Question

Прямокутник ABCD і ABMN лежать в різних площинах. MC перпендикулярне до ABCD. MC=15√2 см, AD=5 см. Знайти кут між площинами. Допоможіть будь-ласка, треба малюнок і розв'язання.Буду дуже вдячна!ДАЮ 25 БАЛІВ!!!​

Answer 1

Объяснение:

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Answer 2

I can help you with your geometry problem. Here is a possible solution:

To find the angle between the planes ABCD and ABMN, we need to find the angle between two lines that are perpendicular to the line of intersection of the planes. One such pair of lines is MC and AB. We can use the cosine formula to find the angle between them:

$$\cos \angle MCB = \frac{MC^2 + BC^2 - MB^2}{2 \cdot MC \cdot BC}$$

We know that MC = 15√2 cm and AD = 5 cm. Since ABCD is a rectangle, we have BC = AD = 5 cm and AB = 10 cm. We can use the Pythagorean theorem to find MB:

$$MB^2 = AB^2 + BM^2 = 10^2 + 5^2 = 125$$

Plugging these values into the cosine formula, we get:

$$\cos \angle MCB = \frac{(15\sqrt{2})^2 + 5^2 - 125}{2 \cdot 15\sqrt{2} \cdot 5} = \frac{150 - 25}{150} = \frac{5}{6}$$

Therefore, the angle between the planes is:

$$\angle MCB = \arccos \frac{5}{6} \approx 33.56^\circ$$

You can see a diagram of the problem here. You can also find more information about the perpendicularity of lines and planes here and here.

I hope this helps you. Have a nice day!