ГДЗ по геометрии 7 класс Атанасян Задание 1029

\[\boxed{\mathbf{1029.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\]

\[BC = a;\ \]

\[\angle B = \alpha;\]

\[\angle C = \beta.\]

\[\mathbf{Найти:}\]

\[CC_{1};BB_{1};AA_{1} - биссектрисы.\]

\[\mathbf{Решение.}\]

\[1)\ Рассмотрим\ \mathrm{\Delta}BB_{1}C:\]

\[\angle BB_{1}C = 180{^\circ} - \frac{\alpha}{2} - \beta.\]

\[По\ теореме\ синусов:\]

\[\frac{\text{BC}}{\sin{\angle BB_{1}C}} = \frac{BB_{1}}{\sin\beta}\]

\[BB_{1} = \frac{{a \bullet sin}\beta}{\sin\left( \beta + \frac{\alpha}{2} \right)}.\]

\[2)\ Рассмотрим\ \mathrm{\Delta}BCC_{1}:\]

\[\angle BC_{1}C = 180{^\circ} - \alpha - \frac{\beta}{2}.\]

\[По\ теореме\ синусов:\]

\[\frac{\text{BC}}{\sin{\angle BC_{1}C}} = \frac{CC_{1}}{\sin\alpha}\]

\[CC_{1} = \frac{{a \bullet sin}\alpha}{\sin\left( \alpha + \frac{\beta}{2} \right)}.\]

\[3)\ Рассмотрим\ \mathrm{\Delta}BAA_{1}:\]

\[\angle BAA_{1} = 90{^\circ} - \frac{\beta + \alpha}{2};\ \]

\[\angle BA_{1}A = 90{^\circ} + \frac{\beta - \alpha}{2}.\]

\[По\ теореме\ синусов:\]

\[\frac{AA_{1}}{\sin{\angle B}} = \frac{\text{AB}}{\sin{\angle A_{1}}}\]

\[AA_{1} = \frac{AB \bullet \sin\alpha}{\sin\left( 90{^\circ} + \frac{\beta - \alpha}{2} \right)};\]

\[AB = \frac{a \bullet \sin\beta}{\sin(\alpha + \beta)};\]

\[AA_{1} = \frac{\frac{a \bullet \sin\beta}{\sin(\alpha + \beta)} \bullet \sin\alpha}{\sin\left( 90{^\circ} + \frac{\beta - \alpha}{2} \right)} =\]

\[= \frac{a \bullet \sin\alpha \bullet \sin\beta}{\sin(\alpha + \beta) \bullet \cos\left( \frac{\beta - \alpha}{2} \right)}.\]

\[Ответ:CC_{1} = \frac{{a \bullet sin}\alpha}{\sin\left( \alpha + \frac{\beta}{2} \right)};\ \]

\[\text{\ B}B_{1} = \frac{{a \bullet sin}\beta}{\sin\left( \beta + \frac{\alpha}{2} \right)};\]

\[AA_{1} = \frac{a \bullet \sin\alpha \bullet \sin\beta}{\sin(\alpha + \beta) \bullet \cos\left( \frac{\beta - \alpha}{2} \right)}.\]