Đáp án:
\[b = 5 \Rightarrow c = 8 \Rightarrow a = 7\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{a^2} = {b^2} + {c^2} - 2bc.\cos A = {b^2} + {c^2} - bc = {\left( {b + c} \right)^2} - 3bc = 169 - 3bc\\
{S_{ABC}} = \frac{1}{2}.b.c.\sin A = \frac{1}{2}bc.\sin 60^\circ = \frac{{\sqrt 3 }}{4}bc\\
{S_{ABC}} = \frac{{a + b + c}}{2}.r = \frac{{a + 13}}{2}.\sqrt 3 = \frac{{\sqrt 3 }}{2}.\left( {a + 13} \right)\\
\Rightarrow \frac{{\sqrt 3 }}{4}.bc = \frac{{\sqrt 3 }}{2}.\left( {a + 13} \right)\\
\Leftrightarrow \frac{{bc}}{2} = a + 13\\
\Leftrightarrow bc - 26 = 2a\\
\Leftrightarrow {b^2}{c^2} - 52bc + {26^2} = 4{a^2}\\
\Leftrightarrow {b^2}{c^2} - 52bc + {26^2} = 4.\left( {169 - 3bc} \right)\\
\Leftrightarrow bc = 40\\
\Leftrightarrow b.\left( {13 - b} \right) = 40\\
\Leftrightarrow {b^2} - 13b + 40 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
b = 8 \Rightarrow c = 5\left( L \right)\\
b = 5 \Rightarrow c = 8 \Rightarrow a = 7
\end{array} \right.
\end{array}\)
