A Домашняя CODAB=1/4 CB=8 9 B Кай ты ABU CH. Решение COLAB; CB

Ответ: AB = 6, CA = \(\frac{32}{3}\)

1. Шаг 1: Находим AB

   \[\cos{\angle B} = \frac{CB}{AB}\]

   \[\frac{4}{9} = \frac{8}{AB}\]

   \[AB = \frac{8 \cdot 9}{4} = 2 \cdot 9 = 18\]

2. Шаг 2: Находим CA

   \[\tan{\angle B} = \frac{CA}{CB}\]

   Т.к. \(\cos^2{\angle B} + \sin^2{\angle B} = 1\), то

   \[\sin^2{\angle B} = 1 - \cos^2{\angle B} = 1 - \left(\frac{4}{9}\right)^2 = 1 - \frac{16}{81} = \frac{81-16}{81} = \frac{65}{81}\]

   \[\sin{\angle B} = \sqrt{\frac{65}{81}} = \frac{\sqrt{65}}{9}\]

   \[\tan{\angle B} = \frac{\sin{\angle B}}{\cos{\angle B}} = \frac{\frac{\sqrt{65}}{9}}{\frac{4}{9}} = \frac{\sqrt{65}}{4}\]

   \[\frac{\sqrt{65}}{4} = \frac{CA}{8}\]

   \[CA = \frac{8\sqrt{65}}{4} = 2\sqrt{65}\]

3. Шаг 1: Находим AB

   \[\cos{\angle B} = \frac{CB}{AB}\]

   \[\frac{4}{9} = \frac{8}{AB}\]

   \[4AB = 72 \Rightarrow AB = \frac{72}{4} = 18\]

4. Шаг 2: Находим \(\sin{\angle B}\)

   \[\sin^2{\angle B} + \cos^2{\angle B} = 1\]

   \[\sin^2{\angle B} = 1 - \cos^2{\angle B} = 1 - \left(\frac{4}{9}\right)^2 = 1 - \frac{16}{81} = \frac{65}{81}\]

   \[\sin{\angle B} = \sqrt{\frac{65}{81}} = \frac{\sqrt{65}}{9}\]

5. Шаг 3: Находим CA

   \[tg{\angle B} = \frac{AC}{BC} \Rightarrow AC = BC \cdot tg{\angle B}\]

   \[tg{\angle B} = \frac{\sin{\angle B}}{\cos{\angle B}} = \frac{\sqrt{65}}{9} : \frac{4}{9} = \frac{\sqrt{65}}{4}\]

   \[AC = 8 \cdot \frac{\sqrt{65}}{4} = 2\sqrt{65}\]

Ответ: AB = 18, CA = \(2\sqrt{65}\)