# Formula for angles $\left({\cot}^{-1} \left(\frac{{AD}-{AC}}{{BD}}\right)-{\cot}^{-1} \left(\frac{{AD}}{{BD}}\right)\right)\cdot \frac{180}{\pi}$

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

• I think this elementary question should be asked on MathStackExchange. Mathoverflow is for questions connected to research. Jun 23 at 9:03
• 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. Jun 23 at 9:36