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: \(\dfrac{1}{MK^{2}}=\dfrac{1}{SM^{2}}+\dfrac{1}{MN^{2}}\)

\(\Rightarrow MK=\dfrac{\sqrt{6}}{3}a\)

\(\dfrac{1}{MH^{2}}=\dfrac{1}{SM^{2}}+\dfrac{1}{MO^{2}}\)

\(\Rightarrow MH=\dfrac{\sqrt{10}}{5}a\)

\(\Rightarrow \cos \widehat{HMK}=\dfrac{HM}{HK}=\dfrac{\dfrac{\sqrt{10}}{5}a}{\dfrac{\sqrt{6}}{3}a}=\dfrac{\sqrt{15}}{5}\)

\(\Rightarrow \widehat{HMK}=39°13'\)