Question

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]}

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]}