Câu hỏi 18

* Nhận xét: \(\Delta A'BC=\Delta A'DC\)

* Bước 1: Vẽ \(DM\bot A'C\Rightarrow \widehat{\left( \left( A'BC \right);\left( A'CD \right) \right)}=\widehat{DMB}\)

* Bước 2: \(CB\bot AB\Rightarrow CB\bot A'B\)

Tính \(\frac{1}{B{{M}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{2{{a}^{2}}}\Rightarrow B{{M}^{2}}=\frac{2{{a}^{2}}}{3}\)

\(\begin{align}  \cos M=\frac{B{{D}^{2}}-M{{B}^{2}}-M{{D}^{2}}}{-2MB.MD}=-\frac{1}{2} \\  \Rightarrow \widehat{DMB}={{120}^{0}}\Rightarrow \widehat{\left( \left( A'BC \right);\left( A'CD \right) \right)}={{60}^{0}} \\ \end{align}\)

Đáp số: \({{60}^{0}}\)