\[\boxed{\mathbf{1199.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\angle BAC = 120{^\circ};\]
\[AB = 5\ см;\]
\[AC = 3\ см;\]
\[S_{грани} = 35\ см^{2}.\]
\[\mathbf{Найти:}\]
\[V - ?\]
\[\mathbf{Решение.}\]
\[1)\ В\ \mathrm{\Delta}BAC,\ по\ теоремме\ \]
\[косинусов:\]
\[BC^{2} =\]
\[= AB^{2} + AC^{2} - 2AB \bullet AC \bullet \cos{\angle BAC}\]
\[BC^{2} =\]
\[= 25 + 9 - 2 \bullet 5 \bullet 3 \bullet \cos{120{^\circ}} =\]
\[= 34 + 15 = 49\]
\[BC = 7\ см.\]
\[2)\ Рассмотрим\ \text{BC}B_{1}C_{1}:\]
\[S = 35\ см^{2};\ \ \ \]
\[S = BC \bullet BB_{1} \Longrightarrow BC \bullet BB_{1} = 35.\]
\[BC = 7 \Longrightarrow BB_{1} = \frac{35}{7} = 5\ см.\]
\[3)\ V = S_{осн} \bullet H =\]
\[= \frac{1}{2}AB \bullet AC \bullet \sin\text{BAC} \bullet BB_{1} =\]
\[= \frac{1}{2} \bullet 5 \bullet 3 \bullet \sin{120{^\circ}} \bullet 5 =\]
\[= \frac{75\sqrt{3}}{4}\ см^{3}.\]
\[Ответ:\frac{75\sqrt{3}}{4}\ см^{3}.\]