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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

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    $\begingroup$ I think this elementary question should be asked on MathStackExchange. Mathoverflow is for questions connected to research. $\endgroup$ Jun 23 at 9:03
  • $\begingroup$ But before sending it to stackexchange, read up on how to ask a good question on that site. In its current form, it is quite unclear. $\endgroup$ Jun 23 at 9:36

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