Cho hình hộp chữ nhật ABCD.A'B'C'D' có AA' = a, AD = a căn bậc hai 3

Ta có: \[AB \bot \left( {ADD'A'} \right) \Rightarrow AB \bot AD'\]

\(\left\{ {\begin{array}{*{20}{l}}{\left( {ABC'D'} \right) \cap \left( {ABCD} \right) = AB}\\{AD \subset \left( {ABCD} \right);{\mkern 1mu} {\mkern 1mu} AD \bot AB}\\{AD' \subset \left( {ABC'D'} \right);{\mkern 1mu} {\mkern 1mu} AD' \bot AB}\end{array}} \right.\)

\( \Rightarrow \angle \left( {\left( {ABC'D'} \right);\left( {ABCD} \right)} \right) = \angle \left( {AD;AD'} \right) = \angle DAD'\)

Xét tam giác vuông \(ADD'\) có: \(\tan \angle DAD' = \frac{{DD'}}{{AD}} = \frac{a}{{a\sqrt 3 }} = \frac{1}{{\sqrt 3 }}\)

\( \Rightarrow \angle DAD' = {30^0}\)

Vậy \(\angle \left( {\left( {ABC'D'} \right);\left( {ABCD} \right)} \right) = {30^0}\).