Cho hình chóp S.ABCD có đáy ABCD cạnh a,$SA = a\sqrt 3 $ và SA vuông góc với đáy ABCD. Gọi BM và DN là 2 đường cao của tam giác SBD. a) CMR: $\left( {AMN} \right) \bot \left( {SBC} \right).$ b) Tính $\widehat {\left( {\left( {SAC} \right),\left( {SBC} \right)} \right)}$
Giải:
$\ a)\,CMR:\left( {SBC} \right) \bot \left( {AMN} \right)\,\,\,\,\,.$
\[\begin{array}{l}
\bullet \left\{ \begin{array}{l}
AD \bot SB\left( {AD \bot \left( {SAB} \right)} \right)\\
DN \bot SB\left( {gt} \right)
\end{array} \right.\\
\Rightarrow SB \bot \left( {DAN} \right) \Rightarrow AN \bot SB\left( 1 \right)\\
\bullet \left\{ \begin{array}{l}
BC \bot AB\\
BC \bot SA
\end{array} \right. \Rightarrow BC \bot \left( {SAB} \right) \Rightarrow BC \bot AN\left( 2 \right)\\
\Rightarrow AN \bot \left( {SBC} \right)\left( {Do\,\left( 1 \right)\& \left( 2 \right)} \right) \Rightarrow AN \bot SC\left( 3 \right)\\
\bullet \left\{ \begin{array}{l}
BA \bot SD\left( {BA \bot \left( {SAD} \right)} \right)\\
BM \bot SD\left( {gt} \right)
\end{array} \right.\\
\Rightarrow SD \bot \left( {BAM} \right) \Rightarrow AM \bot SD\left( 4 \right)\\
\bullet \left\{ \begin{array}{l}
DC \bot AD\\
DC \bot SA
\end{array} \right. \Rightarrow DC \bot \left( {SAD} \right) \Rightarrow DC \bot AM\left( 5 \right)\\
\Rightarrow AM \bot \left( {SDC} \right)\left( {Do\,\left( 4 \right)\& \left( 5 \right)} \right) \Rightarrow AM \bot SC\left( 6 \right)\\
\Rightarrow SC \bot \left( {AMN} \right)\left( {\left( {Do\,\left( 3 \right)\& \left( 6 \right)} \right)} \right) \Rightarrow \left( {SBC} \right) \bot \left( {AMN} \right)
\end{array}\]\[\begin{array}{l}
b)\,\widehat {\left( {\left( {SBC} \right),\left( {SAC} \right)} \right)} = ?\\
AN \bot \left( {SBC} \right)\left( {cmt} \right)\,\,\,\& \,\left\{ \begin{array}{l}
AC \bot BD\\
SA \bot BD
\end{array} \right. \Rightarrow BD \bot \left( {SAC} \right) \Rightarrow \widehat {\left( {\left( {SBC} \right),\left( {SAC} \right)} \right)} = \widehat {\left( {AN,BD} \right)}\\
\bullet \,Do\,\Delta SAB = \Delta SAD \Rightarrow AM = AN \Rightarrow SN = SM\\
\Rightarrow \frac{{SM}}{{SD}} = \frac{{SN}}{{SB}} \Leftrightarrow MN//BD \Rightarrow \widehat {\left( {AN,BD} \right)} = \widehat {\left( {AN,MN} \right)} = \widehat {ANM}\\
\bullet \left\{ \begin{array}{l}
AM = AN = \sqrt {\frac{1}{{\frac{1}{{S{A^2}}} + \frac{1}{{D{A^2}}}}}} = \frac{{a\sqrt 3 }}{2}\\
\frac{{MN}}{{BD}} = \frac{{SN}}{{SD}} = \frac{{\sqrt {S{A^2} - A{N^2}} }}{{\sqrt {S{A^2} + A{D^2}} }} = \frac{{\sqrt {3{a^2} - \frac{{3{a^2}}}{4}} }}{{2a}} = \frac{3}{4} \Rightarrow MN = \frac{3}{4}BD = \frac{{3a\sqrt 2 }}{4}
\end{array} \right.\\
\,\,\,\,\, \Rightarrow cos\widehat {ANM} = \frac{{A{N^2} + M{N^2} - A{M^2}}}{{2AN.MN}} = \frac{{MN}}{{2AN}} = \frac{{\frac{{3a\sqrt 2 }}{4}}}{{2.\frac{{a\sqrt 3 }}{2}}} = \frac{{\sqrt 6 }}{4}
\end{array}\]
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