ГДЗ по геометрии 9 класс Атанасян Задание 1060

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Год:2020-2021-2022
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Задание 1060

\[\boxed{\mathbf{1060.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\textbf{а)}\ AB = 8\ см;\ \angle A = 30{^\circ};\ \]

\[\angle B = 45{^\circ}:\]

\[1)\ \angle C = 180{^\circ} - (30{^\circ} + 45{^\circ}) =\]

\[= 180{^\circ} - 75{^\circ} = 105{^\circ}.\]

\[2)\ По\ теореме\ синусов:\]

\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{AC}}{\sin{\angle B}} = \frac{\text{BC}}{\sin{\angle A}}\]

\[\frac{8}{\sin{105{^\circ}}} = \frac{\text{AC}}{\sin{45{^\circ}}}\]

\[AC = \frac{8\sin{45{^\circ}}}{\sin{105{^\circ}}} = \frac{8 \bullet \frac{\sqrt{2}}{2}}{0,9659} =\]

\[= 5,8564\ см.\]

\[\frac{8}{\sin{105{^\circ}}} = \frac{\text{BC}}{\sin{30{^\circ}}}\]

\[BC = \frac{8\sin{30{^\circ}}}{\sin{105{^\circ}}} = \frac{8 \bullet \frac{1}{2}}{0,9659} =\]

\[= 4,1411\ см.\]

\[\textbf{б)}\ AB = 5\ см;\ \angle C = 60{^\circ};\ \]

\[\angle B = 45{^\circ}:\]

\[1)\ \angle A = 180{^\circ} - (60{^\circ} + 45{^\circ}) =\]

\[= 180{^\circ} - 105{^\circ} = 75{^\circ}.\]

\[2)\ По\ теореме\ синусов:\]

\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{AC}}{\sin{\angle B}} = \frac{\text{BC}}{\sin{\angle A}}\]

\[\frac{5}{\sin{105{^\circ}}} = \frac{\text{AC}}{\sin{45{^\circ}}}\]

\[AC = \frac{5\sin{45{^\circ}}}{\sin{60{^\circ}}} = 4,1\ см.\]

\[\frac{5}{\sin{105{^\circ}}} = \frac{\text{BC}}{\sin{75{^\circ}}}\]

\[BC = \frac{5\sin{75{^\circ}}}{\sin{60{^\circ}}} = 5,58\ см.\]

\[\textbf{в)}\ AB = 3\ см;BC = 3,3\ см;\]

\[\angle A = 48{^\circ}30^{'}:\]

\[1)\ По\ теореме\ синусов:\]

\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{AC}}{\sin{\angle B}} = \frac{\text{BC}}{\sin{\angle A}}\]

\[\frac{3}{\sin{\angle C}} = \frac{3,3}{\sin{48{^\circ}30^{'}}}\]

\[\sin{\angle C} = \frac{3\sin{48{^\circ}30^{'}}}{3,3} =\]

\[= \frac{0,749}{1,1} = 0,6809\]

\[\angle C = 42{^\circ}55^{'}.\]

\[\frac{3,3}{\sin{48{^\circ}30^{'}}} = \frac{\text{AC}}{\sin{88{^\circ}35^{'}}}\]

\[AC = \frac{3,3\sin{88{^\circ}35^{'}}}{\sin{48{^\circ}30^{'}}} =\]

\[= \frac{3,3 \bullet 0,9997}{0,749} = 4,4\ см.\]

\[2)\ \angle B =\]

\[= 180{^\circ} - \left( 42{^\circ}55^{'} + 48{^\circ}30^{'} \right) =\]

\[= 88{^\circ}35^{'}.\]

\[\textbf{г)}\ AC = 10,4\ см;BC = 5,2\ см;\]

\[\angle B = 62{^\circ}48^{'}:\]

\[1)\ По\ теореме\ синусов:\]

\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{AC}}{\sin{\angle B}} = \frac{\text{BC}}{\sin{\angle A}}\]

\[\frac{10,4}{\sin{62{^\circ}48^{'}}} = \frac{5,2}{\sin{\angle A}}\]

\[\sin{\angle A} = \frac{5,2\sin{62{^\circ}48^{'}}}{10,4} =\]

\[= \frac{5,2 \bullet 0,8894}{10,4} = 0,4447\]

\[\angle A = 26{^\circ}24^{'}.\]

\[\frac{10^{'}4}{\sin{62{^\circ}48^{'}}} = \frac{\text{AB}}{\sin{90{^\circ}48^{'}}}\]

\[AB = \frac{10,4\sin{90{^\circ}48^{'}}}{\sin{62{^\circ}48^{'}}} =\]

\[= \frac{10,4 \bullet 0,9999}{0,8894} = 11,69\ см.\]

\[2)\ \angle C =\]

\[= 180{^\circ} - \left( 62{^\circ}48^{'} + 26{^\circ}24^{'} \right) =\]

\[= 90{^\circ}48^{'}.\]

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