Question

已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5,设A...

已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5,设AB＝3,求AB边上的高怎么解

Answer 1

答:AB边上的高=√{√[(15±3√21)/2]}sin(A+B)=3/5,sin(A-B)=1/5,AB=3sin(A+B)+sin(A-B)=2sinA*cosB=3/5+1/5=4/5sinA*cosB=2/5sinC=sin(A+B)=3/5过C点作CD⊥AB,交AB于D点,则CD为AB边上的高,sinA=CD/AC,cosB=BD/ABsinA*cosB=（CD/AC）*（BD/AB）=2/5CD*BD/(AC*AB)=2/5.(1)sin∠ACD=AD/AC,cos∠ACD=CD/ACsin∠BCD=BD/BC,cos∠BCD=CD/BCsinC=sin(∠ACD+∠BCD)=sin∠ACD*cos∠BCD+cos∠ACD*sin∠BCD=(AD/AC)*(CD/BC)+(CD/AC)*(BD/BC)=CD*((AD+BD)/(AC*BC)=CD*AB/(AC*BC)=3CD/(AC*BC)=3/5CD/(AC*BC)=1/5AC*BC=5CD.(2)(2)代入(1)得BD=2AD=3-2=1在RT△ACD和RT△BCD中,根据勾股定理,得CD^2+BD^2=BC^2,CD^2+AD^2=AC^2CD^2+9=BC^2.(3)CD^2+1=AC^2.(4)(3)*(4)得(CD^2+9)*(CD^2+1)=BC^2*AC^2=(AC*BC)^2=(5CD)^2CD^4-15CD^2+9=0△=15*15-4*9=189CD^2=√[(15±√189)/2]=√[(15±3√21)/2]∵CD>0
∴CD=√{√[(15±3√21)/2]}