\[\boxed{\mathbf{697.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Доказать:\]
\[площадь\ описанного\ \]
\[многоугольника\ равна\ \]
\[половине\ произведения\ его\ \]
\[периметра\ на\ радиус\ \]
\[вписанной\ окружности.\]
\[Доказательство.\]
\[1)\ Центр\ вписанной\ \]
\[окружности,\ соединенный\ \]
\[отрезками\ с\ вершинами\]
\[многоугольника,\ разделяет\ его\ \]
\[на\ треугольники,\ в\ каждом\ из\ \]
\[которых\ основание - сторона\ \]
\[многоугольника,\ а\ высота -\]
\[радиус\ r\ вписанной\]
\[окружности.\]
\[2)\ Пусть\ a_{1},\ a_{2},a_{3},\ldots,a_{n} -\]
\[стороны\ многоугольника;\]
\[S_{1},\ S_{2},\ldots,S_{n} - площади\ \]
\[треугольников:\]
\[S_{многоуг} = S_{1} + S_{2} + S_{3} + \ldots + S_{n};\]
\[P_{многоуг} =\]
\[= a_{1} + a_{2} + a_{3} + \ldots + a_{n} =\]
\[Получаем:\]
\[S = \frac{1}{2}r \bullet P.\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{697.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ BC = 8;AB = 17:\]
\[AC = \sqrt{17^{2} - 8^{2}} = \sqrt{289 - 64} =\]
\[= \sqrt{255} = 15;\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{8}{17};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{15}{17};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{15}{17};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{8}{17};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{8}{15};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{15}{8}.\]
\[\textbf{б)}\ BC = 21;AC = 20:\]
\[AB = \sqrt{21^{2} + 20^{2}} =\]
\[= \sqrt{441 + 400} = \sqrt{841} = 29;\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{21}{29};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{20}{29};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{20}{29};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{21}{29};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{21}{20};\ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{20}{21}.\]
\[\textbf{в)}\ BC = 1;AC = 2:\]
\[AC = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5};\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{1}{\sqrt{5}};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{2}{\sqrt{5}};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{2}{\sqrt{5}};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{1}{\sqrt{5}};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{1}{2};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = 2.\]
\[\textbf{г)}\ AC = 24;AB = 25:\]
\[BC = \sqrt{25^{2} - 24^{2}} = \sqrt{49} = 7;\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{7}{25};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{24}{25};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{24}{25};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{7}{25};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{7}{24};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{24}{7}.\]