\[\boxed{\mathbf{1057.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равнобедренный;\]
\[AB = AC = b;\]
\[\angle A = 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[BE,\ AD,\ AE,\ EC,\ BC - ?\]
\[\mathbf{Решение.}\]
\[1)\ Рассмотрим\ \mathrm{\Delta}ABE:\]
\[\angle ABE = 90{^\circ} - 30{^\circ} = 60{^\circ};\ \]
\[\angle A = 30{^\circ} \Longrightarrow \ \]
\[2)\ AE = \sqrt{AB^{2} - BE^{2}} =\]
\[= \sqrt{b^{2} - \frac{b^{2}}{4}} = \frac{b\sqrt{3}}{2}.\]
\[3)\ CE = AC - AE = b - \frac{b\sqrt{3}}{2} =\]
\[= \frac{b\left( 2 - \sqrt{3} \right)}{2}.\]
\[4)\ Рассмотрим\ \mathrm{\Delta}EBC:\]
\[CB = \sqrt{BE^{2} + CE^{2}} =\]
\[= \sqrt{\frac{b^{2}}{4} + \frac{b^{2}\left( 2 - \sqrt{3} \right)^{2}}{4}} =\]
\[= \sqrt{\frac{b^{2} + b^{2}\left( 4 - 2\sqrt{3} + 3 \right)}{4}} =\]
\[= \sqrt{\frac{b^{2} + 4b^{2} - 4\sqrt{3}b^{2} + 3b^{2}}{4}} =\]
\[= \sqrt{\frac{8b^{2} - 4\sqrt{3}b^{2}}{4}} =\]
\[= \sqrt{\frac{4b^{2}\left( 2 - \sqrt{3} \right)}{4}} =\]
\[= \sqrt{b^{2}\left( 2 - \sqrt{3} \right)} = b\sqrt{2 - \sqrt{3}};\]
\[5)\ Рассмотрим\ \mathrm{\Delta}ADC:\]
\[AD = \sqrt{AC^{2} - CD^{2}} =\]
\[= \sqrt{b^{2} - \frac{b^{2}\left( 2 - \sqrt{3} \right)}{4}} =\]
\[= \sqrt{\frac{4b^{2} - 2b^{2} + \sqrt{3}b^{2}}{4}} =\]
\[= \sqrt{\frac{2b^{2} + \sqrt{3}b^{2}}{4}} =\]
\[= \sqrt{\frac{b^{2}\left( 2 + \sqrt{3} \right)}{4}} = \frac{b\sqrt{2 + \sqrt{3}}}{2}.\]
\[\mathbf{Ответ:\ }BE = \frac{b}{2};\ \ \]
\[AD = \frac{b\sqrt{2 + \sqrt{3}}}{2};\ \ AE = \frac{b\sqrt{3}}{2};\]
\[EC = \frac{b\left( 2 - \sqrt{3} \right)}{2};\ \ \]
\[BC = b\sqrt{2 - \sqrt{3}}.\]