\[\boxed{\mathbf{750.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[DABC - тетраэдр;\]
\[DA = DB = DC;\]
\[\angle ADB = 45{^\circ};\]
\[\angle BDC = 60{^\circ}.\]
\[Решение.\]
\[\textbf{а)}\ \angle\left( \overrightarrow{\text{DA}};\overrightarrow{\text{DB}} \right) = 45{^\circ};\]
\[\overrightarrow{\text{DB}} = - \overrightarrow{\text{BD}}:\]
\[\angle\left( \overrightarrow{\text{DA}};\overrightarrow{\text{BD}} \right) = 180{^\circ} - 45{^\circ} =\]
\[= 135{^\circ}.\]
\[\textbf{б)}\ В\ треугольнике\ DBC:\]
\[\angle DBC = 60{^\circ};\ \ DC = DB;\]
\[\angle DCB = \angle CBD.\]
\[\angle DCB + \angle DBC = 120{^\circ}\]
\[\angle DBC = \angle DCB = 60{^\circ}.\]
\[\textbf{в)}\ ⊿DBA - равнобедренный:\]
\[\angle DAB = \angle DBA = \frac{180{^\circ} - 45{^\circ}}{2} =\]
\[= \frac{135}{2} = 67{^\circ}30^{'}.\]