\[\boxed{\mathbf{1037.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\angle\text{CAB} = 12{^\circ}30';\]
\[\angle ABC = 72{^\circ}42';\]
\[AB = 70\ м;\]
\[DC\bot AB.\]
\[\mathbf{Найти:}\]
\[CD - ?\]
\[\mathbf{Решение.}\]
\[1)\ В\ треугольнике\ ADC:\]
\[tg\ 12{^\circ}30^{'} = \frac{\text{DC}}{\text{AD}}\]
\[CD = AD \bullet tg\ 12{^\circ}30^{'}.\]
\[В\ треугольнике\ BDC:\]
\[tg\ 72{^\circ}42^{'} = \frac{\text{CD}}{\text{DB}}\]
\[CD = DB \bullet tg\ 72{^\circ}42^{'}.\]
\[2)\ Пусть\ AD = x;\ DB = 70 - x:\]
\[x \bullet tg\ 12{^\circ}30^{'} =\]
\[= (70 - x) \bullet tg\ 72{^\circ}42^{'}\]
\[x \bullet 0,2217 = (70 - x) \bullet 3,21\]
\[x \bullet 0,2217 = 224,7 - 3,21x\]
\[3,4317x = 224,7\]
\[x = 65,48\ м.\]
\[3)\ CD = 65,48 \bullet 0,2217 =\]
\[= 14,52\ м.\]
\[Ответ:14,52\ м.\]