In a regular triangular prism ABCA1B1C1, the base side is 3√2, the lateral edge is 3√2, M

| February 6, 2021 | education

In a regular triangular prism ABCA1B1C1, the base side is 3√2, the lateral edge is 3√2, M is the center of the CC1B1B face. Find the angle between line AM and the plane of the base.

Point M is the point of intersection of the diagonals BC1 and CB1, which are equal and are halved at the point of intersection. Then the SMB triangle is isosceles. Let’s draw their points M the height MH, the length of which is equal to half of CC1, since MH is the middle line of the triangle CC1B.

MH = CC1 / 2 = 3 * √2 / 2 cm.

In an equilateral triangle ABC, we draw the height AH, which is also the median of the triangle. AH = BC * √3 / 2 = 3 * √2 * √3 / 2 = 3 * √6 / 2.

The AMН triangle is rectangular, then tgMAН = MН / AH = (3 * √2 / 2) / (3 * √6 / 2) = √2 / √6 = 1 / √3.

Angle MAН = arctan (1 / √3) = 30.

Answer: The angle between the straight line AM and the plane of the base is 300.



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