Question

многие сделать геометрию!!!! дам 40 балов ​

Answer 1

Ответ:

1) Теорема косинусов:

$AC^2 = AB^2 + BC^2 -2AB\cdot BC\cdot cosB$

Теорема синусов:

$\frac{BC}{sinA} = \frac{AC}{sinB} =\frac{AB}{sinC} = 2R$

2)

$c^2 > a^2 + b^2\\c^2 = a^2+b^2 - 2ab\cdot cos\alpha,\ \ \ \alpha \in [0;180^{\circ}]\\a^2+b^2 - 2ab\cdot cos\alpha > a^2 + b^2\\- 2ab\cdot cos\alpha > 0\\2ab\cdot cos\alpha < 0, 2ab > 0 \ =>\ cos\alpha < 0\\\alpha \in (90^{\circ}, 180^{\circ}]$

3)

По теореме косинусов:

$MN^2 = NK^2 + MK^2 - 2NK\cdot MK\cdot cosK = 2^2+3^2-2\cdot 2\cdot 3\cdot cos30^{\circ}=\\\\=4+9-12\cdot \frac{\sqrt{3} }{2} = 13-6\sqrt{3}$

4)

Найдем угол С:

$\angle C=180^{\circ} - \angle A - \angle B=180^{\circ}-60^{\circ}-45^{\circ}=120^{\circ}-45^{\circ}=75^{\circ}$

По теореме синусов найдем радиус описанной окружности:

$2R = \frac{AB}{sinC} =\frac{5\sqrt{3} }{sin75^{\circ}} = \frac{5\sqrt{3} }{sin(45^{\circ}+30^{\circ})} = \frac{5\sqrt{3} }{sin45^{\circ}cos30^{\circ}+sin30^{\circ}cos45^{\circ}} = \frac{5\sqrt{3} }{\frac{\sqrt{2} }{2}\cdot \frac{\sqrt{3} }{2} + \frac{1 }{2}\cdot \frac{\sqrt{2} }{2}} =\\= \frac{5\sqrt{3} }{\frac{\sqrt{2}(\sqrt{3}+1 ) }{4} } = \frac{20\sqrt{3} }{\sqrt{2}(\sqrt{3} + 1) }$

$R = \frac{20\sqrt{3} }{2\sqrt{2}(\sqrt{3} + 1) } = \frac{10\sqrt{3} }{\sqrt{2}(\sqrt{3} +1)}=\frac{10\sqrt{3}\sqrt{2}(\sqrt{3} -1) }{\sqrt{2}(\sqrt{3} +1)\sqrt{2}(\sqrt{3} -1) } =\frac{10\sqrt{6}\sqrt{3} -10\sqrt{6} }{2(3-1)}=\frac{15\sqrt{2}-5\sqrt{6} }{2}$