Xét ΔABC có AD là phân giác
nên \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)
=>\(cos\left(\dfrac{BAC}{2}\right)\cdot\dfrac{2\cdot4\cdot12}{4+12}=3\)
=>\(cos\left(\dfrac{BAC}{2}\right)=3:\dfrac{8\cdot12}{16}=3:\dfrac{96}{16}=3:6=\dfrac{1}{2}\)
=>\(\dfrac{\widehat{BAC}}{2}=60^0\)
=.\(\widehat{BAC}=120^0\)