Hello, Parton-Bill!

The term "horizontal" was misused (and highly misleading) here.

I think those three lengths are simply the sides of the triangle. .**

A triangle has sides: .$\displaystyle a= 20.988,\;\;b= 38.011,\;\;c= 25.101$

Determine the angles of the triangle. Code:

B
*
* *
25.101 * * 20.998
* *
* *
* *
A * * * * * * * C
38.011

Use the Law of Cosines . . .

$\displaystyle \cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{38.001^2 + 25.101^2-20.998^2}{2(38.011)(25.101)} \;=\;0.856281392 \quad\Rightarrow$ . . $\displaystyle A \;\approx\;31.1^o$

$\displaystyle \cos B \;=\;\frac{a^2+c^2-b^2}{2ac} \;=\;\frac{20.998^2 + 25.101^2-38.011^2}{2(20.998)(25.101)} \;=\;\text{-}0.354658159 \quad\Rightarrow$ . $\displaystyle B \;\approx\;110.8^o$

$\displaystyle \cos C \;=\;\frac{a^2+b^2-c^2}{2ab} \;=\;\frac{20.998^2 + 38.011^2 - 25.101^2}{2(20.998)(38.011)} \;=\;0.78662162 \quad\Rightarrow$ . . .$\displaystyle C \;\approx\;38.1^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If those were truly *horizontal distances*,

. . then $\displaystyle \Delta ABC$ is "projected" onto a horizontal line. Code:

B
*
*: *
* : *
* : *
* : * C
* : * :
* * :
* * : :
A * : :
: : :
- - * - - - * - - - * - -
A' B' C'

And we see that: .$\displaystyle A'B' + B'C\:\!'\:\text{ must equal }\:A'C\:\!'$

. . . which is not true of the given segments.