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Câu hỏi: Cho hình chóp $S.ABC$ có cạnh $SA$ vuông góc với mặt phẳng $\left( ABC \right)$, biết $AB=AC=a,BC=a\sqrt{3}$ ( tham khảo hình vẽ). Tính góc giữa hai mặt phẳng $\left( SAB \right)$ và $\left( SAC \right).$
A. ${{90}^{0}}$.
B. ${{60}^{0}}$.
C. ${{45}^{0}}$.
D. ${{120}^{0}}$.
B. ${{60}^{0}}$.
C. ${{45}^{0}}$.
D. ${{120}^{0}}$.
Vì $SA\bot \left( ABC \right)$ nên $SA\bot AB$ và $SA\bot AC$
Ta có $\left\{ \begin{aligned}
& \left( SAB \right)\cap \left( SAC \right)=SA \\
& SA\bot AB \\
& SA\bot AC \\
\end{aligned} \right.\Rightarrow \left( \widehat{\left( SAB \right),\left( SAC \right)} \right)=\left( \widehat{AB,AC} \right)=\left[ \begin{aligned}
& \widehat{BAC} ({{0}^{0}}\le \widehat{BAC}\le {{90}^{0}}) \\
& {{180}^{0}}-\widehat{BAC} (\widehat{BAC}>{{90}^{0}}) \\
\end{aligned} \right.$
Xét $\Delta ABC$ có $\cos \widehat{BAC}=\dfrac{A{{B}^{2}}+A{{C}^{2}}-B{{C}^{2}}}{2AB.AC}=\dfrac{{{a}^{2}}+{{a}^{2}}-{{\left( a\sqrt{3} \right)}^{2}}}{2.a.a}=-\dfrac{1}{2}\Rightarrow \widehat{BAC}={{120}^{0}}$
Vậy $\left( \widehat{\left( SAB \right),\left( SAC \right)} \right)={{180}^{0}}-{{120}^{0}}={{60}^{0}}$.
Ta có $\left\{ \begin{aligned}
& \left( SAB \right)\cap \left( SAC \right)=SA \\
& SA\bot AB \\
& SA\bot AC \\
\end{aligned} \right.\Rightarrow \left( \widehat{\left( SAB \right),\left( SAC \right)} \right)=\left( \widehat{AB,AC} \right)=\left[ \begin{aligned}
& \widehat{BAC} ({{0}^{0}}\le \widehat{BAC}\le {{90}^{0}}) \\
& {{180}^{0}}-\widehat{BAC} (\widehat{BAC}>{{90}^{0}}) \\
\end{aligned} \right.$
Xét $\Delta ABC$ có $\cos \widehat{BAC}=\dfrac{A{{B}^{2}}+A{{C}^{2}}-B{{C}^{2}}}{2AB.AC}=\dfrac{{{a}^{2}}+{{a}^{2}}-{{\left( a\sqrt{3} \right)}^{2}}}{2.a.a}=-\dfrac{1}{2}\Rightarrow \widehat{BAC}={{120}^{0}}$
Vậy $\left( \widehat{\left( SAB \right),\left( SAC \right)} \right)={{180}^{0}}-{{120}^{0}}={{60}^{0}}$.
Đáp án B.