<s>[INST] Triangle related problem 
In the figure below if $tan \alpha=\frac{m}{n}$, where $m$ and $n$ are relatively prime, find the value of $m$ and $n$.


My approach is as follow
$\frac{{SinA}}{{14}} = \frac{{SinB}}{{15}} = \frac{{SinC}}{{13}}$
$\Delta APB$
$\angle PBA = B - \alpha $
$\angle APB = 180 - \left( {\alpha  + B - \alpha } \right) = 180 - B$
$\frac{{Sin\left( {180 - B} \right)}}{{13}} = \frac{{Sin\alpha }}{{BP}} \Rightarrow SinB = \frac{{13Sin\alpha }}{{BP}}$
$\Delta APC$
$\angle PAC = A - \alpha $
$\angle APC = 180 - \left( {\alpha  + A - \alpha } \right) = 180 - A$
$\frac{{Sin\left( {180 - A} \right)}}{{15}} = \frac{{Sin\alpha }}{{AP}} \Rightarrow SinA = \frac{{15Sin\alpha }}{{AP}}$
$\Delta BPC$
$\angle PCB = C - \alpha $
$\angle BPC = 180 - \left( {\alpha  + C - \alpha } \right) = 180 - C$
$\frac{{Sin\left( {180 - C} \right)}}{{13}} = \frac{{Sin\alpha }}{{PC}} \Rightarrow SinC = \frac{{13Sin\alpha }}{{PC}}$
$\frac{{SinA}}{{SinB}} = \frac{{\frac{{15Sin\alpha }}{{AP}}}}{{\frac{{13Sin\alpha }}{{BP}}}} = \frac{{14}}{{15}} \Rightarrow \frac{{{{15}^2}}}{{13 \times 14}} = \frac{{AP}}{{BP}}$
How do I approach from here?
 [/INST] P is first Brocard point of the triangle.
We have the relation $$\cot \alpha = \cot A + \cot B + \cot C$$
To quickly find these, we partition the triangle as below

so $$ \cot B = \dfrac{5}{12}, \cot C = \dfrac{3}{4} $$
and in any $\triangle$, $$ \cot A\cot B + \cot B\cot C +\cot C\cot A =1$$

Update :
Following @JeanMarie's link, we see $$ \cot \alpha = \cot A + \cot B + \cot C = \dfrac{a^2+b^2+c^2}{4\triangle} $$
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