\[\boxed{\mathbf{1022.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[S_{\text{ABC}} = 60\ см^{2};\]
\[AC = 15\ см;\]
\[\angle A = 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[AB - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABC}} = \frac{1}{2} \bullet AB \bullet AC \bullet \sin{30{^\circ}}.\]
\[2)\ 60 = \frac{1}{2} \bullet 15 \bullet AB \bullet \frac{1}{2}\text{\ \ }\]
\[60 = \frac{15}{4} \bullet AB\ \ \]
\[AB = 60 \bullet \frac{4}{15} = 16\ см.\]
\[Ответ:16\ см.\]
\[\boxed{\mathbf{1022.еуроки - ответы\ на\ пятёрку}}\]
\[A\] | \[(0;0)\] | \[(x; - 3)\] | \[(6;1,5)\] | \[(a;b)\] | \[(1;2)\] |
---|---|---|---|---|---|
\[B\] | \[(1;1)\] | \[(2; - 7)\] | \[(3;1)\] | \[(a + c;d + b)\] | \[(1;2)\] |
\[\overrightarrow{\text{AB}}\] | \[(1;1)\] | \[\{ 5;y\}\] | \[\{ - 3; - \frac{1}{2}\}\] | \[\{ c;d\}\] | \[\{ 0;0\}\] |
\[\overrightarrow{\text{AB}} = \left\{ 2 - x;\ - 7 - ( - 3) \right\} =\]
\[= \left\{ 5;y \right\}:\]
\[2 - x = 5 \Longrightarrow x = - 3;\]
\[- 4 = y \Longrightarrow y = - 4.\]