Решебник по геометрии 9 класс Мерзляк Задание 95

\[Схематический\ рисунок.\]

\[Дано:\]

\[\mathrm{\Delta}ABC;\]

\[\angle AMC = \varphi;\]

\[AB = c;\]

\[\angle A = \alpha;\]

\[\angle ACB = \gamma.\]

\[Найти:\]

\[\text{CM.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\frac{\text{AB}}{\sin{\angle ACB}} = \frac{\text{BC}}{\sin{\angle A}};\]

\[BC = \frac{AB \bullet \sin{\angle A}}{\sin{\angle ACB}} = \frac{c\sin\alpha}{\sin\gamma};\]

\[\angle BMC = 180{^\circ} - \angle AMC =\]

\[= 180{^\circ} - \varphi;\]

\[\sin{\angle BMC} = \sin(180{^\circ} - \varphi) = \sin\varphi;\]

\[\angle B = 180{^\circ} - \angle A - \angle C =\]

\[= 180{^\circ} - \alpha - \gamma;\]

\[\sin{\angle B} = \sin(180{^\circ} - \alpha - \gamma) =\]

\[= \sin(a + \gamma).\]

\[2)\ В\ \mathrm{\Delta}BMC:\]

\[\frac{\text{BC}}{\sin{\angle BMC}} = \frac{\text{CM}}{\sin{\angle MBC}}\]

\[CM = \frac{BC \bullet \sin{\angle MBC}}{\sin{\angle BMC}} =\]

\[= \frac{c\sin\alpha\sin(a + \gamma)}{\sin\gamma\sin\varphi}.\]

\[Ответ:\ \ \frac{c\sin\alpha\sin(a + \gamma)}{\sin\gamma\sin\varphi}.\]