\[\boxed{\mathbf{654.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ x = \frac{360{^\circ} - 152{^\circ} - 80{^\circ}}{2} =\]
\[\textbf{б)}\ x = 360{^\circ} - 125{^\circ} - 30{^\circ} \bullet 2 =\]
\[= 360{^\circ} - 185{^\circ} = 175{^\circ}.\]
\[\textbf{в)}\ x = \frac{360{^\circ} - 112{^\circ} - 180{^\circ}}{2} =\]
\[= \frac{68{^\circ}}{2} = 34{^\circ}\ \]
\[(по\ теореме\ о\ вписанном\ угле).\]
\[\textbf{г)}\ x = 360{^\circ} - 215{^\circ} - 20{^\circ} \bullet 2 =\]
\[= 360{^\circ} - 255{^\circ} = 105{^\circ}.\]
\[Ответ:а)\ 64{^\circ};б)\ 175{^\circ};в)\ 34{^\circ};\]
\[\textbf{г)}\ 105{^\circ}.\]
\[\boxed{\mathbf{654.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}.\]
\[\mathbf{Доказать:}\]
\[\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} = k.\]
\[\mathbf{Доказательство.}\]
\[1)\ \mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ \]
\[(по\ условию):\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{\text{BC}}{B_{1}C_{1}} = \frac{\text{AC}}{A_{1}C_{1}} = k.\]
\[2)\ AB = k \bullet A_{1}B_{1};\]
\[BC = k \bullet B_{1}C_{1};\]
\[AC = k \bullet A_{1}C_{1}.\]
\[3)\ P_{\text{ABC}} =\]
\[= k \bullet A_{1}B_{1} + k \bullet B_{1}C_{1} + k \bullet A_{1}C_{1};\]
\[P_{\text{ABC}} = k\left( A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1} \right).\]
\[4)\ P_{A_{1}B_{1}C_{1}} = A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1}.\]
\[5)\frac{P_{\text{ABC}}}{P_{A_{1}B_{1}C_{1}}} =\]
\[= \frac{k\left( A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1} \right)}{A_{1}B_{1} + B_{1}C_{1} + A_{1}C_{1}} = k.\]
\[Что\ и\ требовалось\ доказать.\]