Giải thích các bước giải:
\(\begin{array}{l}
\sin x = \sin \left( {180^\circ - x} \right)\\
\cos x = - \cos \left( {180^\circ - x} \right)\\
\cot \left( {B + C} \right) = \frac{1}{5}\\
\Leftrightarrow \frac{{\cos \left( {B + C} \right)}}{{\sin \left( {B + C} \right)}} = \frac{1}{5}\\
\Leftrightarrow \frac{{ - \cos \left( {180^\circ - B - C} \right)}}{{\sin \left( {180 - B - C} \right)}} = \frac{1}{5}\\
\Leftrightarrow \frac{{\cos A}}{{\sin A}} = - \frac{1}{5}\\
\Leftrightarrow \sin A = - 5\cos A\\
0 < \widehat A < 180^\circ \Rightarrow \sin A > 0 \Rightarrow \cos A < 0\\
{\sin ^2}A + {\cos ^2}A = 1\\
\Leftrightarrow {\left( { - 5\cos A} \right)^2} + {\cos ^2}A = 1\\
\Leftrightarrow {\cos ^2}A = \frac{1}{{26}}\\
\Leftrightarrow \cos A = \frac{{ - \sqrt {26} }}{{26}}\,\,\,\,\,\left( {{\mathop{\rm cosA}\nolimits} < 0} \right)\\
B{C^2} = A{B^2} + A{C^2} - 2\cos A.AB.AC\\
= {5^2} + {7^2} - 2.\frac{{ - \sqrt {26} }}{{26}}.5.7\\
\Rightarrow BC = .......
\end{array}\)
