\[\boxed{\mathbf{1138.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[окружность\ (O,r) - вписана\ \]
\[в\ ромб;\]
\[\textbf{а)}\ AC = 6\ см;\]
\[BD = 8\ см;\]
\[\textbf{б)}\ AB = a;\]
\[\angle ABC = \alpha.\]
\[\mathbf{Найти:}\]
\[C - ?\]
\[\mathbf{Решение.}\]
\[1)\ Проведем\ OM\bot AB\ и\ \]
\[OM = r.\]
\[2)\ \mathrm{\Delta}ABO - прямоугольный.\ \]
\[AO = OC\ и\ BO = OD - \ по\ \]
\[свойству\ диагоналей\ ромба:\]
\[AB = \sqrt{AO^{2} + OB^{2}} = \sqrt{9 + 16} =\]
\[= 5\ см.\]
\[3)\ S_{\text{ABO}} = \frac{1}{2}AO \bullet OB = \frac{1}{2} \bullet 4 \bullet 3 =\]
\[= 6\ см.\]
\[4)\ S_{\text{ABO}} = \frac{1}{2}AB \bullet MO = 6\ см\]
\[OM = \frac{2S_{\text{ABO}}}{\text{AB}} = \frac{2 \bullet 6}{5} = \ 2,4\ см.\]
\[5)\ C = 2\pi r = 2 \bullet 3,14 \bullet 2,4 =\]
\[= 15,1\ см.\]
\[\textbf{б)}\ 1)\ \angle ABO = \angle OBC = \frac{\alpha}{2}\ \]
\[(по\ свойству\ диагоналей).\]
\[2)\cos{\angle ABO} = \frac{\text{OB}}{\text{AB}}\]
\[OB = AB \bullet \cos{\angle ABO}\]
\[OB = a \bullet \cos\frac{\alpha}{2}.\]
\[3)\ В\ треугольнике\ BOM:\]
\[\sin{\angle MBO} = \frac{\text{OM}}{\text{OB}}\]
\[OM = OB \bullet \sin{\angle MBO}\]
\[OM = a \bullet \cos\frac{\alpha}{2} \bullet \sin\frac{\alpha}{2} =\]
\[= 2 \bullet \frac{a}{2} \bullet \sin\frac{\alpha}{2} \bullet \cos\frac{\alpha}{2} = \frac{a}{2}\sin\alpha\]
\[r = \frac{a}{2} \bullet \sin\alpha.\]
\[4)\ C = 2\pi r = 2\pi \bullet \frac{a}{2} \bullet \sin\alpha =\]
\[= \pi a \bullet \sin\alpha.\]
\[Ответ:а)\ C = 15,1\ см;\]
\[\textbf{б)}\ C = \pi a \bullet \sin\alpha.\]
\[\boxed{\mathbf{1138.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[A\left( - 1;\sqrt{3} \right);\]
\[B\left( 1; - \sqrt{3} \right);\]
\[C\left( \frac{1}{2};\sqrt{3} \right).\]
\[\mathbf{Найти:}\]
\[\angle A;\angle B;\angle C.\]
\[\mathbf{Решение.}\]
\[1)\ AB =\]
\[= \sqrt{(1 + 1)^{2} + \left( \sqrt{3} + \sqrt{3} \right)^{2}} =\]
\[= \sqrt{4 + 12} = \sqrt{16} = 4;\]
\[BC =\]
\[= \sqrt{\left( 1 - \frac{1}{2} \right)^{2} + \left( - \sqrt{3} - \sqrt{3} \right)^{2}} =\]
\[= \sqrt{\frac{1}{4} + 12} = \sqrt{\frac{49}{4}} = 3,5;\]
\[AC =\]
\[= \sqrt{\left( \frac{1}{2} + 1 \right)^{2} + \left( \sqrt{3} - \sqrt{3} \right)^{2}} =\]
\[= \sqrt{\frac{9}{4}} = 1,5.\]
\[2)\ По\ теореме\ косинусов:\]
\[16 = \frac{49}{4} + \frac{9}{4} - 2 \bullet \frac{7}{2} \bullet \frac{3}{2}\ \bullet \cos{\angle C}\]
\[16 = \frac{58}{4} - \frac{42}{4} \bullet \cos{\angle C}\]
\[\frac{42}{4} \bullet \cos{\angle A} = - \frac{6}{4}\]
\[\cos{\angle C} = - \frac{6}{4} \bullet \frac{4}{42} = - \frac{1}{7} \approx\]
\[\approx - 0,1429 < 0.\]
\[\angle C - тупой \Longrightarrow\]
\[\Longrightarrow \angle C = 180{^\circ} - 81{^\circ}47^{'} =\]
\[= 98{^\circ}13^{'}.\]
\[\frac{49}{4} = 16 + \frac{9}{4} - 2 \bullet 4 \bullet \frac{3}{2}\ \bullet \cos{\angle A}\]
\[\frac{40}{4} - 16 = - 12 \bullet \cos{\angle A}\]
\[- 6 = - 12\cos{\angle A}\]
\[\cos{\angle A} = \frac{6}{12} = \frac{1}{2}\ \]
\[\angle A = 60{^\circ}.\]
\[3)\ \angle B =\]
\[= 180{^\circ} - \left( 98{^\circ}13^{'} + 60{^\circ} \right) \approx\]
\[\approx 21{^\circ}47^{'}.\]
\[\mathbf{Ответ:}\angle A = 60{^\circ};\angle B = 21{^\circ}47^{'};\]
\[\angle C = 98{^\circ}13'\mathbf{.}\]