1.在三角形ABC中,D为边BC上一点,BD=1/2DC,角ADB=120度,AD=2,若三角形ADC的面积为3-√￣3,则角ABC=?

∵∠ADB=120°∴∠ADC=60°∵AD=2,S△ADC=1/2·AD·DC·sin∠ADC=3-√3∴DC=2S△ADC/AD·sin∠ADC=2√3-2∵BD=DC/2∴BD=√3-1由余弦定理AB=√(AD²+BD²-2AD·BD·cos∠ADB)=√6由正弦定理sin∠ADB/AB=sin∠...角BAC等于多少？求解题过程！(上面回答不要“由正弦定理”及以下部分)由余弦定理AC=√(AD^2+CD^2-2AD·CD·cos∠ADC)=3√2-√6∵S△ADC=3-√3，BD=DC/2∴S△ABC=3/2·S△ADC=(9-3√3)/2∵S△ABC=1/2·AB·AC·sin∠BAC∴sin∠BAC=2S△ABC/AB·AC=√3/2∴∠BAC=60°或者求得DC=2√3-2，BC=3√3-3，AC=3√2-√6 后∵AC^2=(3√2-√6)^2=6(√3-1)^2BC·DC=(3√3-3)(2√3-2)=6(√3-1)^2∴AC^2=BC·DC∴AC/BC=DC/AC∴△ABC∽△DAC∴∠BAC=∠ADC=60°