\[\boxed{\mathbf{788.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\mathrm{\Delta}ABC - правильный;\]
\[AB = a;\]
\[BD \uparrow \uparrow CE;\]
\[BD\bot ABC;\]
\[CE\bot ABC;\]
\[BD = \frac{a}{\sqrt{2}};\]
\[CE = a\sqrt{2}.\]
\[Доказать:\]
\[\mathrm{\Delta}ADE - прямоугольный.\]
\[Найти:\]
\[\angle(ABC,ADE).\]
\[Решение.\]
\[1)\ Построим\ в\ плоскости\ DCE:\ \ \]
\[DF\bot EC\ \ и\ \ DF \parallel BC.\]
\[Достроим\ ED\ и\ \text{BC\ }до\ \]
\[пересечения\ в\ точке\ H:\]
\[ED = BD = FC = \frac{a\sqrt{2}}{2};\]
\[EF = CE - FC = EC - BD =\]
\[= a\sqrt{2} - \frac{a\sqrt{2}}{2} = \frac{a\sqrt{2}}{2};\]
\[HD = DE;\ \ \ HB = BC - a.\]
\[2)\ В\ \mathrm{\Delta}ABD - прямоугольном:\ \]
\[AD = \sqrt{a + \frac{a^{2}}{2}} = a\sqrt{\frac{3}{2}}.\]
\[В\ \mathrm{\Delta}ACE - прямоугольном:\]
\[AE = \sqrt{a^{2} + 2a^{2}} = a\sqrt{3}.\]
\[\ В\ \mathrm{\Delta}DFE - прямоугольном:\]
\[DE = \sqrt{a^{2} + \frac{a^{2}}{2}} = a\sqrt{\frac{3}{2}}.\]
\[3)\ AE^{2} = AD^{2} + DE^{2};\ \ \ \]
\[3a^{2} = \frac{3a^{2}}{2} + \frac{3a^{2}}{2}:\]
\[\angle ADE = 90{^\circ};\]
\[\mathrm{\Delta}ADE - прямоугольный.\ \]
\[Что\ и\ требовалось\ доказать.\]
\[4)\ В\ \mathrm{\Delta}HBA:\ \ \]
\[\angle HBA = 180{^\circ} - 60{^\circ} = 120{^\circ};\]
\[HB = BA = a;\ \ \]
\[\angle HAB = \angle AHB = 30{^\circ}.\]
\[Значит:\]
\[\angle HAC = 30{^\circ} + 60{^\circ} = 90{^\circ};\]
\[\angle HAE = 90{^\circ}\ (по\ теореме\ о\ трех\ \]
\[перпендикулярах);\ \ \]
\[\angle EAC - искомый\ угол\ между\ \]
\[плоскостями\ \text{ABC\ }и\ \text{ADE.}\]
\[5)\ В\ \mathrm{\Delta}ACE:\ \]
\[\cos{\angle EAC} = \frac{\text{AC}}{\text{AE}} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}} =\]
\[= \frac{\sqrt{3}}{3}\]
\[\angle(ABC,ADE) = \arccos\frac{\sqrt{3}}{3}.\]
\[\mathbf{Ответ}\mathbf{:\ \ }\arccos\frac{\sqrt{3}}{3}.\]