Решебник по геометрии 7 класс Атанасян ФГОС Задание 592

 

\[\boxed{\mathbf{592.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\textbf{а)}\ tg\ \alpha = \frac{1}{2}:\ \ \ \]

\[tg\ \alpha = \frac{\text{BC}}{\text{AC}} = \frac{1}{2}\]

\[BC = 1;\ AC = 2.\]

\[\textbf{б)}\ tg\ \alpha = \frac{3}{4}:\ \ \ \]

\[tg\ \alpha = \frac{\text{BC}}{\text{AC}} = \frac{3}{4}\]

\[BC = 3;\ AC = 4.\]

\[\textbf{в)}\cos\alpha = \frac{2}{10}:\]

\[\cos\alpha = \frac{\text{AC}}{\text{AB}} = \frac{1}{5}\]

\[AC = 1;\ AB = 5.\]

\[\textbf{г)}\cos\alpha = \frac{2}{3}:\]

\[\cos\alpha = \frac{\text{AC}}{\text{AB}} = \frac{2}{3}\]

\[\ AC = 2;\ AB = 3.\]

\[\textbf{д)}\sin\alpha = \frac{1}{2}:\]

\[\sin\alpha = \frac{\text{BC}}{\text{AB}} = \frac{1}{2}\]

\[BC = 1;\ AB = 2.\]

\[\textbf{е)}\sin\alpha = \frac{4}{10}:\ \ \]

\[\sin\alpha = \frac{\text{BC}}{\text{AB}} = \frac{2}{5}\]

\[BC = 2;\ AB = 5.\]

\[\boxed{\mathbf{592.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - ромб;\]

\[AB = 10\ см;\]

\[AC = 12\ см.\]

\[\mathbf{Найти:}\]

\[S_{\text{ABCD}} - ?\]

\[DB - ?\]

\[\mathbf{Решение.}\]

\[1)\ По\ свойству\ ромба:\]

\[DO = OB;\]

\[AO = OC = \frac{\text{AC}}{2} = 6\ см.\]

\[2)\ AC\bot DB \Longrightarrow ⊿AOB -\]

\[прямоугольный.\]

\[3)\ По\ теореме\ Пифагора:\ \]

\[OB^{2} = AB^{2} - AO^{2}\]

\[OB^{2} = 100 - 36 = 64\]

\[OB = 8\ см.\]

\[DB = OB \bullet 2 = 16\ см.\]

\[4)\ S_{\text{ABCD}} = \frac{1}{2} \bullet AC \bullet BD =\]

\[= \frac{1}{2} \bullet 16 \bullet 12 = 96\ см^{2}.\]

\[\mathbf{Ответ}:S_{\text{ABCD}} = 96\ см^{2};\]

\[DB = 16\ см\mathbf{.}\]