已知点A(1,0),B(0,1),C(2,sinθ) (1)若|
(2)若(
|
(1)由A(1,0),B(0,1),C(2,sinθ),得到
=(1,sinθ),AC
=(2,sinθ-1),BC
因为|
|=|AC
|,所以BC
=1+sin2θ
,4+(sinθ-1)2
两边平方得:1+sin2θ=4+sin2θ-2sinθ+1,解得sinθ=
;1 2
(2)
=(1,0),OA
=(0,1),OB
=(2,sinθ),代入(OC
+OA
)•OB
=OC
中,13 5
化简得:2+sinθ=
,解得:sinθ=13 5
,又0<θ<π,所以cosθ=-3 5
,4 5
则tanθ=-
.3 4