tag{2}\label{2}
,
\end{align} 
\begin{align} 
a\cos(120^\circ-\theta)&=
\frac {a\sqrt3}2\,\sin\theta
-\frac a2\cos\theta
\\
&=
\frac {a\sqrt3}2\,\sin\theta
-1
,
\end{align} 
so the system \eqref{1},\eqref{2}
changes to
\begin{align}
a\cos\theta&=2
\tag{3}\label{3}
,\\
a\,\sin\theta
&=\frac{4\sqrt3}3
\tag{4}\label{4}
,
\end{align} 
which can be easily solved for $a$:
\begin{align}
a^2\cos^2\theta+
a^2\sin^2\theta
&=2^2+\left(\frac{4\sqrt3}3\right)^2
,\\
a^2&=\frac {28}3
,\\
a&=\frac 23\,\sqrt{21}
.
\end{align}
</s>