# In a regular quadrangular pyramid MABCD, all edges of which are equal to 1

In a regular quadrangular pyramid MABCD, all edges of which are equal to 1, find the angle between line AK and plane MDC, where K is the midpoint of MD.

Since the lengths of all edges are 1 cm, the side faces of the pyramid are equilateral triangles. The point K by condition is the middle of the MD edge, then AK is the median and the height of the ADM triangle. The segment CK is also the height and median of the triangle CDM, then the angles AKD and CKD are straight, and the angle AKC is our desired angle.

The height of the AK of an equilateral triangle ADK is determined by the formula: AK = DM * √3 / 2 = √3 / 2 cm.

CK = AK = √3 / 2 cm.

At the base of the pyramid is the square ABCD, then AC ^ 2 = AD ^ 2 + CD ^ 2 = 2.

AC = √2 cm.

In triangle ACK we apply the cosine theorem.

AC ^ 2 = AK ^ 2 + CK ^ 2 – 2 * AK * CK * CosAKC.

2 = (3/4) + (3/4) – 2 * (√3 / 2) * (√3 / 2) * CosAKC.

6/4 – 2 = (3/2) * CosAKC.

-1/2 = 3/2 * CosAKC.

CosAKC = -1/3.

Angle AKC = arcos (-1/3).

Answer: The angle between the straight line AK and the MDC plane is equal to arcos (-1/3).

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