Решебник по геометрии 9 класс Атанасян ФГОС Задание 1016

 

\[\boxed{\mathbf{1016.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\textbf{а)}\sin{120{^\circ}} = \sin(180{^\circ} - 60{^\circ}) =\]

\[= \sin{60{^\circ}} = \frac{\sqrt{3}}{2};\]

\[\cos{120{^\circ}} = \cos(180{^\circ} - 60{^\circ}) =\]

\[= - \cos{60{^\circ}} = - \frac{1}{2};\]

\[\text{tg\ }120{^\circ} = \frac{\sin{120{^\circ}}}{\cos{120{^\circ}}} =\]

\[= \frac{\sqrt{3}}{2} \bullet \left( - \frac{2}{1} \right) = - \sqrt{3}.\]

\[\textbf{б)}\sin{150{^\circ}} = \sin(150{^\circ} - 30{^\circ}) =\]

\[= \sin{30{^\circ}} = \frac{1}{2};\]

\[\cos{150{^\circ}} = \cos(150{^\circ} - 30{^\circ}) =\]

\[= - \cos{30{^\circ}} = - \frac{\sqrt{3}}{2};\]

\[\text{tg\ }150{^\circ} = \frac{\sin{150{^\circ}}}{\cos{150{^\circ}}} =\]

\[= \frac{1}{2} \bullet \left( - \frac{2}{\sqrt{3}} \right) = - \frac{\sqrt{3}}{3}.\]

\[\textbf{в)}\sin{135{^\circ}} = \sin(135{^\circ} - 45{^\circ}) =\]

\[= \sin{45{^\circ}} = \frac{\sqrt{2}}{2};\]

\[\cos{135{^\circ}} = \cos(135{^\circ} - 45{^\circ}) =\]

\[= - \cos{45{^\circ}} = - \frac{\sqrt{2}}{2};\]

\[\text{tg\ }135{^\circ} = \frac{\sin{135{^\circ}}}{\cos{135{^\circ}}} =\]

\[= \frac{\sqrt{2}}{2} \bullet \left( - \frac{2}{\sqrt{2}} \right) = - 1.\]

\[\boxed{\mathbf{1016.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[A \in O_{x}\ ;\]

\[B \in O_{y};\]

\[OACB - прямоугольник;\]

\[\textbf{а)}\ OA = 6,5;OB = 3;\]

\[\textbf{б)}\ OA = a;OB = b.\]

\[\mathbf{Найти:}\]

\[координаты\ A\ ,B\ и\ \text{C.}\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ OA = 6,5;OB = 3:\ \]

\[\text{A\ }(6,5;0);\ \text{B\ }(0;3);\ C\ (6,5;3).\]

\[\textbf{б)}\ OA = a;OB = b:\]

\[\text{A\ }(a;0);\ \text{B\ }(0;b);\text{\ C}(a;b).\]