В треугольнике \(\displaystyle ABC\) угол \(\displaystyle A\) равен \(\displaystyle 30^{\circ} {\small,}\) угол \(\displaystyle B\) равен \(\displaystyle 45^{\circ} {\small,}\) \(\displaystyle BC=11\sqrt{2} {\small.}\) Найдите \(\displaystyle AC {\small.}\)
По теореме синусов
Теорема синусов
В треугольнике \(\displaystyle ABC\) \(\displaystyle \frac{\color {blue}{BC}}{\sin \angle \color {blue}{A}}=\frac{\color {#339900}{CA}}{\sin \angle \color {#339900}{B}}=\frac{\red{AB}}{\sin \angle \red{C}}=2R{\small ,}\) где \(\displaystyle R\) – радиус описанной окружности. |  |
\(\displaystyle \frac{BC}{\sin \angle A}=\frac{AC}{\sin \angle B}{\small.}\)
Так как \(\displaystyle \angle A=30^{\circ} {\small,}\) \(\displaystyle \angle B=45^{\circ} {\small,}\) \(\displaystyle BC=11\sqrt{2} {\small ,}\) то
\(\displaystyle \frac{11\sqrt{2} }{\sin 30^{\circ} }=\frac{AC}{\sin 45^{\circ} }{\small.}\)
Значит,
\(\displaystyle AC= \sin 45^{\circ} \cdot \frac{11\sqrt{2} }{\sin 30^{\circ} }{\small.}\)
Поскольку \(\displaystyle \sin 30^{\circ} =\frac{1 }{2}{\small}\) и \(\displaystyle \sin 45^{\circ} =\frac{\sqrt{2} }{2}{\small ,}\) то
\(\displaystyle AC= \frac{\sqrt{2} }{2} \cdot \frac{11\sqrt{2} }{\frac{1 }{2} }=\frac{11\sqrt{4} }{1 }=11\cdot 2=22{\small.}\)
Ответ: \(\displaystyle 22{\small.}\)