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Cho hình chóp S.ABC có tam giác ABC cân đỉnh A,(BAC) ̂=120° và AB=2a.Biết SA=SB=SC=(a√39)/3
8 months ago
35
Cho hình chóp S.ABC có tam giác ABC cân đỉnh A,(BAC) ̂=120° và AB=2a.Biết
SA=SB=SC=(a√39)/3.Đặt khoảng cách từ điểm S đến mặt phẳng (ABC) là h và
α=[S,BC,A].Khi đó,(90h^2)/a^2 = và α=
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