\[\boxed{\mathbf{1057.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равнобедренный;\]
\[AB = AC = b;\]
\[\angle A = 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[BE,\ AD,\ AE,\ EC,\ BC - ?\]
\[\mathbf{Решение.}\]
\[1)\ Рассмотрим\ \mathrm{\Delta}ABE:\]
\[\angle ABE = 90{^\circ} - 30{^\circ} = 60{^\circ};\ \]
\[\angle A = 30{^\circ} \Longrightarrow \ \]
\[2)\ AE = \sqrt{AB^{2} - BE^{2}} =\]
\[= \sqrt{b^{2} - \frac{b^{2}}{4}} = \frac{b\sqrt{3}}{2}.\]
\[3)\ CE = AC - AE = b - \frac{b\sqrt{3}}{2} =\]
\[= \frac{b\left( 2 - \sqrt{3} \right)}{2}.\]
\[4)\ Рассмотрим\ \mathrm{\Delta}EBC:\]
\[CB = \sqrt{BE^{2} + CE^{2}} =\]
\[= \sqrt{\frac{b^{2}}{4} + \frac{b^{2}\left( 2 - \sqrt{3} \right)^{2}}{4}} =\]
\[= \sqrt{\frac{b^{2} + b^{2}\left( 4 - 2\sqrt{3} + 3 \right)}{4}} =\]
\[= \sqrt{\frac{b^{2} + 4b^{2} - 4\sqrt{3}b^{2} + 3b^{2}}{4}} =\]
\[= \sqrt{\frac{8b^{2} - 4\sqrt{3}b^{2}}{4}} =\]
\[= \sqrt{\frac{4b^{2}\left( 2 - \sqrt{3} \right)}{4}} =\]
\[= \sqrt{b^{2}\left( 2 - \sqrt{3} \right)} = b\sqrt{2 - \sqrt{3}};\]
\[5)\ Рассмотрим\ \mathrm{\Delta}ADC:\]
\[AD = \sqrt{AC^{2} - CD^{2}} =\]
\[= \sqrt{b^{2} - \frac{b^{2}\left( 2 - \sqrt{3} \right)}{4}} =\]
\[= \sqrt{\frac{4b^{2} - 2b^{2} + \sqrt{3}b^{2}}{4}} =\]
\[= \sqrt{\frac{2b^{2} + \sqrt{3}b^{2}}{4}} =\]
\[= \sqrt{\frac{b^{2}\left( 2 + \sqrt{3} \right)}{4}} = \frac{b\sqrt{2 + \sqrt{3}}}{2}.\]
\[\mathbf{Ответ:\ }BE = \frac{b}{2};\ \ \]
\[AD = \frac{b\sqrt{2 + \sqrt{3}}}{2};\ \ AE = \frac{b\sqrt{3}}{2};\]
\[EC = \frac{b\left( 2 - \sqrt{3} \right)}{2};\ \ \]
\[BC = b\sqrt{2 - \sqrt{3}}.\]
\[\boxed{\mathbf{1057.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[Окружность\ (O;R);\]
\[A( - 3;0);\]
\[B(0;9);\ \]
\[A \in (O;R);\]
\[B \in (O;R);\]
\[O \in OY.\]
\[\mathbf{Найти:}\]
\[уравнение\ окружности.\]
\[\mathbf{Решение.}\]
\[1)\ Пусть\ точка\ O\ имеет\ \]
\[координаты\ (0;y).\]
\[2)\ Так\ как\ \text{A\ }и\ \text{B\ }принадлежат\ \]
\[окружности:\]
\[OA = OB = R.\]
\[3)\ OA = \ \]
\[= \sqrt{( - 3 - 0)^{2} + (0 - y)^{2}} =\]
\[= \sqrt{9 + y^{2}};\]
\[OB = \sqrt{(0 - 0)^{2} + (9 - y)^{2}} =\]
\[= \sqrt{(9 - y)^{2}}.\]
\[4)\ \sqrt{9 + y^{2}} = \sqrt{(9 - y)^{2}}\]
\[9 + y^{2} = (9 - y)^{2}\]
\[9 + y^{2} = 81 - 18y + y^{2}\]
\[18y = 72\]
\[y = 4 \Longrightarrow O(0;4).\]
\[5)\ OA = R = \sqrt{9 + 4^{2}} =\]
\[= \sqrt{9 + 16} = \sqrt{25} = 5;\]
\[x^{2} + {(y - 4)}^{2} = 25.\ \ \]
\[Ответ:\ x^{2} + {(y - 4)}^{2} = 25.\]