Cho hình lập phương $ABCD.{A}'{B}'{C}'{D}'$. Góc giữa hai mặt...

Gọi $H={B}'C\cap B{C}',K=A{D}'\cap {A}'D$.
Khi đó $\left( AB{C}'{D}' \right)\cap \left( {A}'{B}'CD \right)=HK$.
Có $\left\{ \begin{aligned}
& {D}'{C}'\bot {B}'{C}' \\
& {D}'{C}'\bot C{C}' \\
\end{aligned} \right.\Rightarrow {D}'{C}'\bot \left( BC{C}'{B}' \right)\Rightarrow \left\{ \begin{aligned}
& {D}'{C}'\bot {B}'C \\
& {D}'{C}'\bot B{C}' \\
\end{aligned} \right.$.
Mà $HK,{D}'{C}'$ song song nhau nên $\left\{ \begin{aligned}
& HK\bot {B}'C \\
& HK\bot B{C}' \\
\end{aligned} \right.$.
Ta có:
$\left\{ \begin{aligned}
& \left( AB{C}'{D}' \right)\cap \left( {A}'{B}'CD \right)=HK \\
& HK\bot B{C}',B{C}'\subset \left( AB{C}'{D}' \right) \\
& HK\bot {B}'C,{B}'C\subset \left( {A}'{B}'CD \right) \\
\end{aligned} \right.\Leftrightarrow \left( \left( AB{C}'{D}' \right),\left( {A}'{B}'CD \right) \right)=\left( B{C}',{B}'C \right)=90{}^\circ $.