Решебник по геометрии 9 класс Мерзляк Задание 93

\[Рисунок\ в\ учебнике.\]

\[Дано:\]

\[CD = a;\]

\[\angle BAC = \gamma;\]

\[\angle DBA = \beta.\]

\[Найти:\]

\[\text{AD.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABD:\]

\[\angle D = 180{^\circ} - \angle A - \angle B =\]

\[= 180{^\circ} - \gamma - \beta;\]

\[\angle CDB = 180{^\circ} - \angle D = \gamma + \beta.\]

\[2)\ В\ \mathrm{\Delta}BCD:\]

\[\angle C = 90{^\circ};\ \ \ \]

\[\cos{\angle CDB} = \frac{\text{CD}}{\text{BD}};\]

\[BD = \frac{\text{CD}}{\cos{\angle CDB}} = \frac{a}{\cos(\beta + \gamma)}.\]

\[3)\ В\ \mathrm{\Delta}ABD:\]

\[\frac{\text{AD}}{\sin{\angle B}} = \frac{\text{BD}}{\sin{\angle A}}\]

\[AD = \frac{BD \bullet \sin{\angle B}}{\sin{\angle A}}\]

\[AD = \frac{a\sin\beta}{\cos{(\beta + \gamma)\sin\gamma}}.\]

\[Ответ:\ \ \frac{a\sin\beta}{\cos{(\beta + \gamma)\sin\gamma}}.\]