Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по геометрии 8 класс Атанасян ФГОС Задание 1102
Задание 1102
\[\boxed{\mathbf{1102.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[C = 2\pi R.\]
\[\textbf{а)}\ R_{1} = 3R \Longrightarrow C_{1} = 2\pi \bullet 3R =\]
\[= 6\pi R:\]
\[\frac{C_{1}}{C} = \frac{6\pi R}{2\pi R} = 3 - увеличится\ \]
\[в\ три\ раза;\]
\[\textbf{б)}\ R_{1} = \frac{R}{2} \Longrightarrow C_{1} = 2\pi \bullet \frac{R}{2} = \pi R:\]
\[\frac{C_{1}}{C} = \frac{\text{πR}}{2\pi R} = \frac{1}{2} - уменьшится\ \]
\[в\ два\ раза;\]
\[\textbf{в)}\ R_{1} = kR \Longrightarrow C_{1} = 2\pi \bullet kR =\]
\[= 2k\pi R:\]
\[\frac{C_{1}}{C} = \frac{2k\pi R}{2\pi R} = k - увеличится\ \]
\[в\ k\ раз.\]
\[\textbf{г)}\ R_{1} = \frac{R}{k} \Longrightarrow C_{1} = 2\pi \bullet \frac{R}{k} = \frac{2}{k}\pi R:\]
\[\frac{C_{1}}{C} = \frac{2\pi R}{k \bullet 2\pi R} = \frac{1}{k} - уменьшится\ \]
\[в\ \text{k\ }раз.\]
\[\boxed{\mathbf{1102.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ cos\ \alpha = 1:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\sin^{2}\alpha = 1 - \cos^{2}\alpha = 1 - 1 =\]
\[= 0 \Longrightarrow \sin\alpha = 0;\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{0}{1} = 0.\]
\[\textbf{б)}\ cos\ \alpha = - \frac{\sqrt{3}}{2}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\sin^{2}\alpha = 1 - \cos^{2}\alpha = 1 - \frac{3}{4} = \frac{1}{4}\]
\[\sin\alpha = \pm \frac{1}{2}.\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \pm \frac{1}{2}\ :\left( - \frac{\sqrt{3}}{2} \right) =\]
\[= \pm \frac{\sqrt{3}}{3}.\]
\[\textbf{в)}\ sin\ \alpha = \frac{\sqrt{2}}{2};\ 0{^\circ} < \alpha < 90{^\circ}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{2}{4} =\]
\[= \frac{2}{4} \Longrightarrow \cos\alpha = \pm \frac{\sqrt{2}}{2};\]
\[0{^\circ} < \alpha < 90{^\circ};\cos\alpha > 0 \Longrightarrow\]
\[\Longrightarrow \cos\alpha = \frac{\sqrt{2}}{2};\ \]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{2}}{2}\ :\frac{\sqrt{2}}{2} = 1.\]
\[\textbf{г)}\ sin\ \alpha = \frac{3}{5};\ 90{^\circ} < \alpha < 180{^\circ}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{9}{25} =\]
\[= \frac{16}{25} \Longrightarrow \cos\alpha = \pm \frac{4}{5};\]
\[90{^\circ} < \alpha < 180{^\circ};\cos\alpha < 0 \Longrightarrow\]
\[\Longrightarrow \cos\alpha = - \frac{4}{5};\ \]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{3}{5}\ :\left( - \frac{4}{5} \right) = - \frac{3}{4}.\]
