\[\boxed{\mathbf{1024.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\ \]
\[\angle A = \alpha;\]
\[BB_{1}\ и\ CC_{1} - высоты;\]
\[BB_{1} = h_{B};\ \]
\[CC_{1} = h_{C}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ Рассмотрим\ \mathrm{\Delta}ABB_{1}:\]
\[AB = \frac{BB_{1}}{\sin\alpha} = \frac{h_{B}}{\sin\alpha}.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}AC_{1}C:\]
\[AC = \frac{CC_{1}}{\sin\alpha} = \frac{h_{C}}{\sin\alpha}.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2} \bullet AB \bullet AC \bullet \sin{\alpha;}\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet \frac{h_{B}}{\sin\alpha} \bullet \frac{h_{C}}{\sin\alpha} \bullet \sin\alpha =\]
\[= \frac{h_{B} \bullet h_{C}}{2\sin\alpha}.\]
\[Ответ:\frac{h_{B} \bullet h_{C}}{2\sin\alpha}.\]
\[\mathbf{б)\ Рисунок\ по\ условию\ задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\ \]
\[\angle A = \alpha;\]
\[\angle B = \beta;\]
\[BB_{1} = h;\]
\[BB_{1} - высота.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ Рассмотрим\ \mathrm{\Delta}ABB_{1}:\]
\[AB = \frac{BB_{1}}{\sin\alpha} = \frac{h}{\sin\alpha}.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}BB_{1}C:\]
\[BC = \frac{BB_{1}}{\sin{\angle C}} =\]
\[= \frac{BB_{1}}{\sin{(180{^\circ} - (\alpha + \beta)}} =\]
\[= \frac{h}{\sin(\alpha + \beta)}.\]
\[3)\ S_{\text{ABC}} = \frac{1}{2} \bullet AB \bullet BC \bullet \sin{\beta;}\]
\[S_{\text{ABC}} =\]
\[= \frac{1}{2} \bullet \frac{h}{\sin\alpha} \bullet \frac{h}{\sin(\alpha + \beta)} \bullet \sin\beta =\]
\[= \frac{h^{2}\sin\beta}{2\sin{\alpha \bullet \sin{(\alpha + \beta)}}}.\]
\[Ответ:\ \frac{h^{2}\sin\beta}{2\sin{\alpha \bullet \sin{(\alpha + \beta)}}}.\]
\[\boxed{\mathbf{1024.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\text{A\ }(0;1);\]
\[\text{B\ }(5; - 3);\]
\[AB = BC;\]
\[BD = DC;\]
\[B \in AC;\ \]
\[D \in BC.\]
\[\mathbf{Найти:}\]
\[координаты\ C\ и\ \text{D.}\]
\[\mathbf{Решение.}\]
\[1)\ \left\{ \begin{matrix} x_{B} = \frac{x_{A} + x_{C}}{2} \\ y_{B} = \frac{y_{A} + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 5 = \frac{0 + x_{C}}{2}\text{\ \ \ \ } \\ - 3 = \frac{1 + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} \frac{x_{C}}{2} = 5\ \ \ \ \ \ \ \\ \frac{y_{C}}{2} = - 3,5 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x_{C} = 10\ \\ y_{C} = - 7 \\ \end{matrix} \Longrightarrow C\ (10; - 7). \right.\ \]
\[2)\ \left\{ \begin{matrix} x_{D} = \frac{x_{B} + x_{C}}{2} \\ y_{D} = \frac{y_{B} + y_{c}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{D} = \frac{5 + 10}{2}\text{\ \ \ \ \ \ \ \ } \\ y_{D} = \frac{- 3 + ( - 7)}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{D} = 7,5 \\ y_{D} = - 5 \\ \end{matrix}\ \Longrightarrow D\ (7,5;\ - 5). \right.\ \]
\(Ответ:C\ (10; - 7);D\ (7,5; - 5).\)