Решебник по геометрии 10 класс Атанасян ФГОС 717

\[\boxed{\mathbf{717.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\angle\text{BA}C_{1} = 60{^\circ};\]

\[\angle DAC_{1} = 60{^\circ}.\]

\[Найти:\]

\[\angle CAC_{1}.\]

\[Решение.\]

\[\overrightarrow{AC_{1}}\left\{ \cos{60{^\circ}};\cos{60{^\circ}};\sin{\angle CAC_{1}} \right).\]

\[Пусть\ \overrightarrow{AC_{1}} - единичный\ \]

\[вектор.\]

\[\overrightarrow{AC_{1}} =\]

\[= \left( \frac{1}{2} \right)^{2} + \left( \frac{1}{2} \right)^{2} + \sin^{2}{\angle CAC_{1}} =\]

\[= 1\]

\[\frac{1}{4} + \frac{1}{4} + \sin^{2}{\angle CAC_{1}} = 1\]

\[\sin^{2}{\angle CAC_{1}} = 1 - \frac{1}{2} = \frac{1}{2}\]

\[\sin{\angle CAC_{1}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2};\]

\[\angle CAC_{1} = \arcsin\frac{\sqrt{2}}{2} = 45{^\circ}.\]

\[Ответ:45{^\circ}.\]