ГДЗ по геометрии 9 класс Атанасян Задание 1051

\[\boxed{\mathbf{1051.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунки\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\left| \overrightarrow{a} \right| = 1;\]

\[\left| \overrightarrow{b} \right| = \left| \overrightarrow{c} \right| = 2;\]

\[\widehat{\overrightarrow{a}\overrightarrow{c}} = \widehat{\overrightarrow{b}\overrightarrow{c}} = 60{^\circ}.\]

\[\mathbf{Найти:}\]

\[\left( \overrightarrow{a} + \overrightarrow{b} \right) \bullet \overrightarrow{c}.\]

\[\mathbf{Решение.}\]

\[1)\ \mathrm{\Delta}ABK\ и\ \mathrm{\Delta}AFK -\]

\[прямоугольные:\]

\[\angle DAB = \angle DAK + \angle CAB = 120{^\circ};\]

\[\angle BAK = 120{^\circ} - 90{^\circ} = 30{^\circ}.\]

\[По\ свойству\ прямоугольного\ \]

\[треугольника:\]

\[BK = \frac{1}{2}AB = \frac{1}{2}.\]

\[2)\ FK = BF - BK = 2 - \frac{1}{2} = 1\frac{1}{2}.\]

\[3)\ \left. \ \frac{AK = \sqrt{AB^{2} - KB^{2}}}{AK = \sqrt{AF^{2} - FK^{2}}} \right| \Longrightarrow\]

\[\Longrightarrow AB^{2} - KB^{2} = AF^{2} - FK^{2}:\]

\[1 - \frac{1}{4} = AF^{2} - \frac{9}{4}\]

\[AF^{2} = 1 - \frac{1}{4} + \frac{9}{4} = 3\]

\[AF = \sqrt{3}.\]

\[Отсюда:\]

\[\left| \overrightarrow{a} + \overrightarrow{b} \right| = \sqrt{3}.\]

\[4)\ \overrightarrow{\text{AF}} = \frac{1}{2}\overrightarrow{\text{AE}} = > \left| \overrightarrow{\text{AE}} \right| -\]

\[биссектрисса:\]

\[угол\ между\ \overrightarrow{c}\ и\ \left( \overrightarrow{a} + \overrightarrow{b} \right)\ \]

\[равен\ 30{^\circ}.\]

\[5)\ \left( \overrightarrow{a} + \overrightarrow{b} \right) \bullet \overrightarrow{c} =\]

\[= \left| \overrightarrow{a} + \overrightarrow{b} \right| \bullet \left| \overrightarrow{c} \right| \bullet \cos{30{^\circ}} =\]

\[= \sqrt{3} \bullet 2 \bullet \frac{\sqrt{3}}{2} = 3.\]

\[\mathbf{Ответ:}3.\]