题目内容

已知点A(cos10°,sin10°)、B(sin40°,cos40°),则直线AB的倾斜角等于(  )
A.135°B.120°C.105°D.95°
设直线AB的倾斜角为α,则tanα=
cos40°-sin10°
sin40°-cos10°

=
cos(30°+10°)-sin10°
sin(30°+10°)-cos10°
=
cos30°cos10°-sin30°sin10°-sin10°
sin30°cos10°+cos30°sin10°-cos10°

=
3
2
cos10°-
3
2
sin10°
3
2
sin10°-
1
2
cos10°
=
3
(cos60°cos10°-sin60°sin10°)
cos30°sin10°-sin30°cos10°

=
3
cos70°
sin(-20°)
=-
3

因为0°≤α<180°,所以α=120°.
故选B.
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