\(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=-\dfrac{1}{4}\)
\(cosABC=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}=\dfrac{11}{16}\)
\(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}=\dfrac{7}{8}\)
\(cos\left(\widehat{ABC}+2\cdot\widehat{ACB}\right)\)
\(=cosB\cdot cos2C-sinB\cdot sin2C\)
\(=\dfrac{11}{16}\cdot\left(2\cdot cos^2C-1\right)-\dfrac{3\sqrt{15}}{16}\cdot2\cdot sinC\cdot cosC\)
\(=\dfrac{11}{16}\cdot\left[2\cdot\left(\dfrac{7}{8}\right)^2-1\right]-\dfrac{6\sqrt{15}}{16}\cdot\dfrac{7}{8}\cdot\dfrac{\sqrt{15}}{8}\)
\(=-\dfrac{1}{4}=cosA\)
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