C
分析:做出輔助線,連接AF并延長交BC于H,取線段AF的中點G,連接EG,證出線面角,把線面角放到一個直角三角形中,根據(jù)三角函數(shù)的定義得到結果,
解答:連接SF,則SF⊥平面ABC.連接AF并延長交BC于H,取線段AF的中點G,連接EG,由E為SA的中點,則EG∥SF,
∴EG⊥平面ABC,
∴∠EFG即為EF與平面ABC所成的角.
設正四面體的邊長為a,則AH=
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a,且AF=
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,
在Rt△AGE中,AE=
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a,AG=
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AF=
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a,∠EGA=90°,
∴EG=AE
2-AG
2=
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a.
在Rt△EGF中,F(xiàn)G=
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AF=
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a,EG=
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a,∠EGF=90°,
∴tan∠EFG=
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即EF與平面ABC所成的角的正切值是
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,
故選C.
點評:本題考查直線與平面所成的角,本題解題的關鍵是先做出線面角,再證出線面角,最后把角放到一個三角形中解出結果.