Решебник по геометрии 9 класс Мерзляк Задание 94

\[Рисунок\ в\ учебнике.\]

\[Дано:\]

\[BD = m;\]

\[\angle ABC = \alpha;\]

\[\angle ADC = \beta.\]

\[Найти:\]

\[\text{AC.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle ABD = 180{^\circ} - \angle B = 180{^\circ} - \alpha.\]

\[2)\ В\ \mathrm{\Delta}ABD:\]

\[\angle BAD = 180{^\circ} - \angle ABD - \angle ADB\]

\[\angle BAD = \alpha - \beta;\]

\[\frac{\text{BD}}{\sin{\angle BAD}} = \frac{\text{AB}}{\sin{\angle ADB}}\]

\[AB = \frac{BD \bullet \sin{\angle ADB}}{\sin{\angle BAD}} =\]

\[= \frac{m\sin\beta}{\sin(\alpha - \beta)}.\]

\[3)\ В\ \mathrm{\Delta}ABC:\]

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

\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}};\]

\[AC = AB \bullet \sin{\angle B} = \frac{m\sin\alpha\sin\beta}{\sin(\alpha - \beta)}.\]

\[Ответ:\ \ \frac{m\sin\alpha\sin\beta}{\sin(\alpha - \beta)}.\]