\[\boxed{\mathbf{643.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[окружность\ (O,\ R).\]
\[AB\ и\ \text{AC} - касательные;\]
\[AB = 5\ см;\]
\[\angle OAB = 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[BC - ?\]
\[\mathbf{Решение.}\]
\[1)\ AB - касательная\ \]
\[(по\ условию):\]
\[OB\bot AB \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}AOB - прямоугольный.\]
\[2)\ В\ \mathrm{\Delta}AOB:\]
\[OB = \frac{1}{2}\text{OA\ }(так\ как\ \angle A = 30{^\circ}).\]
\[3)\ Пусть\ BO = x;\ AO = 2x.\ \]
\[По\ теореме\ Пифагора:\]
\[AO^{2} = AB^{2} = OB^{2}\ \]
\[4x^{2} - x^{2} = 25\]
\[3x^{2} = 25\]
\[x^{2} = \frac{25}{3}\]
\[x = BO = \frac{5\sqrt{3}}{3}.\]
\[4)\ \angle BOA = 90{^\circ} - \angle BAC =\]
\[= 90{^\circ} - 30{^\circ} = 60{^\circ};\]
\[\angle BOE = 60{^\circ} \Longrightarrow \angle OBE =\]
\[= 90{^\circ} - 60{^\circ} = 30{^\circ}.\]
\[5)\ В\ \mathrm{\Delta}BOE:\]
\[BE = BO \bullet \cos{\angle B} =\]
\[= \frac{5\sqrt{3}}{3} \bullet \cos{30{^\circ}} = \frac{5\sqrt{3}}{3} \bullet \frac{\sqrt{3}}{2} =\]
\[= \frac{5}{2}\ см.\]
\[6)\ BC = 2BE = \frac{5}{2} \bullet 2 = 5\ см.\]
\[Ответ:BC = 5\ см.\]
\[\boxed{\mathbf{643.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[BD - биссектриса\ \angle B;\]
\[\textbf{а)}\ BC = 9\ см;\]
\[AD = 7,5\ см;\]
\[DC = 4,5\ см;\]
\[\textbf{б)}\ AB = 30;\]
\[AD = 20;\]
\[BC = 16.\]
\[\mathbf{Найти:}\]
\[\textbf{а)}\ AB - ?\]
\[\textbf{б)}\ DC - ?\]
\[\mathbf{Решение.}\]
\[1)\ BH - высота,\ общая\ для\ \]
\[\mathrm{\Delta}\text{ABD\ }и\ \mathrm{\Delta}BCD:\ \]
\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{\text{AD}}{\text{CD}}.\]
\[2)\ \angle ABD = \angle DBC:\]
\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{AB \bullet BD}{BD \bullet BC} = \frac{\text{AB}}{\text{BC}}.\]
\[3)\ \frac{\text{AD}}{\text{CD}} = \frac{\text{AB}}{\text{BC}}\]
\[AD \bullet BC = CD \bullet AB\]
\[\frac{\text{AB}}{\text{AD}} = \frac{\text{BC}}{\text{CD}}.\]
\[\textbf{а)}\ AB = \frac{AD \bullet BC}{\text{CD}} = \frac{7,5 \bullet 9}{4,5} =\]
\[= 15\ см.\]
\[\textbf{б)}\ DC = \frac{AD \bullet BC}{\text{AB}} = \frac{20 \bullet 16}{30} =\]
\[= 10\frac{2}{3}.\]
\[\mathbf{Ответ:}\mathbf{а)}\ 15\ см;б)\ 10\frac{2}{3}.\]