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Câu hỏi: Cho hình chóp $S.ABCD$ có đáy là hình vuông tâm $O$ cạnh bằng $a$ và $SA\bot \left( ABCD \right),SA=a\sqrt{3}$. Tính góc giữa hai mặt phẳng $\left( SAB \right)$ và $\left( SDC \right)$.
A. ${{30}^{o}}$.
B. ${{90}^{o}}$.
C. ${{60}^{o}}$.
D. ${{45}^{o}}$.
Ta có:
$\begin{aligned}
& \left\{ \begin{aligned}
& \left( SAB \right)\bot \left( SAD \right)\left( AB\subset \left( SAB \right),AB\bot SA,AB\bot AD \right) \\
& \left( SDC \right)\bot \left( SAD \right)\left( DC\subset \left( SDC \right),DC\bot SA,DC\bot AD \right) \\
& \left( SAB \right)\bigcap \left( SAD \right)=SA \\
& \left( SDC \right)\bigcap \left( SAD \right)=SD \\
\end{aligned} \right. \\
& \Rightarrow \widehat{\left( \left( SAB \right),\left( SDC \right) \right)}=\widehat{ASD} \\
\end{aligned}$
Trong $\Delta SAD$ vuông tại $A$ có:
$\tan \widehat{ASD}=\dfrac{AD}{AS}=\dfrac{a}{a\sqrt{3}}=\dfrac{1}{\sqrt{3}}\Rightarrow \widehat{ASD}={{30}^{o}}$.
Vậy $\widehat{\left( \left( SAB \right),\left( SDC \right) \right)}={{30}^{o}}$.
A. ${{30}^{o}}$.
B. ${{90}^{o}}$.
C. ${{60}^{o}}$.
D. ${{45}^{o}}$.
$\begin{aligned}
& \left\{ \begin{aligned}
& \left( SAB \right)\bot \left( SAD \right)\left( AB\subset \left( SAB \right),AB\bot SA,AB\bot AD \right) \\
& \left( SDC \right)\bot \left( SAD \right)\left( DC\subset \left( SDC \right),DC\bot SA,DC\bot AD \right) \\
& \left( SAB \right)\bigcap \left( SAD \right)=SA \\
& \left( SDC \right)\bigcap \left( SAD \right)=SD \\
\end{aligned} \right. \\
& \Rightarrow \widehat{\left( \left( SAB \right),\left( SDC \right) \right)}=\widehat{ASD} \\
\end{aligned}$
Trong $\Delta SAD$ vuông tại $A$ có:
$\tan \widehat{ASD}=\dfrac{AD}{AS}=\dfrac{a}{a\sqrt{3}}=\dfrac{1}{\sqrt{3}}\Rightarrow \widehat{ASD}={{30}^{o}}$.
Vậy $\widehat{\left( \left( SAB \right),\left( SDC \right) \right)}={{30}^{o}}$.
Đáp án A.
