题目内容
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A. B.
C. D.
D 过A向l作相交的垂线AC,过B向l作相交垂线BD,则AC=a,BD=b,
因为“α⊥β,α∩β=l,ACα,AC⊥l”这些已知条件可得AC⊥β,同理可得出BD⊥α.
因此,AB在平面α内的射影是AD;AB在平面β内的射影是BC,进而有:AB与平面α所成角是θ=∠BAD,AB与平面β所成角是φ=∠ABC.
根据题意有:AD=m,BC=n,
在Rt△ABD中,AB2=AD2+BD2=m2+b2,
在Rt△ABC中,AB2=AC2+BC2=a2+n2,
∴m2+b2=a2+n2.
∵a>b,∴m>n.
又∵sinθ=sin∠BAD=,sinφ=sin∠ABC=
,
∴sinθ<sinφ.∴θ<φ.
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