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?
Best Answer
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 $$\begin{vmatrix} a & a_x & a_y=b_y & b & b_x & c= a_x+b_x\\ 13 & 5 & 12 & 15 & 9 & 14\\ 13 & 5 & 12 & 20 & 16 & 21\\ 26 & 10 & 24 & 25 & 7 & 17\\ 25 & 7 & 24 & 30 & 18 & 25\\ 5 & \dfrac75 & \dfrac{24}5 & 6 & \dfrac{18}5 & 5\\ \end{vmatrix}$$
$$\color{brown}{\textbf{About triangle calculation}}.$$
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.