Решебник по геометрии 9 класс Мерзляк Задание 159

\[Схематический\ рисунок.\]

\[Дано:\]

\[AC = b;\]

\[\angle A = \alpha;\]

\[\angle B = \beta.\]

\[Найти:\]

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

\[Решение.\]

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

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

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

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

\[BC = \frac{AC \bullet \sin{\angle A}}{\sin{\angle B}} = \frac{b\sin\alpha}{\sin\beta}.\]

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

\[= \frac{b^{2}\sin\alpha\sin(\alpha + \beta)}{2\sin\beta}.\]

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