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Câu hỏi: Cho hình chóp tứ giác đều $S.ABCD$ có cạnh đáy bằng $a\sqrt{2}$ và chiều cao bằng $\dfrac{a\sqrt{2}}{2}$. Góc giữa hai mặt phẳng $\left( SCD \right)$ và $\left( ABCD \right)$ bằng
A. $90{}^\circ .$
B. $45{}^\circ .$
C. $30{}^\circ .$
D. $60{}^\circ .$
A. $90{}^\circ .$
B. $45{}^\circ .$
C. $30{}^\circ .$
D. $60{}^\circ .$
Gọi $O=AC\cap BD\Rightarrow SO\bot \left( ABC\text{D} \right)$.
Kẻ $OP\bot C\text{D}$, ta có $\left\{ \begin{aligned}
& C\text{D}\bot \text{S}O \\
& C\text{D}\bot OP \\
\end{aligned} \right.\Rightarrow C\text{D}\bot \left( SOP \right)\Rightarrow C\text{D}\bot \text{S}P$.
Mà $C\text{D}\bot OP\Rightarrow \widehat{\left( (SC\text{D});(ABC\text{D}) \right)}=\widehat{SPO}$.
$\tan \widehat{SPO}=\dfrac{SO}{OP}=\dfrac{\dfrac{a\sqrt{2}}{2}}{\dfrac{a\sqrt{2}}{2}}=1\Rightarrow \widehat{SPO}=45{}^\circ $.
Kẻ $OP\bot C\text{D}$, ta có $\left\{ \begin{aligned}
& C\text{D}\bot \text{S}O \\
& C\text{D}\bot OP \\
\end{aligned} \right.\Rightarrow C\text{D}\bot \left( SOP \right)\Rightarrow C\text{D}\bot \text{S}P$.
Mà $C\text{D}\bot OP\Rightarrow \widehat{\left( (SC\text{D});(ABC\text{D}) \right)}=\widehat{SPO}$.
$\tan \widehat{SPO}=\dfrac{SO}{OP}=\dfrac{\dfrac{a\sqrt{2}}{2}}{\dfrac{a\sqrt{2}}{2}}=1\Rightarrow \widehat{SPO}=45{}^\circ $.
Đáp án B.