Toán 11

D

drmssi

Ta có: SB=SC gọi K là trung điểm BC
[TEX]\Rightarrow SK \perp BC, AK \perp BC[/TEX]
[TEX]BC \subset (SBC), (AK,SK) \subset(SAK) =>(SBC) \perp (SAK)[/TEX]
Gọi H là h chiếu A lên SK.
[TEX](SAK) \subset (SBC) = SK, AH\perp SK \rightarrow AH\perp(SBC) [/TEX]
[TEX]\Rightarrow \widehat{(SA),(SBC)}= \widehat{ASH}[/TEX]
Xét [TEX]\triangle ABC:\frac{1}{AK^2}=\frac{1}{AC^2}+\frac{1}{AB^2} \Rightarrow AK= [/TEX]
Xét [TEX]\triangle SAK: tan\widehat{ASH}=\frac{AK}{SA}[/TEX]
Cho đ M?
 
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