Решебник по геометрии 10 класс Атанасян ФГОС 714

\[\boxed{\mathbf{714.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\(OA;OB;OC - лучи,\ \)

\[отложенные\ на\ осях\ \text{Ox};Oy;Oz;\]

\[OA = OB = OC = 1;\]

\[OM;ON - биссектрисы\ углов;\]

\[\angle AOD = \angle DOC;\]

\[\angle AOE = \angle EOD.\]

\[Найти:\]

\[\angle DOE.\]

\[Решение.\]

\[1)\ AM = MC = \frac{1}{2}AC;\]

\[AN = NB = \frac{1}{2}\text{AB.}\]

\[2)\ A(1;0;0);B(0;1;0);\]

\[C(0;0;1);\]

\[M\left( \frac{1}{2};0;\frac{1}{2} \right);\ \ N\left( \frac{1}{2};\frac{1}{2};0 \right).\]

\[3)\ \overrightarrow{\text{OM}}\left\{ \frac{1}{2};0;\frac{1}{2} \right\};\ \ \overrightarrow{\text{ON}}\left\{ \frac{1}{2};\frac{1}{2};0 \right\};\]

\[\cos{\angle\left( \overrightarrow{\text{OM}};\overrightarrow{\text{ON}} \right)} =\]

\[= \frac{\left( \frac{1}{2} \right)^{2} + 0 + 0}{\sqrt{\frac{1}{4} + 0 + \frac{1}{4}} \cdot \sqrt{\frac{1}{4} + \frac{1}{4} + 0}} =\]

\[= \frac{\frac{1}{4}}{\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}}} = \frac{1}{2};\]

\[\angle\left( \overrightarrow{\text{OM}};\overrightarrow{\text{ON}} \right) = \arccos\frac{1}{2} = 60{^\circ};\]

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

\[Ответ:60{^\circ}.\]