Решебник по геометрии 9 класс Мерзляк Задание 157

\[Схематический\ рисунок.\]

\[Дано:\]

\[BC = a;\]

\[\angle B = \beta;\]

\[\angle C = \gamma.\]

\[Найти:\]

\[S_{\text{ABC}}.\]

\[Решение.\]

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

\[\sin{\angle A} = \sin(\angle B + \angle C) =\]

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

\[\frac{\text{BC}}{\sin{\angle A}} = \frac{\text{AB}}{\sin{\angle C}}\]

\[AB = \frac{BC \bullet \sin{\angle C}}{\sin{\angle A}} = \frac{a\sin\gamma}{\sin(\beta + \gamma)}\]

\[S_{\text{ABC}} = \frac{1}{2}AB \bullet BC\sin{\angle B} =\]

\[= \frac{a^{2}\sin\beta\sin\gamma}{2\sin(\beta + \gamma)}.\]

\[Ответ:\ \ \frac{a^{2}\sin\beta\sin\gamma}{2\sin(\beta + \gamma)}.\]