[tex]\it \mathcal{A}_{D EF}=\mathcal{A}_{ABC}\cdot cos u \Rightarrow cosu=\dfrac{\mathcal{A}_{D EF}}{\mathcal{A}_{ABC}}[/tex]
[tex]\bf a)\ \ \it cosu=\dfrac{6\sqrt2}{4\sqrt6}=\dfrac{2\cdot3\sqrt2}{2\cdot2\cdot\sqrt3\cdot\sqrt2}=\dfrac{^{\sqrt3)}3}{\ \ 2\sqrt3}=\dfrac{3\sqrt3}{2\cdot3}=\dfrac{\sqrt3}{2} \Rightarrow u=30^o\\ \\ \\ \bf\ b)\ \ \it cosu=\dfrac{\not9\sqrt3}{\not9\sqrt6}=\dfrac{\ \ \sqrt3\ ^{(\sqrt3}}{\sqrt6} =\dfrac{^{\sqrt2)}1}{\ \sqrt2}=\dfrac{\sqrt2}{2} \Rightarrow u=45^o[/tex]
