\[\boxed{\mathbf{1064.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\angle ACB = \alpha;\]
\[BC = a;\]
\[AC = b.\]
\[\mathbf{Найти:}\]
\[AB - ?\]
\[\mathbf{Решение.}\]
\[По\ теореме\ косинусов:\]
\[AB^{2} =\]
\[= AC^{2} + BC^{2} - 2AC \bullet BC \bullet \cos{\angle C}\]
\[AB^{2} = a^{2} + b^{2} - 2ab \bullet \cos\alpha\]
\[AB = \sqrt{a^{2} + b^{2} - 2ab \bullet \cos\alpha}.\]
\[Ответ:\ \sqrt{a^{2} + b^{2} - 2ab \bullet \cos\alpha}.\]
\[\boxed{\mathbf{1064.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[AC \in OX;\]
\[BD \in OY;\]
\[AC = 4\ см;\]
\[BD = 10\ см.\]
\[\mathbf{Найти:}\]
\[уравнения\ прямых\]
\[AB,\ BC,\ CD\ и\ \text{AD.}\]
\[\mathbf{Решение.}\]
\[1)ABCD - ромб.\ По\ свойству\ \]
\[диагоналей\ ромба:\]
\[AO = OC = 2\ см;\]
\[BO = OD = 5см.\]
\[2)\ A( - 2;0);C(2;0);B(0;5);\]
\[D(0; - 5).\]
\[3)\ A( - 2;0)\ и\ B(0;5):\]
\[\left\{ \begin{matrix} - 2a + c = 0 \\ 5b + c = 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} a = \frac{1}{2}\text{c\ \ \ \ } \\ b = - \frac{1}{5}c \\ \end{matrix} \right.\ \]
\[\left. \ \frac{1}{2}cx - \frac{1}{5}cy + c = 0 \right|:\frac{10}{c}\]
\[5x - 2y + 10 = 0 - прямая\ \text{AB.}\]
\[4)\ CD \parallel AB.\ \ \ \]
\[CD:\]
\[y = \frac{5}{2}x + b\]
\[y(2) = 0;\]
\[5 + b = 0\]
\[- 5 = b\]
\[y = \frac{5}{2}x - 5.\]
\[5x - 2y - 10 = 0 -\]
\[прямая\ \text{CD.}\]
\[5)\ B(0;5)\ и\ C(2;0):\]
\[\left\{ \begin{matrix} 5b + c = 0 \\ 2a + c = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} b = - \frac{1}{5}c \\ a = - \frac{1}{2}c \\ \end{matrix} \right.\ \]
\[\left. \ - \frac{1}{2}cx = - \frac{1}{5}cy + c = 0 \right|:\left( - \frac{10}{c} \right)\]
\[5x + 2y - 10 = 0 - прямая\ \text{BC.}\]
\[6)\ BC \parallel AD.\ \ \ \]
\[AD:\]
\[y = - \frac{5}{2}x + b\]
\[y(0) = - 5 \Longrightarrow b = - 5\]
\[y = - \frac{5}{2}x - 5.\]
\[5x + 2y + 10 = 0 - прямая\ \text{AD.}\]