\[\boxed{\mathbf{1029.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[BC = a;\ \]
\[\angle B = \alpha;\]
\[\angle C = \beta.\]
\[\mathbf{Найти:}\]
\[CC_{1};BB_{1};AA_{1} - биссектрисы.\]
\[\mathbf{Решение.}\]
\[1)\ Рассмотрим\ \mathrm{\Delta}BB_{1}C:\]
\[\angle BB_{1}C = 180{^\circ} - \frac{\alpha}{2} - \beta.\]
\[По\ теореме\ синусов:\]
\[\frac{\text{BC}}{\sin{\angle BB_{1}C}} = \frac{BB_{1}}{\sin\beta}\]
\[BB_{1} = \frac{{a \bullet sin}\beta}{\sin\left( \beta + \frac{\alpha}{2} \right)}.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}BCC_{1}:\]
\[\angle BC_{1}C = 180{^\circ} - \alpha - \frac{\beta}{2}.\]
\[По\ теореме\ синусов:\]
\[\frac{\text{BC}}{\sin{\angle BC_{1}C}} = \frac{CC_{1}}{\sin\alpha}\]
\[CC_{1} = \frac{{a \bullet sin}\alpha}{\sin\left( \alpha + \frac{\beta}{2} \right)}.\]
\[3)\ Рассмотрим\ \mathrm{\Delta}BAA_{1}:\]
\[\angle BAA_{1} = 90{^\circ} - \frac{\beta + \alpha}{2};\ \]
\[\angle BA_{1}A = 90{^\circ} + \frac{\beta - \alpha}{2}.\]
\[По\ теореме\ синусов:\]
\[\frac{AA_{1}}{\sin{\angle B}} = \frac{\text{AB}}{\sin{\angle A_{1}}}\]
\[AA_{1} = \frac{AB \bullet \sin\alpha}{\sin\left( 90{^\circ} + \frac{\beta - \alpha}{2} \right)};\]
\[AB = \frac{a \bullet \sin\beta}{\sin(\alpha + \beta)};\]
\[AA_{1} = \frac{\frac{a \bullet \sin\beta}{\sin(\alpha + \beta)} \bullet \sin\alpha}{\sin\left( 90{^\circ} + \frac{\beta - \alpha}{2} \right)} =\]
\[= \frac{a \bullet \sin\alpha \bullet \sin\beta}{\sin(\alpha + \beta) \bullet \cos\left( \frac{\beta - \alpha}{2} \right)}.\]
\[Ответ:CC_{1} = \frac{{a \bullet sin}\alpha}{\sin\left( \alpha + \frac{\beta}{2} \right)};\ \]
\[\text{\ B}B_{1} = \frac{{a \bullet sin}\beta}{\sin\left( \beta + \frac{\alpha}{2} \right)};\]
\[AA_{1} = \frac{a \bullet \sin\alpha \bullet \sin\beta}{\sin(\alpha + \beta) \bullet \cos\left( \frac{\beta - \alpha}{2} \right)}.\]
\[\boxed{\mathbf{1029.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\text{A\ }(0;1);\]
\[B(1; - 4);\]
\[C(5;2).\]
\[\mathbf{Найти:}\]
\[AM - ?\]
\[\mathbf{Решение.}\]
\[\left\{ \begin{matrix} x_{M} = \frac{x_{B} + x_{C}}{2} \\ y_{M} = \frac{y_{B} + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{M} = \frac{1 + 5}{2}\text{\ \ \ \ } \\ y_{M} = \frac{- 4 + 2}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{M} = 3\ \ \ \\ y_{M} = - 1 \\ \end{matrix} \right.\ \ \Longrightarrow M(3; - 1).\]
\[AM =\]
\[= \sqrt{(3 - 0)^{2} + ( - 1 - 1)^{2}} =\]
\[= \sqrt{9 + 4} = \sqrt{13}.\]
\[\mathbf{Ответ:}AM = \sqrt{13}\mathbf{.}\]