Lets start off with a simple right angled triangle 'abc'. (ie: use cartesian coordinates, we mark 'a' and 'b' on x,y axis, 'c' is calculated from Pythagoras therom). Now pick an arbitrary point 'o' to the right of 'c'. We then measure lengths e and f.

Question: Can 'q' be calculated without using the Sine or Cosine Rule, using only lengths 'a' through f?

UPDATE: I found another solution (no Sine law, only squares & roots) after posting part 2 (links in comments). Also added some numbers to make it easier to verify the answers. enter image description here

Here we replaced $a = \sqrt{33}, b = 4, c = 7, f = 5, e = \sqrt{32}$ and $B(x,y)$ will touch $Origin(0,0)$ when rotated. Using Heron's area formula, we can deduce $o = A(0,4)$. Similarly, we can deduce $B(1,71,-3.28)$. From there, a pythagoras formula to obtain $q = 7.479$

  • $\begingroup$ Big picture: Triangulation for a kid. We cant see O(x,y) from origin (0,0), but we can walk along the X and Y axis until we see O, then measure (estimate) distance. From there, calculate 'q' (wondering if I skip sine law would simplify calculations). $\endgroup$ – Alvin K. Jun 14 '14 at 5:56
  • $\begingroup$ Is there a reason not to use law of cosines? The solution is very easy with it. $\endgroup$ – Jean-Claude Arbaut Jun 14 '14 at 19:00
  • $\begingroup$ math.stackexchange.com/questions/834327/… is a similar construct. Notice top figure has simple solution but lower figure involves quadratic formula. @jean: Am trying to find a simple solution which doesn't involve Trig functions. Meaning any kids can take a normal calculator and have some fun with triangulations. $\endgroup$ – Alvin K. Jun 14 '14 at 21:46

We can use coordinate geometry. Let the origin be at the right angle, and let the axes be drawn in the natural way. Then the other vertices of the triangle have coordinates $(b,0)$ and $(0,a)$. Let $O$ have coordinates $(x,y)$.

We have $(x-b)^2+y^2=e^2$ and $x^2+(y-a)^2=f^2$. Solve.

  • 1
    $\begingroup$ Note that this system has two solutions. You should pick the right one. Here are the solutions: i.imgur.com/1xOCqGN.png $\endgroup$ – Martijn Courteaux Jun 14 '14 at 9:33
  • $\begingroup$ @MartijnCourteaux. Are you sure that there is not a typo (joke for sure) ? Cheers :) $\endgroup$ – Claude Leibovici Jun 14 '14 at 15:38
  • $\begingroup$ @ClaudeLeibovici: Oh god! How did that happen? I copy pasted his expressions... $\endgroup$ – Martijn Courteaux Jun 14 '14 at 21:13
  • $\begingroup$ Note that this system has two solutions. You should pick the right one. Here are the FIXED solutions: i.imgur.com/1b1oHN4.png $\endgroup$ – Martijn Courteaux Jun 14 '14 at 21:16
  • $\begingroup$ Oooh, he edited his post :) $\endgroup$ – Martijn Courteaux Jun 15 '14 at 8:43

Let us consider points $A$ of coordinates $(0,a)$, $B$ of coordinates $(b,0)$ and $O$ of coordinates $(x,y)$ we do not know. So the known distances write $$f^2=x^2+(y-a)^2$$ $$e^2=(x-b)^2+y^2$$ so two equations for two unknowns $x$ and $y$.

Assume that we have the solution. Then $q^2=x^2+y^2$

For the solution of the equation, compute $f^2-e^2$; this will give you a linear relation between $x$ and $y$. Say, express $y$ as a function of $x,a,b,f^2,e^2$ and replace it in the second equation. Develop to get a quadratic equation in $x$; solve it; compute $y$ and you are done.


If I am understanding this correctly, we are given "lengths $a$ through $f$" such that the lengths in your diagram $a$, $b$, $c$, $e$, and $f$ are known. In that case, the length $q$ in your diagram can be calculated by $\sqrt{a^2 + f^2}$. I am relatively new to posting here and am not sure how to get Tex commands to render so my apologies for the below-par formatting, but I hope I could help.

Regards, A

  • $\begingroup$ Updated pic, the angle between 'a' and 'f' is not right angle, sorry for confusion. $\endgroup$ – Alvin K. Jun 14 '14 at 5:47

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