\[\boxed{\mathbf{1027.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AD\bot CB;\]
\[\angle A = 45{^\circ};\]
\[\angle C = 30{^\circ};\]
\[AD = 3\ м.\]
\[\mathbf{Найти:}\]
\[AB;BC;AC - ?\]
\[\mathbf{Решение.}\]
\[1)\ \ \mathrm{\Delta}ADC - прямоугольный:\]
\[\angle A = 90{^\circ} - 30{^\circ} = 60{^\circ}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[AC = 2AD = 2 \bullet 3 = 6\ м.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}ACB:\]
\[\angle CBA = 180{^\circ} - (30{^\circ} + 45{^\circ}) =\]
\[= 105{^\circ}.\]
\[3)\ По\ теореме\ синусов:\]
\[\frac{\text{CB}}{\sin{45{^\circ}}} = \frac{\text{AB}}{\sin{30{^\circ}}} = \frac{\text{AC}}{\sin{105{^\circ}}};\]
\[\frac{\text{AB}}{\sin{30{^\circ}}} = \frac{6}{\sin{105{^\circ}}} \Longrightarrow AB =\]
\[= \frac{6\sin{30{^\circ}}}{\sin{105{^\circ}}} = \frac{3}{0,96} \approx 3,1\ м;\]
\[\frac{\text{CB}}{\sin{45{^\circ}}} = \frac{6}{\sin{105{^\circ}}} \Longrightarrow CB =\]
\[= \frac{6\sin{45{^\circ}}}{\sin{105{^\circ}}} = \frac{4,24}{0,96} \approx 4,4\ м.\]
\[Ответ:AB = 3,1\ м;AC = 6\ м;\]
\[CB = 4,4\ м.\]
\[\boxed{\mathbf{1027.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{а)\ }\text{A\ }(2;7)\ и\ \text{B\ }( - 2;7):\]
\[AB = \sqrt{( - 2 - 2)^{2} + (7 - 7)^{2}} =\]
\[= \sqrt{16} = 4.\]
\[\mathbf{б}\mathbf{)\ }\text{A\ }( - 5;1)\ и\ \text{B\ }( - 5; - 7):\ \ \ \]
\[AB = \sqrt{( - 5 + 5)^{2} + ( - 7 - 1)^{2}} =\]
\[= \sqrt{64} = 8.\]
\[\mathbf{в)\ }\text{A\ }( - 3;0)\ и\ \text{B\ }(0;4):\ \ \ \]
\[AB = \sqrt{(0 + 3)^{2} + (4 - 0)^{2}} =\]
\[= \sqrt{25} = 5.\]
\[\mathbf{г)\ }\text{A\ }(0;3)\ и\ \text{B\ }( - 4;0):\ \ \ \]
\[AB = \sqrt{( - 4 - 0)^{2} + (0 - 3)^{2}} =\]
\[= \sqrt{25} = 5.\]