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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 591
Задание 591
\[\boxed{\mathbf{591.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ BC = 8;AB = 17:\]
\[AC = \sqrt{17^{2} - 8^{2}} = \sqrt{289 - 64} =\]
\[= \sqrt{255} = 15;\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{8}{17};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{15}{17};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{15}{17};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{8}{17};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{8}{15};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{15}{8}.\]
\[\textbf{б)}\ BC = 21;AC = 20:\]
\[AB = \sqrt{21^{2} + 20^{2}} =\]
\[= \sqrt{441 + 400} = \sqrt{841} = 29;\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{21}{29};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{20}{29};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{20}{29};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{21}{29};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{21}{20};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{20}{21}.\]
\[\textbf{в)}\ BC = 1;AC = 2:\]
\[AC = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5};\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{1}{\sqrt{5}};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{2}{\sqrt{5}};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{2}{\sqrt{5}};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{1}{\sqrt{5}};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{1}{2};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = 2.\]
\[\textbf{г)}\ AC = 24;AB = 25:\]
\[BC = \sqrt{25^{2} - 24^{2}} = \sqrt{49} = 7;\]
\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{7}{25};\ \ \]
\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{24}{25};\]
\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{24}{25};\ \]
\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{7}{25};\]
\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{7}{24};\ \ \ \]
\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{24}{7}.\]
\[\boxed{\mathbf{591.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[AC = 10\ см;\]
\[BD = 24\ см.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[AB - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABCD}} = \frac{1}{2} \bullet AC \bullet BD =\]
\[= \frac{1}{2} \bullet 10 \bullet 24 = 120\ см^{2}.\]
\[2)\ По\ свойству\ ромба:\]
\[DO = OB = 12\ см;\]
\[AO = OC = 5\ см.\]
\[3)\ AC\bot DB \Longrightarrow ⊿AOB -\]
\[прямоугольный.\]
\[4)\ По\ теореме\ Пифагора:\]
\[AB^{2} = AO^{2} + OB^{2}\]
\[AB^{2} = 25 + 144 = 169\]
\[AB = 13\ см.\]
\[\mathbf{Ответ}:S_{\text{ABCD}} = 120\ см^{2};\]
\[AB = 13\ см\mathbf{.}\]
