Решебник по геометрии 9 класс Атанасян ФГОС Задание 1103

 

\[\boxed{\mathbf{1103.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[C = 2\pi R \Longrightarrow R = \frac{C}{2\pi}.\]

\[\textbf{а)}\ C_{1} = kC \Longrightarrow R_{1} = \frac{\text{kC}}{2\pi}:\]

\[\frac{R_{1}}{R} = \frac{kC \bullet 2\pi}{2\pi \bullet C} = k - увеличится\ \]

\[в\ \text{k\ }раз;\]

\[\textbf{б)}\ C_{1} = \frac{C}{k} \Longrightarrow R_{1} = \frac{C}{k \bullet 2\pi}:\]

\[\frac{R_{1}}{R} = \frac{C \bullet 2\pi}{k \bullet 2\pi \bullet C} = \frac{1}{k} -\]

\[уменьшится\ в\ \text{k\ }раз.\]

\[\boxed{\mathbf{1103.еуроки - ответы\ на\ пятёрку}}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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