By the given condition the trihedrons with vertices $A$ and $B$ are equal and also $C$ and $D$ are equal. Also we have two equal planar angles in each trihedron. Hence the triangles $ABC$ and $ABD$ are equal and the same is true for the triangles $CDA$ and $CDB$. Thus $AD=BC$ and $BD=AC$. It's easy to see that the triangles $ABC$ and $BCD$ are equal. Let $k$ be the ratio of their similitude. Note that $AD/DB=BD/DC$ so $BD^2=AD\cdot DC$. Similarly $AD^2=AB\cdot BD$. Combining these two relationships we obtain $BD^3=AB\cdot CD^2$. Thus $$k=\frac{BD}{DC}=\frac{\sqrt[3]{AB\cdot CD^2}}{CD}=\sqrt[3]{\frac{AB}{CD}}.$$Using the formula $[ABCD]=\frac{2}{3}\frac{[ABC][ABD]}{AB}\sin (AB)$ and its similar formula for $CD$ we get $$\frac{[ABC]^2}{AB}\sin\beta=\frac{[BCD]^2}{CD}\sin\alpha$$$\longrightarrow$ $$\frac{AB}{CD}=\frac{\sin\beta}{\sin\alpha}\cdot k^4=\frac{\sin\beta}{\sin\alpha}\cdot \sqrt[3]{\frac{AB^4}{CD^4}}$$$\longrightarrow$ $$\frac{AB}{CD}=\frac{\sin^3\alpha}{\sin^3\beta}.$$