Looking for a rational sine function for an acute triangle

Question

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?

Answer 1

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.

Answer 2

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

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

Answer 3

Call a triangle rational if the sines and cosines of all three of its angles are rational.

What follows shows that you build rational triangles by gluing together rational right triangles (scaled Pythagorean triples) along a common edge. For example, $13,14,15$ and $9,10,17$:

Lemma. Rationality is scale invariant. The sides of rational triangle are commensurate. A rational triangle is always similar to a rational triangle with integer sides;

Proof: Scale invariance is obvious. For the second assertion use the law of sines. For the third, find the least common multiple of the denominators.

Lemma. A triangle is rational if two of its three angles have rational sine and cosine.

Proof: Use the fact that the sum of the angles is $\pi$ and the addition formulas for sine and cosine to calculate the sine and cosine of the third angle.

Corollary. A right triangle is rational if and only if it's similar to a Pythagorean triangle (a right triangle with integer sides).

Theorem. A triangle is rational if and only if each of the triangles formed by dropping an altitude from a vertex is a Pythagorean triangle.

Proof. Put together the pieces. Note that you can use any one of the three altitudes.

So there are lots of rational triangles, with many shapes, obtuse as well as acute, scalable if you don't require integer side lengths. I am pretty sure that rational triangles are dense in the space of all triangles, with just about any measure of the distance between triangles.

Note. It was fun to take up this question after more than $60$ years: Area of an acute triangle, given two sides and an altitude