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)