+) Gọi độ dài cạnh đáy là \(x\), gọi \(M\) là trung điểm của \(C D, O \equiv A C \cap B D\).

\(\Rightarrow((S C D) ;(A B C D))=S M O=\alpha^{\circ}\).

+) Có \(O M=\frac{x}{2} \Rightarrow S O=O M \cdot \tan \alpha^{\circ}=\frac{x}{2} \cdot \tan \alpha^{\circ}\).

+) \(V=\frac{4 a^{3} \sqrt{3}}{3} \Rightarrow \frac{1}{3} x^{2} \cdot \frac{x}{2} \cdot \tan \alpha^{\circ}=\frac{4}{3} a^{3} \sqrt{3} \Rightarrow x^{3} \cdot \tan \alpha^{\circ}=8 a^{3} \sqrt{3}(1)\).

+) Theo giả thiết \(S_{x q}=4 \mathrm{~S}_{\triangle S C \mathrm{D}}=4 \cdot \frac{1}{2} \cdot S M \cdot C \mathrm{D}=2 \cdot \frac{x}{2 \cdot \cos \alpha^{\circ}}=\frac{x^{2}}{\cos \alpha^{\circ}}=8 a^{2}\) ( giả thiết) \(\Rightarrow x^{2}=8 a^{2} \cdot \cos \alpha^{\circ}\) (2).

+) Từ (1) và (2) ta có hệ: \(\left\{\begin{array}{l}x^{3} \cdot \tan \alpha^{\circ}=8 a^{3} \sqrt{3} \\ x^{2}=8 a^{2} \cdot \cos \alpha^{\circ}\end{array} \Rightarrow x^{6}=\left(8 \sqrt{3} \cdot a^{3} \cdot \frac{\cos \alpha^{\circ}}{\sin \alpha^{\circ}}\right)^{2}=\left(8 a^{2} \cdot \cos \alpha^{\circ}\right)^{3}\right.\).

\(\Rightarrow 3 \cdot \frac{\cos ^{2} \alpha^{\circ}}{\sin ^{2} \alpha^{\circ}}=8 \cdot \cos ^{3} \alpha^{\circ} \Rightarrow \frac{3}{\sin ^{2} \alpha^{\circ}}=8 \cos \alpha^{\circ} \Leftrightarrow 3=8\left(1-\cos ^{2} \alpha^{\circ}\right) \cos \alpha^{\circ}\).

\(\Rightarrow 8 \cdot \cos ^{3} \alpha^{o}-8 \cos \alpha^{o}+3=0 \Rightarrow\left(2 \cos \alpha^{o}-1\right)\left(4 \cos ^{2} \alpha^{o}+2 \cos \alpha^{o}-3\right)=0\)

\(\Rightarrow\left[\begin{array}{l}\cos \alpha^{\circ}=\frac{1}{2} \\ \cos \alpha^{\circ}=\frac{-1+\sqrt{13}}{4} \\ \cos \alpha^{\circ}=\frac{-1-\sqrt{13}}{4}\lt -1\end{array} \Rightarrow \alpha \notin Z \Rightarrow \alpha^{\circ}=60^{\circ}\right.\).