Ту задача 18

[antoniy]

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[ammornil]

Screenshot 2025-06-05 010529.png (14.65 KiB) Прегледано 295 пъти

$\\[12pt] \tg{\theta}= \tg{\begin{bmatrix}180^{\circ} -(90^{\circ}-\sigma) -(90^{\circ}-\phi)\end{bmatrix}}= \tg{(\sigma+\phi)}= \frac{\tg{\sigma}+\tg{\phi}}{1-\tg{\sigma}\tg{\phi}} \\[12pt] \tg{\sigma}=\dfrac{BC}{AB}=\dfrac{1}{2} \quad \tg{\phi}=\dfrac{CM}{CD}= \dfrac{1}{4} \cdots $ Отговор: (В).