Đáp án:
\(39°13'\)
Giải thích các bước giải:
Ta có: $\begin{cases}NP \perp MN\\NP \perp SM\end{cases}$
\(\Rightarrow NP \perp (SMN)\)
\(\Rightarrow NP \perp MK\)
Từ M kẻ \(MK \perp SN\):
Ta có: $\begin{cases}MK \perp NP\\MK \perp SM\end{cases}$
\(\Rightarrow MK \perp (SNP)\) (1)
Ta có: $\begin{cases}NQ \perp MP\\NQ \perp SM\end{cases}$
\(\Rightarrow NQ \perp (SMO)\)
\(\Rightarrow NQ \perp MH\)
Từ M kẻ \(MH \perp SO\):
Ta có: $\begin{cases}MH \perp SO\\MH \perp NQ\end{cases}$
\(\Rightarrow MH \perp (SNQ)\) (2)
Từ (1)(2) Suy ra: \(\widehat{[(SNQ),(SNP)]}=\widehat{HMK}\)
Ta có: \(\dfrac{1}{MK^{2}}=\dfrac{1}{SM^{2}}+\dfrac{1}{MN^{2}}\)
\(\Rightarrow MK=\dfrac{\sqrt{6}}{3}a\)
\(\dfrac{1}{MH^{2}}=\dfrac{1}{SM^{2}}+\dfrac{1}{MO^{2}}\)
\(\Rightarrow MH=\dfrac{\sqrt{10}}{5}a\)
\(\Rightarrow \cos \widehat{HMK}=\dfrac{HM}{HK}=\dfrac{\dfrac{\sqrt{10}}{5}a}{\dfrac{\sqrt{6}}{3}a}=\dfrac{\sqrt{15}}{5}\)
\(\Rightarrow \widehat{HMK}=39°13'\)
