Задача 57645 Дан треугольник ABC <A = 45 ° см, <C =...
Условие
Решение
[blue][m]\frac{AC}{sin105 ° }=\frac{BC}{sin45 ° }[/m][/blue]
[m]sin105 ° =sin(45 ° +60 ° )=sin45 ° cos60 ° +cos45 ° sin60 ° =\frac{\sqrt{2}}{2}\cdot (\frac{\frac{1}{2}+\sqrt{3}}{2})[/m]
[m]BC=\frac{AC\cdot sin45 ° }{sin105 ° }=\frac{4 \sqrt{2}\cdot\frac{\sqrt{2}}{2} }{\frac{\sqrt{2}}{2}\cdot (\frac{\sqrt{3}}{2}+\frac{1}{2}) }[/m]
[m]=\frac{8\sqrt{2}}{\sqrt{3}+1}=4\cdot \sqrt{2}\cdot(\sqrt{3}-1)[/m]
[blue][m]\frac{AB}{sin30 ° }=\frac{BC}{sin45 ° }[/m][/blue]
[m]AB=\frac{4\sqrt{2}\cdot (\sqrt{3}-1)}{2sin45 °}=4\cdot(\sqrt{3}-1)[/m]
[m]P=AB+BC+AC=4\cdot(\sqrt{3}-1)+4\cdot \sqrt{2}\cdot(\sqrt{3}-1)+4\sqrt{2}=4\cdot\sqrt{3}-4+4\cdot \sqrt{6}=[/m]
[m]=4\cdot (\sqrt{3}+\sqrt{6}-1)[/m]