\[\boxed{\mathbf{594.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC -\]
\[прямоугольный;\]
\[\angle C = 90{^\circ};\]
\[AC = b;\ \angle B = \beta;\]
\[b = 10\ см;\]
\[\beta = 50{^\circ}.\]
\[\mathbf{а)\ Выразить:}\]
\[\text{BC\ },AB\ \angle A\ через\ b\ и\ \text{β.}\]
\[\textbf{б)}\ Найти:\]
\[\angle A;BC\ и\ AB - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ tg\ \beta = \frac{\text{AC}}{\text{BC}} \Longrightarrow BC = \frac{b}{\text{tg\ β}};\]
\[\sin\beta = \frac{\text{AC}}{\text{AB}} \Longrightarrow AB = \frac{b}{\sin\beta}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[\angle A = 90{^\circ} - \text{β.}\]
\[\textbf{б)}\ \angle A = 90{^\circ} - \beta = 90{^\circ} - 50{^\circ} =\]
\[= 40{^\circ}.\]
\[BC = \frac{b}{\text{tg\ β}} = \frac{10}{tg\ 50{^\circ}} = \frac{10}{1,1918} =\]
\[= 8,39\ см.\]
\[AB = \frac{b}{\sin\beta} = \frac{10}{\sin{50{^\circ}}} = \frac{10}{0,766} =\]
\[= 13,05\ см.\]
\[\mathbf{Ответ:}\mathbf{а)}\ BC = \frac{b}{\text{tg\ β}};\ \]
\[AB = \frac{b}{\sin\beta}\ ;\ \angle A = 90{^\circ} - \beta;\]
\[\textbf{б)}\ \angle A = 40{^\circ};BC = 8,39\ см;\]
\[AB = 13,05\ см.\]
\[\boxed{\mathbf{594.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[CD\bot AB;\]
\[AD = BC;\]
\[AB = 3;\]
\[CD = \sqrt{3}.\]
\[\mathbf{Найти:}\]
\[AC - ?\]
\[\mathbf{Решение.}\]
\[Пусть\ BC = AD = x.\]
\[1)\ ⊿CBD - прямоугольный.\ \]
\[По\ теореме\ Пифагора:\]
\[BC^{2} = DC^{2} + DB^{2}\]
\[x^{2} = \left( \sqrt{3} \right)^{2} + (3 - x)^{2}\]
\[x^{2} = 3 + 9 - 6x + x^{2}\]
\[x^{2} = 12 - 6x + x^{2}\]
\[12 - 6x = 0\]
\[x = 2.\]
\[BC = AD = 2\ см.\]
\[3)\ ⊿ADC - прямоугольный.\ \]
\[По\ теореме\ Пифагора:\]
\[AC^{2} = DC^{2} + AD^{2}\]
\[AC^{2} = 3 + 4 = 7;\]
\[AC = \sqrt{7}.\]
\[\mathbf{Ответ}:\sqrt{7}\mathbf{.}\]