In triangle ABC, Find $\tan(A)$. 
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.


$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \tan(A)=\frac{4\triangle}{b^2+c^2-a^2}$
$\implies \tan(A)=\frac{4\triangle}{16\triangle-2bc}$. But the answer is in the form, $\frac{x}{y}$ where $x$ and $y$ are integers.
Any help is appreciated. Thanks in advance.

You have: $\dfrac{\sin A}{\cos A} =\tan A = \dfrac{4\triangle}{16\triangle - 2bc} = \dfrac{2bc\sin A}{8bc\sin A-2bc} = \dfrac{\sin A}{4\sin A-1} \Rightarrow \cos A = 4\sin A -1 \Rightarrow \cos^2A = \left(1-4\sin A\right)^2 = 1 - 8\sin A + 16\sin^2A \Rightarrow 1 - \sin^2A = 1 - 8\sin A + 16\sin^2A \Rightarrow \sin A\left(8 - 17\sin A\right) = 0 \Rightarrow \sin A = 0 \text{ or} \text{ }\sin A = \dfrac{8}{17}$. But $\sin A > 0$, so $\sin A = \dfrac{8}{17} \Rightarrow \cos A = 4\sin A -1  = 4\cdot \dfrac{8}{17} - 1 = \dfrac{15}{17} \Rightarrow \tan A = \dfrac{\dfrac{8}{17}}{\dfrac{15}{17}} = \dfrac{8}{15}$.
