If we have triangle $\varDelta ABC$, with base $AC$ and height $BD$,we can calculate the angle $\angle ABC$ expressed in degrees,with the formula
$$ \left({\cot}^{-1} \left(\frac{{AD}-{AC}}{{BD}}\right)-{\cot}^{-1} \left(\frac{{AD}}{{BD}}\right)\right)\cdot \frac{180}{\pi}$$
For example if ${AC}=10$,and ${BD}=5$,and ${AD}=5$ we will get
$$ \left({\cot}^{-1} \left(\frac{{5}-{10}}{{5}}\right)-{\cot}^{-1} \left(\frac{{5}}{{5}}\right)\right)\cdot \frac{180}{\pi}=90$$
for these values is the value of the angle expressed in degrees. If we connect the values of ${AD}$ and ${BD}$ in some relation for example ${x^2}$ we will get graph with values of angles expressed in degrees, for any point ${x^2}$ with points on the coordinate system with values on the x axis $0$ and $10$
$$ \left({\cot}^{-1} \left(\frac{{x}-{10}}{{x^2}}\right)-{\cot}^{-1} \left(\frac{{x}}{{x^2}}\right)\right)\cdot \frac{180}{\pi}$$ and now the question of whether my formula works and whether this result can be reached in a simpler way
