ГДЗ по геометрии 9 класс Атанасян Задание 591

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Год:2020-2021-2022
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Задание 591

\[\boxed{\mathbf{591.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\textbf{а)}\ BC = 8;AB = 17:\]

\[AC = \sqrt{17^{2} - 8^{2}} = \sqrt{289 - 64} =\]

\[= \sqrt{255} = 15;\]

\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{8}{17};\ \ \]

\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{15}{17};\]

\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{15}{17};\ \]

\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{8}{17};\]

\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{8}{15};\ \ \ \]

\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{15}{8}.\]

\[\textbf{б)}\ BC = 21;AC = 20:\]

\[AB = \sqrt{21^{2} + 20^{2}} =\]

\[= \sqrt{441 + 400} = \sqrt{841} = 29;\]

\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{21}{29};\ \ \]

\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{20}{29};\]

\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{20}{29};\ \]

\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{21}{29};\]

\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{21}{20};\ \ \ \]

\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{20}{21}.\]

\[\textbf{в)}\ BC = 1;AC = 2:\]

\[AC = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5};\]

\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{1}{\sqrt{5}};\ \ \]

\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{2}{\sqrt{5}};\]

\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{2}{\sqrt{5}};\ \]

\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{1}{\sqrt{5}};\]

\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{1}{2};\ \ \ \]

\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = 2.\]

\[\textbf{г)}\ AC = 24;AB = 25:\]

\[BC = \sqrt{25^{2} - 24^{2}} = \sqrt{49} = 7;\]

\[\sin{\angle A} = \frac{\text{CB}}{\text{AB}} = \frac{7}{25};\ \ \]

\[\sin{\angle B} = \frac{\text{AC}}{\text{AB}} = \frac{24}{25};\]

\[\cos{\angle A} = \frac{\text{AC}}{\text{AB}} = \frac{24}{25};\ \]

\[\cos{\angle B} = \frac{\text{BC}}{\text{AB}} = \frac{7}{25};\]

\[tg\ \angle A = \frac{\text{CB}}{\text{AC}} = \frac{7}{24};\ \ \ \]

\[tg\ \angle B = \frac{\text{AC}}{\text{BC}} = \frac{24}{7}.\]

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