Geometry: triangulation for measuring a lot/yard

[Thomas Klemm]

Using \(SW=(0, 0)\), \(SE=(u, 0)\) and \(NE = (x, y)\) we conclude:
\[ \begin{eqnarray}
v^2&=&|NE-SW|^2&=&(x-0)^2+(y-0)^2&=&x^2+y^2 \\
w^2&=&|NE-SE|^2&=&(x-u)^2+(y-0)^2&=&x^2-2ux+u^2+y^2
\end{eqnarray} \]
Taking the difference:
\[ \begin{eqnarray}
v^2-w^2=2ux-u^2 \\
2ux=u^2+v^2-w^2
\end{eqnarray} \]
This leads to:
\[ \begin{eqnarray}
x&=&\frac{u^2+v^2-w^2}{2u} \\
y&=&\sqrt{v^2-x^2}
\end{eqnarray} \]

Example:
u = 8
v = 11
w = 7

\[ \begin{eqnarray}
x&=&\frac{64+121-49}{16}=\frac{136}{16}=8.5 \\
y&=&\sqrt{121-72.25}=\sqrt{48.75}\approx6.9821
\end{eqnarray} \]

We can use the same equations for \(NW=(x,y)\) but use instead:
u = 8
v = 9
w = 12

\[ \begin{eqnarray}
x&=&\frac{64+81-144}{16}=\frac{1}{16}=0.0625 \\
y&=&\sqrt{81-0.00390625}=\sqrt{80.99609375}\approx8.9998
\end{eqnarray} \]

(07-21-2018 03:59 PM)roadrunner Wrote:  I used the brute force method

As we can see this isn't always needed. A little bit of algebra and maybe a four-banger to calculate the square roots is often enough.

Kind regards
Thomas