Looking for a rational sine function for an acute triangle I have a triangle $(5,6,7)$, that is $a=5$, $b=6$, $c=7$ and I want to find the coordinates of the vertices on the  $xy$ plane in a manner that I can implement in a program or a spreadsheet. I let $(x_{ac},y_{ac})=(0,0)$. I let $(x_{bc},y_{bc}=(7,0$). Then, from the law of cosines, to get $(x_{ab},y_{ab})$, I let

$$x_{ab}=a*\cos⁡(\alpha)=a*\frac{a^2+c^2-b^2}{2ab}=5*\frac{25+49-36}{2*5*7}=\frac{38}{14}=2.714285714285714$$
and this appears to be the correct value according to answers to my earlier question. 

I can use this sine identity to get the value for $y_{ab}$:
$$y_{ab}=a\sin⁡(\alpha)=a\sqrt{1-\cos^2(\alpha)}=a\sqrt{1-\left(\frac{a^2+c^2-b^2}{2ab}\right)^2}$$
 but I would like to find a $rational$ solution (without the square root) like the cosine function I used for $x_{ab}$ (if one exists).
Can the law(s) of sines (or cosines) be manipulated to produce a sine value using only side lengths without the radical?
This is possible only for acute triangles of a special type.

For example, a triangle with sides $a=10,\ b=17,\ c=21$ can be constructed as follows.

Let $n = 2$ and $c\in Ox.$ Then:
$$a = 2(2^{2n-2}+1) = 10,\quad a_x = 2(2^{2n-2}-1) = 6,\quad a_y= 2^{n+1}=8,$$
$$b = 2^{2n}+1 = 17,\quad b_x = 2^{2n}-1 = 15,\quad b_y = a_y=8,\quad c = a_x+b_x = 21,$$
wherein
$$\sin(\alpha)=\dfrac {a_y}{a} = \dfrac45,\quad \cos(\alpha)= \dfrac{a_x}{a} = \dfrac35,$$
$$\quad \sin(\beta) = \dfrac{b_y}{b} = \dfrac8{17},\quad \cos(\beta)=\dfrac{b_x}{b} = \dfrac{15}{17}.$$

The reason is that for any $n\geq 2,\quad k=0\dots n-1$ there are Pythagorean triples
with the sides $2^k(2^{2n-2k}\pm1)$ and with the common branch $2^{n+1}.$

More interesting examples can be constructed using known Pythagorean triples.
Some of them are given in the table below
$$\begin{vmatrix}
a &amp; a_x &amp; a_y=b_y &amp; b &amp; b_x &amp; c= a_x+b_x\\
13 &amp; 5  &amp; 12 &amp; 15 &amp; 9 &amp; 14\\
13 &amp; 5  &amp; 12 &amp; 20 &amp; 16 &amp; 21\\
26 &amp; 10 &amp; 24 &amp; 25 &amp; 7 &amp; 17\\
25 &amp; 7 &amp; 24 &amp; 30 &amp; 18 &amp; 25\\
5 &amp; \dfrac75 &amp; \dfrac{24}5 &amp; 6 &amp; \dfrac{18}5 &amp; 5\\
\end{vmatrix}$$

$$\color{brown}{\textbf{About triangle calculation}}.$$

Let $a=5,\ b=6,\ c=7,\ t=a_x,$ then
$$a^2-a_x^2 = b^2-b_x^2,$$
$$a^2-t^2 = b^2-(c-t)^2,$$
$$a_x=t=\dfrac{a^2-b^2+c^2}{2c} = \dfrac{19}7,\quad b_x= c-t = \dfrac{30}7,$$
$$a_y=b_y=\sqrt{a^2-a_x^2}=\sqrt{b^2-b_x^2}=\dfrac{12}7\sqrt6,$$
$$A=(0,0),\ B=(a_x, a_y) = \left(\dfrac{19}7,\dfrac{12}7\sqrt6\right),\ C=(0,c)=(0,7).$$


$$\color{brown}{\textbf{About triangular pyramid calculation (earlier question)}}.$$

Let 
$$l_A=AD=9,\ l_B=BD=8,\ l_C=CD=10,\ x=D_x,\ y=D_y,\ h=D_z.$$
Then
$$l_A^2-(x^2+y^2) = l_B^2 - ((x-a_x)^2+(y-a_y)^2) = l_C^2 - ((c-x)^2+y^2)=h^2,$$
$$\begin{cases}
2a_xx+2a_yy = l_A^2 -l_B^2 + a^2\\[4pt]
2cx=l_A^2-l_C^2+c^2\\[4pt]
h=\sqrt{l_A^2-(x^2+y^2)}
\end{cases}\rightarrow
\begin{cases}
x=\dfrac{l_A^2-l_C^2+c^2}{2c} = \dfrac{15}7\\[4pt]
y = \dfrac{l_A^2 -l_B^2 + a^2 -2a_xx}{2a_y} = \dfrac{81-64+25-570/49}{\dfrac{24}7\sqrt6} = \dfrac{31}{21}\sqrt6\\[4pt]
h=\sqrt{l_A^2-x^2-\dfrac{\left(l_A^2 -l_B^2 + a^2 -2a_xx\right)^2}{4(a^2-a_x^2)}}  = \color{brown}{\sqrt{\dfrac{190}{3}}}
\end{cases}$$
(see also Wolfram Alpha).

And trigonometry is not needed.
I think I would look for the intersection of the cricles

$x^2 + y^2 = 25\\
(x-7)^2 + y^2 = 36\\
x^2 -14x + 49 + y^2 = 36\\
x^2 - 14x + y^2 = -13$

Subtract the last line from the first line.

$14x = 38\\
x = \frac {38}{14}$

And substitute into either equation

$\frac {38^2}{14^2} + y^2 = 25\\
y^2 = \frac {4900 - 1444}{196}\\
y = \frac {\sqrt {3456}}{14}\\
y = \frac {12\sqrt {6}}{7}$