Решебник по геометрии 10 класс Атанасян ФГОС 235

\[\boxed{\mathbf{235.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\angle ABC = 90{^\circ};\]

\[\angle CAB = \varphi;\]

\[\mathrm{\Delta}A_{1}CB - сечение;\]

\[\angle ACBA_{1} = \theta.\]

\[Найти:\]

\[\frac{S_{бок}}{S_{сеч}}.\]

\[Решение.\]

\[1)\ C_{1}B_{1}\bot B_{1}A_{1};\ \ CB_{1}\bot B_{1}A_{1}:\]

\[\angle CB_{1}C_{1} = \theta - линейный\ угол\ \]

\[двугранного\ угла\ C_{1}A_{1}B_{1}\text{C.}\]

\[2)\ Из\ треугольника\ A_{1}B_{1}C_{1}:\]

\[A_{1}B_{1} = C_{1}B_{1} \cdot tg\ \varphi;\]

\[A_{1}C_{1} = \frac{C_{1}B_{1}}{\cos\varphi}\text{.\ }\]

\[3)\ Из\ треугольника\ CB_{1}C_{1}:\]

\[CC_{1} = C_{1}B_{1} \cdot tg\ \theta;\]

\[B_{1}C = \frac{C_{1}B_{1}}{\cos\theta}.\]

\[3)\ ⊿A_{1}B_{1}C - прямоугольный;\ \ \]

\[A_{1}B_{1}\bot B_{1}C_{1}:\]

\[\angle B_{1} = 90{^\circ}.\]

\[4)\ S_{A_{1}B_{1}C} = \frac{1}{2}A_{1}B_{1} \cdot B_{1}C =\]

\[= \frac{C_{1}B_{1} \cdot tg\varphi \cdot C_{1}B_{1}\ }{2\cos\theta} =\]

\[= \frac{C_{1}B_{1}^{2} \cdot tg\ \varphi}{2\cos\theta};\]

\[S_{бок} = S_{1} + S_{2} + S_{3}.\]

\[5)\ Все\ боковые\ грани\ являются\ \]

\[прямоугольниками:\]

\[S_{1} = S_{AA_{1}B_{1}B} = A_{1}B_{1} \cdot B_{1}B =\]

\[= C_{1}B_{1} \cdot tg\ \varphi \cdot C_{1}B_{1} \cdot tg\ \theta =\]

\[= \left( C_{1}B_{1} \right)^{2} \cdot tg\varphi \cdot tg\theta;\]

\[S_{2} = S_{B_{1}C_{1}\text{CB}} = B_{1}C_{1} \cdot CC_{1} =\]

\[= B_{1}C_{1} \cdot C_{1}B_{1} \cdot tg\theta =\]

\[= \left( C_{1}B_{1} \right)^{2} \cdot tg\theta;\]

\[S_{3} = S_{C_{1}\text{CA}A_{1}} = CC_{1} \cdot A_{1}C_{1} =\]

\[= C_{1}B_{1} \cdot tg\theta \cdot \frac{C_{1}B_{1}}{\cos\varphi} =\]

\[= \frac{C_{1}B_{1}^{2} \cdot tg\theta}{\cos\varphi};\]

\[S_{бок} =\]

\[= \left( C_{1}B_{1} \right)^{2} \cdot tg\theta \cdot \left( tg\varphi + 1 + \frac{1}{\cos\varphi} \right) =\]

\[= \left( C_{1}B_{1} \right)^{2} \cdot tg\theta \cdot \left( \frac{\sin\varphi + \cos\varphi + 1}{\cos\varphi} \right).\]