Решебник по геометрии 10 класс Атанасян ФГОС 325

\[\boxed{\mathbf{325.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\frac{S_{осн}}{S_{сеч}} = \frac{\sqrt{3}\pi}{4}.\]

\[Решение.\]

\[S_{осн} = \pi R^{2};\ \ S_{сеч} = 2Rh:\]

\[\frac{\pi R^{2}}{2Rh} = \frac{\sqrt{3}\pi}{4}\]

\[\frac{R}{h} = \frac{\sqrt{3}}{2}.\]

\[\textbf{а)}\ tg\ \alpha = \frac{2R}{h} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3};\]

\[\alpha = arctg\sqrt{3}\]

\[\alpha = 60{^\circ}.\]

\[\gamma = 90{^\circ} - \alpha = 30{^\circ}.\]

\[\textbf{б)}\ \angle\beta = 180{^\circ} - 2 \cdot 30{^\circ} = 120{^\circ}\]

\[\angle B_{1}AB = 30{^\circ};\]

\[\angle A_{1}BA = 30{^\circ};\]

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

\[Ответ:а)\ 30{^\circ};\ \ б)\ 60{^\circ}.\]