Đáp án:
\(49°53'\)
Giải thích các bước giải:
Ta có: $\begin{cases}BD \perp AO\\BD \perp SA\end{cases}$
\(\Rightarrow BD \perp (SAO)\)
\(\Rightarrow BD \perp AF\)
Từ A kẻ \(AF \perp SO\):
Ta có: $\begin{cases}AF \perp SO\\AF \perp BD\end{cases}$
\(\Rightarrow AF \perp (SBD)\) (1)
Ta có: $\begin{cases}BC \perp AB\\BC \perp SA\end{cases}$
\(\Rightarrow BC \perp (SAB)\)
\(\Rightarrow BC \perp AH\)
Từ A kẻ \(AH \perp SB\):
Ta có: $\begin{cases}AH \perp SB\\AH \perp BC\end{cases}$
\(\Rightarrow AH \perp (SBC)\) (2)
Từ (1)(2) Suy ra: \(\widehat{[(SBC),(SBD)]}=\widehat{HAF}\)
Ta có: \(\dfrac{1}{AH^{2}}=\dfrac{1}{SA^{2}}+\dfrac{1}{AB^{2}}\)
\(\Rightarrow AH=\dfrac{\sqrt{3}}{2}a\)
\(\dfrac{1}{AF^{2}}=\dfrac{1}{SA^{2}}+\dfrac{1}{AO^{2}}\)
\(\Rightarrow AF=\dfrac{\sqrt{21}}{7}a\)
\(\Rightarrow \cos \widehat{HAF}=\dfrac{AF}{AH}=\dfrac{\dfrac{\sqrt{21}}{7}a}{\dfrac{\sqrt{3}}{2}a}=\dfrac{2\sqrt{7}}{7}\)
\(\Rightarrow \widehat{HAF}=40°53'\)
