已知∠ABC=60°,p为∠ABC内一定点,且p点到边AB,AC的距离分别为1,2,着p到顶点B的距离为

3个回答

  • 设P到顶点B的距离为x,∠ABP=a,∠CBP=b

    sina=1/x cosa=√(x^2-1)/x

    sinb=2/x cosb=√(x^2-4)/x

    sin∠ABC=sin(a+b)=sin60°=√3/2

    sinacosb+cosasinb=√3/2

    1/x*√(x^2-4)/x+√(x^2-1)/x*2/x=√3/2

    2√(x^2-4)+4√(x^2-1)=√3x^2

    4x^2-16+16x^2-16+16√(x^2-1)(x^2-4)=3x^4

    16√(x^2-1)(x^2-4)=3x^4-20x^2+32

    256x^4-1280x^2+1024=9x^8+400x^4+1024-120x^6+192x^4-1280x^2

    9x^8-120x^6+336x^4=0

    3x^4-40x^2+112=0

    △=256

    x^2=(40±16)/6=28/3或4

    x=2/3*√21或2(舍去)

    所以P到顶点B的距离为2/3*√21