Решебник по геометрии 9 класс Мерзляк Задание 86

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - параллелограмм;\]

\[\angle BAC = \alpha;\]

\[\angle DAC = \beta;\]

\[AC = d.\]

\[Найти:\]

\[AB;\ AD.\]

\[Решение\]

\[1)\ \angle A = \angle BAC + \angle DAC = \alpha + \beta;\]

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

\[\sin{\angle D} = \sin{\angle B} = \sin(\alpha + \beta).\]

\[2)\ В\ \mathrm{\Delta}ABC:\]

\[\frac{\text{AC}}{\sin{\angle B}} = \frac{\text{BC}}{\sin{\angle BAC}}\]

\[\frac{d}{\sin(\alpha + \beta)} = \frac{\text{BC}}{\sin\alpha}\]

\[AD = BC = \frac{d\sin\alpha}{\sin(\alpha + \beta)}.\]

\[3)\ В\ \mathrm{\Delta}ADC:\]

\[\frac{\text{AC}}{\sin{\angle D}} = \frac{\text{CD}}{\sin{\angle DAC}}\]

\[\frac{d}{\sin(\alpha + \beta)} = \frac{\text{CD}}{\sin\beta}\]

\[AB = CD = \frac{d\sin\beta}{\sin(\alpha + \beta)}.\]

\[Ответ:\ \ \frac{d\sin\alpha}{\sin(\alpha + \beta)};\ \frac{d\sin\beta}{\sin(\alpha + \beta)}.\]