Решебник по геометрии 10 класс Атанасян ФГОС 545

\[\boxed{\mathbf{545.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[конус;\]

\[R_{осн} = r;\]

\[l - образующая;\]

\[O - центр\ вписанной\ сферы.\]

\[Найти:\]

\[\text{c.}\]

\[Решение.\]

\[1)\ MB - высота\ конуса;\ \ \]

\[OP = OC = OH = R_{сферы};\]

\[HA = HB = r;\ \ \]

\[MA = MB = l.\]

\[Точки\ D,C,H - касания\ сферы\ \]

\[и\ конуса.\]

\[2)\ \mathrm{\Delta}AHO - прямоугольный:\]

\[\frac{\text{HO}}{\text{AH}} = tg\ \angle OAH\]

\[HO = AH \bullet tg\ \angle OAH;\]

\[R_{сферы} = r \bullet tg\ \angle OAH.\]

\[3)\ AO - биссектриса\ \angle MAB:\]

\[\angle MAB = 2 \bullet \angle OAH.\]

\[4)\ \mathrm{\Delta}MAH - прямоугольный:\]

\[\cos(2 \bullet \angle OAH) = \frac{\text{AH}}{\text{AM}} = \frac{r}{l}\]

\[2\cos^{2}{\angle OAH} - 1 = \frac{r}{l}\]

\[\cos^{2}{\angle OAH} = \frac{r + l}{2l}\]

\[\cos{\angle OAH} = \sqrt{\frac{r + l}{2l}}.\]

\[Отсюда:\]

\[\sin^{2}{\angle OAH} = 1 - \frac{r + l}{2l}\]

\[\sin{\angle OAH} = \sqrt{\frac{l - r}{2l}};\]

\[tg\ \angle OAH = \frac{\sin{\angle OAH}}{\cos{\angle OAH}} =\]

\[= \sqrt{\frac{l - r}{2l} \bullet \frac{2l}{r + l}} = \sqrt{\frac{l - r}{r + l}}.\]

\[5)\ \angle MDO = \angle MAH = 2 \bullet \angle AOH;\]

\[\angle AMH = 90{^\circ} - 2 \bullet \angle AOH;\]

\[\angle ADC = 180{^\circ} - 2 \bullet \angle AOH;\]

\[\angle ADO = 90{^\circ},\ отсюда\ \angle ADC =\]

\[= 180{^\circ} - 2\angle AOH =\]

\[= 90{^\circ} - 2\angle AOH.\]

\[6)\ DK = AO \bullet \cos{\angle ADC} =\]

\[= AO \bullet \cos(90{^\circ} - 2\angle AOH)\]

\[DK = AO \bullet \sin(2\angle AOH) =\]

\[= AO \bullet \sin{\angle AOH} \bullet \cos{\angle AOH}\]

\[DK =\]

\[= r \bullet \sqrt{\frac{l - r}{l + r}} \bullet 2\sqrt{\frac{l - r}{2l}} \bullet \sqrt{\frac{l + r}{2l}} =\]

\[= \frac{2r(l - r)}{2l} = r\frac{l - r}{l}.\]

\[7)\ DK - радиус\ окружности,\ по\ \]

\[которой\ сфера\ касается\ конуса:\ \]

\[c = 2\pi \bullet DK = 2\pi r\frac{l - r}{l}.\]

\[\mathbf{Отв}ет:\ \ 2\pi r\frac{l - r}{l}.\]