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МатематикаГДЗ по геометрии 9 класс Атанасян Задание 376
Задание 376
\[\boxed{\mathbf{376.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунки\ по\ условию\ задачи:\]
\[\textbf{а)} - г)\]
\[\textbf{д)}\ \]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[\textbf{а)}\ \angle A = 84{^\circ}\]
\[\textbf{б)}\ \angle A - \angle B = 55{^\circ}\]
\[\textbf{в)}\ \angle A + \angle C = 142{^\circ}\]
\[\textbf{г)}\ \angle A = 2\angle B\]
\[\textbf{д)}\ \angle CAD = 16{^\circ}\]
\[\angle ACD = 37{^\circ}.\]
\[\mathbf{Найти:}\ \]
\[\angle A,\angle B,\angle C,\angle D.\]
\[\mathbf{Решение.}\]
\[ABCD - праллелограмм:\]
\[\angle A + \angle B =\]
\[= 180{^\circ}\ (как\ односторонние).\]
\[\textbf{а)}\ 1)\ \angle A = \angle C = 84{^\circ}.\]
\[2)\ \angle B = 180{^\circ} - 84{^\circ} = 96{^\circ};\ \ \]
\[\angle B = \angle D = 96{^\circ}.\]
\[\textbf{б)}\ 1)\ \angle A - \angle B = 55{^\circ}\]
\[\angle A = \angle B + 55{^\circ}.\]
\[2)\ \angle B + 55{^\circ} + \angle B = 180{^\circ}\]
\[2\angle B = 125\]
\[\angle B = 62{^\circ}30^{'}\]
\[\angle B = \angle D = 62{^\circ}30^{'}.\]
\[3)\ \angle A = 62{^\circ}30^{'} + 55{^\circ} = 117{^\circ}30^{'};\ \ \]
\[\ \angle A = \angle C = 117{^\circ}30^{'}.\]
\[\textbf{в)}\ 1)\ \angle A + \angle C = 142{^\circ}\]
\[2\angle A = 142{^\circ}\ \]
\[\angle A = 71{^\circ};\]
\[\angle A = \angle C = 71{^\circ}.\]
\[2)\ \angle B = 180{^\circ} - 71{^\circ} = 109{^\circ};\ \ \ \]
\[\angle B = \angle D = 109{^\circ}.\]
\[\textbf{г)}\ 1)\ 2\angle B + \angle B = 180{^\circ}\]
\[3\angle B = 180{^\circ}\]
\[\angle B = 60{^\circ}.\]
\[\angle B = \angle D = 60{^\circ}.\]
\[2)\ \angle A = 2\angle B = 2 \bullet 60{^\circ} = 120{^\circ};\ \]
\[\angle A = \angle C = 60{^\circ}.\]
\[\textbf{д)}\ 1)\ \angle ACD + \angle CDA + \angle DAC =\]
\[= 180{^\circ}\]
\[\angle CDA = 180 - (37 + 16) = 127\]
\[\angle D = \angle B = 127{^\circ}.\]
\[2)\ \angle A = 180{^\circ} - 127{^\circ} = 53{^\circ};\ \ \ \]
\[\angle A = \angle C = 53{^\circ}.\]
\[Ответ:а)\ \angle A = \angle C = 84{^\circ};\ \]
\[\angle B = \angle D = 96{^\circ};\]
\[\textbf{б)}\angle A = \angle C = 117{^\circ}30^{'};\ \]
\[\angle B = \angle D = 62{^\circ}30';\]
\[\textbf{в)}\angle A = \angle C = 71{^\circ};\]
\[\angle B = \angle D = 109{^\circ};\]
\[\textbf{г)}\ \angle A = \angle C = 120{^\circ};\]
\[\angle B = \angle D = 60{^\circ};\ \]
\[\textbf{д)}\angle A = \angle C = 53{^\circ};\]
\[\angle B = \angle D = 127{^\circ}.\]
