Question

В правильной четырёхугольной пирамиде SABCD высота SO равна 9, диагональ основания BD равна 8. Точки K и M — середины рёбер CD и BC соответственно. Найдите тангенс угла между плоскостью SMK и плоскостью основания

Answer 1

Пусть ОС∩КМ=Н. Поскольку ОН =правсSH и КМ ⊥ ОН по теореме о трех перпендикулярах SH ⊥ КМ. Поскольку КМ ⊥ ОН, SH ⊥ КМугол ОНС яв­ля­ет­ся ли­ней­ным углом дву­гран­но­го угла между плос­ко­стя­ми  SМК и АВС, тогда тангенс ОНS = SО/ОН=SО/0,5ОС=9/2=4,5.
Ответ: 4,5 см.

Answer 2

Problem Analysis

We are given a right quadrilateral pyramid SABCD, where the height SO is 9 and the diagonal of the base BD is 8. Points K and M are the midpoints of the edges CD and BC, respectively. We need to find the tangent of the angle between the plane SMK and the base plane.

Solution

To find the tangent of the angle between the plane SMK and the base plane, we need to find the angle between the two planes first. Let's denote the angle between the two planes as θ.

To find θ, we can use the dot product of the normal vectors of the two planes. The normal vector of the plane SMK can be found by taking the cross product of the vectors SM and SK. Similarly, the normal vector of the base plane can be found by taking the cross product of the vectors SB and SD.

Let's calculate the normal vectors of the two planes:

- Vector SM = Vector SK = Vector MK = (M - S) = (K - S) - Vector SB = (B - S) - Vector SD = (D - S)

Now, let's calculate the cross products:

- Normal vector of plane SMK = Vector SM x Vector SK - Normal vector of base plane = Vector SB x Vector SD

Finally, we can find the angle θ between the two planes using the dot product of the normal vectors:

θ = arccos((Normal vector of plane SMK) · (Normal vector of base plane) / (|Normal vector of plane SMK| · |Normal vector of base plane|))

To find the tangent of θ, we can use the trigonometric identity:

tan(θ) = sin(θ) / cos(θ)

Let's calculate the values step by step.

Calculation

Given: - Height SO = 9 - Diagonal of base BD = 8

First, let's find the lengths of the vectors:

- |Vector SM| = |Vector SK| = |Vector MK| = |M - S| = |K - S| = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) = sqrt((0 - 0)^2 + (0 - 0)^2 + (9 - 0)^2) = 9 - |Vector SB| = |B - S| = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) = sqrt((0 - 0)^2 + (0 - 0)^2 + (8 - 0)^2) = 8 - |Vector SD| = |D - S| = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) = sqrt((0 - 0)^2 + (0 - 0)^2 + (8 - 9)^2) = 1

Next, let's calculate the cross products:

- Normal vector of plane SMK = Vector SM x Vector SK = (SM) x (SK) = (0, 0, 9) x (0, 0, 9) = (0 - 0, 0 - 0, 0 - 0) = (0, 0, 0) - Normal vector of base plane = Vector SB x Vector SD = (SB) x (SD) = (0, 0, 8) x (0, 0, 1) = (0 - 0, 0 - 0, 0 - 0) = (0, 0, 0)

Since the cross products of the vectors are both zero vectors, the two planes are parallel and the angle between them is 0 degrees. Therefore, the tangent of the angle between the two planes is also 0.

Answer

The tangent of the angle between the plane SMK and the base plane is 0.

Note: The given information and calculations indicate that the two planes are parallel, which means they do not intersect and the angle between them is 0 degrees.