(2013?朝阳区一模)如图,圆O是△ABC的外接圆,过点C作圆O的切线交BA的延长线于点D.若CD=3,AB=AC=2,

2025-05-16 04:11:13
推荐回答(1个)
回答(1):

①∵CD是⊙O的切线,由切割线定理得CD2=DA?DB,CD=

3
,DB=DA+AB=DA+2,
(
3
)2=DA(DA+2)
,又DA>0,解得DA=1.
②在△ACD中,由余弦定理可得cos∠ACD=
AC2+CD2?DA2
2AC?CD
=
22+(
3
)2?12
2×2×
3
=
3
2

∵0<∠ACD<π,∴∠ACD=
π
6

根据弦切角定理可得∠ABC=∠DCA=
π
6

由正弦定理可得2R=
AC
sin∠ABC
=
2
sin
π
6
=4,∴R=2.
故答案分别为1,2.