Measuring diaognals without Sine Law

Question

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.

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 $q = 7.479$

Answer 1

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.

Answer 2

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

Answer 3

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.