Đặt \(BC=x\Rightarrow p=\dfrac{x+13}{2}\)
Áp dụng định lý hàm cos:
\(x^2=AB^2+AC^2-2AB.AC.cos60^0=AB^2+AC^2-AB.AC\)\(\Rightarrow x^2=\left(AB+AC\right)^2-3AB.AC=169-3AB.AC\)
\(\Rightarrow AB.AC=\dfrac{169-x^2}{3}\)
\(S_{ABC}=\dfrac{1}{2}AB.AC.sinA=\dfrac{1}{2}\left(\dfrac{169-x^2}{3}\right).\dfrac{\sqrt{3}}{2}\)
\(S=pr\Rightarrow\dfrac{1}{2}\left(\dfrac{169-x^2}{3}\right).\dfrac{\sqrt{3}}{2}=\left(\dfrac{x+13}{2}\right).\sqrt{3}\)
\(\Leftrightarrow169-x^2=6\left(x+13\right)\Rightarrow x^2+6x-91=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-13\left(loại\right)\end{matrix}\right.\)
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