Задание 8, стр 60

Question

Плоскость $\alpha$ проведена через стороны BC треугольника CAB , а из вершины A к этой плоскости проведен перпендикуляр AD. Если SΔCAB = 48 см$^2$, S ΔCDB = 24 см2, найти угол между плоскостью ΔBCA и плоскостью $\alpha$.

Gimli · Accepted Answer

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Из формулы, полученной в 5ом упражнении, можем заключить, что
$$S_{ABC}\cdot \cos \phi=S_{BDC}$$
$$\cos \phi =\frac{S_{BDC}}{S_{ABC}}=\frac{24}{48}=0.5$$
$$\cos \phi = 0.5$$
$$\phi = 60$$

Задание 8, стр 60

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Задание 8, стр 60

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