ГДЗ по геометрии 11 класс. Атанасян ФГОС 750

\[\boxed{\mathbf{750.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[DABC - тетраэдр;\]

\[DA = DB = DC;\]

\[\angle ADB = 45{^\circ};\]

\[\angle BDC = 60{^\circ}.\]

\[Решение.\]

\[\textbf{а)}\ \angle\left( \overrightarrow{\text{DA}};\overrightarrow{\text{DB}} \right) = 45{^\circ};\]

\[\overrightarrow{\text{DB}} = - \overrightarrow{\text{BD}}:\]

\[\angle\left( \overrightarrow{\text{DA}};\overrightarrow{\text{BD}} \right) = 180{^\circ} - 45{^\circ} =\]

\[= 135{^\circ}.\]

\[\textbf{б)}\ В\ треугольнике\ DBC:\]

\[\angle DBC = 60{^\circ};\ \ DC = DB;\]

\[\angle DCB = \angle CBD.\]

\[\angle DCB + \angle DBC = 120{^\circ}\]

\[\angle DBC = \angle DCB = 60{^\circ}.\]

\[\textbf{в)}\ ⊿DBA - равнобедренный:\]

\[\angle DAB = \angle DBA = \frac{180{^\circ} - 45{^\circ}}{2} =\]

\[= \frac{135}{2} = 67{^\circ}30^{'}.\]