По теореме синусов:
\dfrac{AB}{\sin{C}} = \dfrac{BC}{\sin{A}}
sinC
AB
=
sinA
BC
Отсюда, выразим BC:
BC = AB \cdot \dfrac{\sin{A}}{\sin{C}} = 3\sqrt{2} \cdot \dfrac{\sin{120^\text{o}}}{\sin{45^\text{o}}} = 3\sqrt{2} \cdot \dfrac{\sin{60^\text{o}}}{\sin{45^\text{o}}} = 3\sqrt{2} \cdot \dfrac{\sqrt{3}/2}{1/\sqrt{2}} =BC=AB⋅
sin120
o
sin45
sin60
ответ: 3√3.
По теореме синусов:
\dfrac{AB}{\sin{C}} = \dfrac{BC}{\sin{A}}
sinC
AB
=
sinA
BC
Отсюда, выразим BC:
BC = AB \cdot \dfrac{\sin{A}}{\sin{C}} = 3\sqrt{2} \cdot \dfrac{\sin{120^\text{o}}}{\sin{45^\text{o}}} = 3\sqrt{2} \cdot \dfrac{\sin{60^\text{o}}}{\sin{45^\text{o}}} = 3\sqrt{2} \cdot \dfrac{\sqrt{3}/2}{1/\sqrt{2}} =BC=AB⋅
sinC
sinA
sin120
o
sin45
o
sin60
o
ответ: 3√3.