Thi Hk1 08-09 (12nc) 2

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Cách 1 Tính trực tiếp • Gọi E là trung điểm BC, chứng minh AE ⊥ BC • Gọi H là hình chiếu A lên SE, chứng minh AH ⊥ ( SBC ) 2a.cos α • Trong ∆ABC, tính được AE = 1 + sin α 6a.cos α • Trong ∆SAE, tính được AH = 9(1 + sin α ) 2 + 16 cos 2 α Cách 2 Tính gián tiếp qua thể tích S.ABCD 3.V • d ( A;( SBC ) ) = S . ABCD S ∆SBC 2a.cos α • Tính được AE = 1 + sin α 1 • Tính được S ∆SBC = SE.BC = ... 2 • Suy ra d ( A;( SBC ) )

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