Câu hỏi 26

Gọi \(O = AC \cap BD\), trong \(\left( {SBD} \right)\) gọi \(I = HK \cap SO\), trong \(\left( {SAC} \right)\) gọi \(M = AI \cap SC\).

Khi đó ta có \(\left( {AHK} \right) \equiv \left( {AHMK} \right)\).

Ta có:

\(\begin{array}{l}\left\{ \begin{array}{l}BC \bot AB\\BC \bot SA\end{array} \right. \Rightarrow BC \bot \left( {SAB} \right) \Rightarrow BC \bot AH\\\left\{ \begin{array}{l}AH \bot BC\\AH \bot SB\end{array} \right. \Rightarrow AH \bot \left( {SBC} \right) \Rightarrow AH \bot SC\end{array}\).

Hoàn toàn tương tự ta chứng minh được \(AK \bot SC \Rightarrow SC \bot \left( {AHMK} \right) \Rightarrow \left\{ \begin{array}{l}SC \bot HM\\SC \bot KM\end{array} \right.\)

\(\left\{ \begin{array}{l}\left( {SBC} \right) \cap \left( {SCD} \right) = SC\\\left( {SBC} \right) \supset HM \bot SC\\\left( {SCD} \right) \supset KM \bot SC\end{array} \right. \Rightarrow \angle \left( {\left( {SBC} \right);\left( {SCD} \right)} \right) = \angle \left( {HM;KM} \right) = \angle HMK\)

Ta có: \(AH \bot \left( {SBC} \right) \Rightarrow AH \bot HM \Rightarrow \angle AHM = {90^0}\). Tương tự ta có \(\angle AKM = {90^0}\).

Xét tứ giác \(AHMK\) có :

\(\angle HAK + \angle AHM + \angle AKM + \angle HMK = {360^0} \Leftrightarrow \angle HMK = {360^0} - {40^0} - {90^0} - {90^0} = {140^0} > {90^0}\).

Vậy \(\angle \left( {HM;KM} \right) = {180^0} - {140^0} = {40^0} \Rightarrow \angle \left( {\left( {SBC} \right);\left( {SCD} \right)} \right) = {40^0}\).

Chọn A.