Question

等腰三角形ABC的三个顶点在半径为5cm的圆上,AB=AC,tan B=1\3,

Answer 1

设圆心为O,连接AO交BC于D,

则AO=5cm,BC=2BD=2DC,

AB=AC,所以BC边上的高为AD,

tan B=1/3=AD/BD

BD=3AD

BD=3(AO+DO),

BD=3(5+DO),

BD²=9(5+DO)²,...(1)

连接BO,BO=5cm,

BO²=BD²+DO²,

BD²=BO²-DO²=(BO+DO)(BO-DO)=(5+DO)(5-DO),代入(1):

(5+DO)(5-DO)=9(5+DO)²,

5-DO=9(5+DO)

10DO=-40,

DO=-4(cm),{说明D在O的上面,AD=AO+DO=5-4=1cm,D距A为1cm};

BC边上的高AD=AO+DO=5-4=1(cm),

BC=2BD=2(3AD)=6AD=6(cm)