\[\boxed{\mathbf{595.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - прямоугольный;\]
\[\angle C = 90{^\circ};\]
\[AC = b;\ \]
\[\angle A = \alpha.\]
\[b = 12\ см;\]
\[\alpha = 42{^\circ}\]
\[\mathbf{а)\ Выразить:}\]
\[\text{BC\ },AB\ \angle B\ через\ \alpha\ и\ b;\]
\[\textbf{б)}\ Найти:\]
\[\angle B;BC\ и\ AB - ?\]
\[\mathbf{Решение.}\]
\[\mathbf{а)\ }tg\ \alpha = \frac{\text{BC}}{\text{AC}} \Longrightarrow BC = b \bullet tg\ \alpha;\]
\[\cos\alpha = \frac{\text{AC}}{\text{AB}} \Longrightarrow AB = \frac{b}{\cos\alpha}.\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[\angle B = 90{^\circ} - \text{α.}\]
\[\textbf{б)}\ \angle B = 90{^\circ} - \alpha = 90{^\circ} - 42{^\circ} =\]
\[= 48{^\circ}.\]
\[BC = b \bullet tg\ \alpha = 12 \bullet tg\ 42{^\circ} =\]
\[= 12 \bullet 0,9004 = 10,8\ см.\]
\[AB = \frac{b}{\cos\alpha} = \frac{12}{\cos{42{^\circ}}} =\]
\[= \frac{12}{0,7431} = 16,15\ см.\]
\[\mathbf{Ответ:}\mathbf{а)}\ BC = b \bullet tg\ \alpha;\ \]
\[AB = \frac{b}{\cos\alpha}\ ;\ \angle B = 90{^\circ} - \alpha;\]
\[\textbf{б)}\ \angle B = 48{^\circ};BC = 10,8\ см;\]
\[AB = 16,15\ см.\]