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