\[\boxed{\mathbf{594.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC -\]
\[прямоугольный;\]
\[\angle C = 90{^\circ};\]
\[AC = b;\ \angle B = \beta;\]
\[b = 10\ см;\]
\[\beta = 50{^\circ}.\]
\[\mathbf{а)\ Выразить:}\]
\[\text{BC\ },AB\ \angle A\ через\ b\ и\ \text{β.}\]
\[\textbf{б)}\ Найти:\]
\[\angle A;BC\ и\ AB - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ tg\ \beta = \frac{\text{AC}}{\text{BC}} \Longrightarrow BC = \frac{b}{\text{tg\ β}};\]
\[\sin\beta = \frac{\text{AC}}{\text{AB}} \Longrightarrow AB = \frac{b}{\sin\beta}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[\angle A = 90{^\circ} - \text{β.}\]
\[\textbf{б)}\ \angle A = 90{^\circ} - \beta = 90{^\circ} - 50{^\circ} =\]
\[= 40{^\circ}.\]
\[BC = \frac{b}{\text{tg\ β}} = \frac{10}{tg\ 50{^\circ}} = \frac{10}{1,1918} =\]
\[= 8,39\ см.\]
\[AB = \frac{b}{\sin\beta} = \frac{10}{\sin{50{^\circ}}} = \frac{10}{0,766} =\]
\[= 13,05\ см.\]
\[\mathbf{Ответ:}\mathbf{а)}\ BC = \frac{b}{\text{tg\ β}};\ \]
\[AB = \frac{b}{\sin\beta}\ ;\ \angle A = 90{^\circ} - \beta;\]
\[\textbf{б)}\ \angle A = 40{^\circ};BC = 8,39\ см;\]
\[AB = 13,05\ см.\]