Show there are 1977 non-similar triangles such that$\frac{\sin X+\sin Y+\sin Z}{\cos X+\cos Y+\cos Z}=\frac{12}7$and$\sin X\sin Y\sin Z=\frac{12}{25}$ 
Show that there are 1977 non-similar triangles whose angles $X$, $Y$, $Z$ satisfy the conditions
$$\begin{align}
\frac{\sin{X}+ \sin{Y}+ \sin{Z}}{\cos{X}+\cos{Y}+ \cos{Z}}=\frac{12}{7} \tag1\\[4pt]
\sin{X}\sin{Y} \sin{Z}=\frac{12}{25} \tag2
\end{align}$$

My attempt: $$\begin{align} \sin{X}+ \sin{Y}+ \sin{Z}=4\cos{\frac{X}{2}} \cos{\frac{Y}{2}} \cos{\frac{Z}{2}} \\ \cos{X}+ \cos{Y}+ \cos{Z}=1+4\sin{\frac{X}{2}} \sin{\frac{Y}{2}} \sin{\frac{Z}{2}}\\ X+Y+Z=\pi \\ \sin{X} \sin{Y}\sin{Z}=8\cos{\frac{X}{2}} \cos{\frac{Y}{2}} \cos{\frac{Z}{2}}\sin{\frac{X}{2}} \sin{\frac{Y}{2}} \sin{\frac{Z}{2}}=\frac{12}{25}\end{align}$$ Here after what to do to find the required results.
How to show such points which satisfy these conditions?
Please, help. Thanks in advance.

Using standard notation,
given $\triangle ABC$ with angles $\alpha,\beta,\gamma$,
side lengths $a,b,c$, semiperimeter $\rho$,
radius  $r$ of the inscribed circle,
and radius  $R$ of the circumscribed circle,
\begin{align}
\frac{\sin\alpha+ \sin\beta+ \sin\gamma}
{\cos\alpha+ \cos\beta+ \cos\gamma}
&=\frac{12}{7}
\tag{1}\label{1}
,\\
\sin\alpha\sin\beta\sin\gamma
&=\frac{12}{25}
\tag{2}\label{2}
.
\end{align}
Using known identities,
\begin{align}
\sin\alpha+ \sin\beta+ \sin\gamma
&=
\frac\rho R=u
\tag{3}\label{3}
,\\
\cos\alpha+ \cos\beta+ \cos\gamma
&=
\frac rR+1=v+1
\tag{4}\label{4}
,\\
\sin\alpha\sin\beta\sin\gamma
&=
\frac{\rho r}{2R^2}=\tfrac12\,uv
\tag{5}\label{5}
,
\end{align}
we ca rewrite \eqref{1}-\eqref{2}
in terms of parameters $u=\rho/R,\,v=r/R$ as
\begin{align}
\frac u{v+1}&=\frac{12}{7}
\tag{6}\label{6}
,\\
\tfrac12\,uv
&=\frac{12}{25}
\tag{7}\label{7}
.
\end{align}
The system \eqref{6}-\eqref{7}
has just two solutions,
\begin{align} 
u &= -\frac{24}{35},\quad v = -\frac75
\tag{8}\label{8}
,\\
u &= \frac{12}5,\quad v =\frac25
\tag{9}\label{9}
,
\end{align}
and obviously, only positive is valid,
so, there is only one type of triangle with given properties.
Solution of the cubic equation
\begin{align} 
x^3-2u\,x^2+(u^2+v^2+4v)\,x-4uv
&=0
\tag{10}\label{10}
,\\
x^3-\frac{24}5\,x^2+\frac{188}{25}\,x-\frac{96}{25}
&=0
\tag{11}\label{11}
\end{align}
gives a unique triplet of side lengths of triangle
with $R=1$,
which satisfies \eqref{1} and \eqref{2}
\begin{align} 
a&=\frac65,\quad b=\frac85,\quad c=2
\tag{12}\label{12}
.
\end{align}
As we can see, this triangle is similar to the famous $3-4-5$
right-angled triangle.
Indeed, we have
\begin{align} 
\sin\alpha&=\frac35,\quad\sin\beta=\frac45,\quad\sin\gamma=1
\tag{13}\label{13}
,\\
\cos\alpha&=\frac45,\quad\cos\beta=\frac35,\quad\cos\gamma=0
\tag{14}\label{14}
,
\end{align}
\begin{align}
\sin\alpha+\sin\beta+\sin\gamma
&=
\frac{12}5
\tag{15}\label{15}
,\\
\cos\alpha+ \cos\beta+ \cos\gamma
&=
\frac75
\tag{16}\label{16}
,\\
\frac{\sin\alpha+ \sin\beta+ \sin\gamma}
{\cos\alpha+ \cos\beta+ \cos\gamma}
&=\frac{12}{7}
\tag{17}\label{17}
,\\
\sin\alpha\sin\beta\sin\gamma
&=
\frac{12}{25}
\tag{18}\label{18}
.
\end{align}
