Решебник по геометрии 9 класс Мерзляк Задание 96

\[Схематический\ рисунок.\]

\[Дано:\]

\[\angle A = \alpha;\]

\[\angle B = \beta;\]

\[\angle ADB = \varphi;\]

\[AD = m.\]

\[Найти:\]

\[\text{BC.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABD:\]

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

\[= 180{^\circ} - \varphi;\]

\[\sin{\angle ADC} = \sin(180{^\circ} - \varphi) =\]

\[= \sin\varphi.\]

\[2)\ В\ \mathrm{\Delta}ABC:\]

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

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

\[\sin{\angle C} = \sin(180{^\circ} - \alpha - \beta) =\]

\[= \sin(\alpha + \beta).\]

\[3)\ В\ \mathrm{\Delta}ACD:\]

\[\frac{\text{AC}}{\sin{\angle ADC}} = \frac{\text{AD}}{\sin{\angle ACD}}\]

\[AC = \frac{AD \bullet \sin{\angle ADC}}{\sin{\angle ACD}} =\]

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

\[4)\ В\ \mathrm{\Delta}ABC:\]

\[\frac{\text{BC}}{\sin{\angle A}} = \frac{\text{AC}}{\sin{\angle B}}\]

\[BC = \frac{AC \bullet \sin{\angle A}}{\sin{\angle B}} =\]

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

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