Question

在三角形ABC中,D为变BC上一点,BD=1/2DC,∠ADB=120°,AD=2,若三角形ADC的面积为3-根号3,则

Answer 1

作AO⊥BC

AO=√3 ==>

DC=2√3-2,DO=1 ==>

BC=3√3-3,OC=2√3-3 ==>

BO=√3 ==>

AB=√6,∠ABC=45°

则AC^2=AB^2+BC^2-2*AB*BC*cos∠ABC

=(3√2-√6)^2

则cos∠BAC=(AB^2+AC^2-BC^2)/(2AB*AC)=-1/2

则∠BAC=120°