题目

三角形中,已知sin(A+B)=3/5 sina(A-B)=1/5 tanA=2tanB AB=3,求AB上的高

提问时间：2021-03-03

答案

答:AB边上的高=√{√[(15±3√21)/2]}
解:
sin(A+B)=3/5,sin(A-B)=1/5,AB=3
sin(A+B)+sin(A-B)=2sinA*cosB=3/5+1/5=4/5
sinA*cosB=2/5
sinC=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/AB
sinA*cosB=（CD/AC）*（BD/AB）=2/5
CD*BD/(AC*AB)=2/5.(1)
sin∠ACD=AD/AC,cos∠ACD=CD/AC
sin∠BCD=BD/BC,cos∠BCD=CD/BC
sinC=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)
=3/5
CD/(AC*BC)=1/5
AC*BC=5CD.(2)
(2)代入(1)得
BD=2
AD=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^2
CD^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)^2
CD^4-15CD^2+9=0
△=15*15-4*9=189
CD^2=√[(15±√189)/2]=√[(15±3√21)/2]
∵CD>0
∴CD=√{√[(15±3√21)/2]