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Câu hỏi: Cho hình chóp $S.ABC$ có $ABC$ là tam giác vuông tại $A,AB=a,AC=a\sqrt{3}$. Cạnh bên $SA=\dfrac{a}{2}$ vuông góc với mặt đáy. Tính góc tạo bởi hai mặt phẳng $(SBC)$ và $(ABC)$.
A. $30{}^\circ $.
B. $45{}^\circ $.
C. $60{}^\circ $.
D. $90{}^\circ $.
▪ Trong tam giác $ABC$ kẻ đường cao $AH.$
▪ $\Delta ABC \left( \widehat{A}=90{}^\circ \right), AH\bot BC\Rightarrow AH=\dfrac{AB.AC}{\sqrt{A{{B}^{2}}+A{{C}^{2}}}}$ $\Rightarrow AH=\dfrac{a.a\sqrt{3}}{\sqrt{{{a}^{2}}+3{{a}^{2}}}}=\dfrac{a\sqrt{3}}{2}$.
▪ Ta có:
$\left. \begin{aligned}
& BC=\left( SBC \right)\bigcap \left( ABC \right) \\
& BC\bot AH \\
& BC\bot SH \left( \text{do} \text{BC}\bot \left( SAH \right) \right) \\
\end{aligned} \right\}\Rightarrow \left( \widehat{\left( SBC \right),\left( ABC \right)} \right)=\widehat{SHA}$.
▪ $\Delta SAH \left( \widehat{A}=90{}^\circ \right)\Rightarrow \tan \widehat{SHA}=\dfrac{SA}{AH}=\dfrac{\dfrac{a}{2}}{\dfrac{a\sqrt{3}}{2}}=\dfrac{\sqrt{3}}{3}$. Suy ra: $\widehat{SHA}=30{}^\circ $.
A. $30{}^\circ $.
B. $45{}^\circ $.
C. $60{}^\circ $.
D. $90{}^\circ $.
▪ $\Delta ABC \left( \widehat{A}=90{}^\circ \right), AH\bot BC\Rightarrow AH=\dfrac{AB.AC}{\sqrt{A{{B}^{2}}+A{{C}^{2}}}}$ $\Rightarrow AH=\dfrac{a.a\sqrt{3}}{\sqrt{{{a}^{2}}+3{{a}^{2}}}}=\dfrac{a\sqrt{3}}{2}$.
▪ Ta có:
$\left. \begin{aligned}
& BC=\left( SBC \right)\bigcap \left( ABC \right) \\
& BC\bot AH \\
& BC\bot SH \left( \text{do} \text{BC}\bot \left( SAH \right) \right) \\
\end{aligned} \right\}\Rightarrow \left( \widehat{\left( SBC \right),\left( ABC \right)} \right)=\widehat{SHA}$.
▪ $\Delta SAH \left( \widehat{A}=90{}^\circ \right)\Rightarrow \tan \widehat{SHA}=\dfrac{SA}{AH}=\dfrac{\dfrac{a}{2}}{\dfrac{a\sqrt{3}}{2}}=\dfrac{\sqrt{3}}{3}$. Suy ra: $\widehat{SHA}=30{}^\circ $.
Đáp án A.