Question

Hallar la tan (teta)

Answer 1

P es un punto de Brocard que cumple la propiedad siguiente

                $\cot \theta = \cot A+\cot B+ \cot C$

Hallemos cada ángulo

1) 
$\cos A = \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{26^2+3^2-25^2}{2(26)(3)}=\dfrac{5}{13}\\ \\ \\ \to \sin A =\dfrac{12}{13}\to \boxed{\cot A=\dfrac{5}{12}}$

2) 
$\cos B = \dfrac{a^2+c^2-b^2}{2ac}=\dfrac{25^2+3^2-26^2}{2(25)(3)}=-\dfrac{7}{25}\\ \\ \\ \to \sin B =\dfrac{24}{25}\to \boxed{\cot B=-\dfrac{7}{5}}$

3) 
$\cos C = \dfrac{a^2+b^2-c^2}{2ab}=\dfrac{25^2+26^2-3^2}{2(25)(26)}=\dfrac{323}{325}\\ \\ \\ \to \sin C =\dfrac{36}{325}\to \boxed{\cot C=\dfrac{323}{36}}$

por ende

                       $\cot \theta =\dfrac{5}{12}-\dfrac{7}{5}+\dfrac{323}{36}\\ \\ \\ \boxed{\cot \theta =\dfrac{719}{90}}$