·
=1,
·
=-2.(1)求AB边的长度;
(2)证明tanA=2tanB;
(3)若|
|=2,求|
|.
·=1,·=-2.(1)求AB边的长度;
(2)证明tanA=2tanB;
(3)若||=2,求||.
解
:(1)∵=-,∴·=
·(
-
)=
·
-|
|2=-2.∵
·
=1,∴|
|2=3,|
|=
,即
边的长度为3.
(2)由
·
=1,
·
=-2,得|
||
|cosA=1,①
|
||
|cos(π-B)=-2,
即|
||
|cosB=2.②
由①②得
·
=
,
由正弦定理得
=
,
∴
·
=
=
.
∴tanA=2tanB.
(3)∵|
|=2,由(2)中①得cosA=
=
.
由余弦定理得|
|2=|
|2+|
|2-2|
||
|cosA=3+4-4
·
=5.
∴|
|=
.