In a regular quadrangular pyramid SABCD (S-vertex) M-midpoint of SA

| September 4, 2021 | education

In a regular quadrangular pyramid SABCD (S-vertex) M-midpoint of SA, K-midpoint of SC. Find the angle between planes BMK and ABC? AB = 6, SC = 8.

Let’s connect the points K, M and B. Since K and M are the middle of the lateral edges SC and SA, the triangles SVK and ABM are equal, and then the triangle BKM is isosceles.

Let’s draw the height BH of the triangle BKM, which is also its median. Since KM is the middle line of the ACS triangle, point H divides the SO height in half.

OH = SO / 2.

Let us determine the length of the AC diagonal.

AC ^ 2 = 2 * AB ^ 2 = 2 * 36.

AC = 6 * √2 cm.

Then AO = AC / 2 = 3 * √2 cm.

In a right-angled triangle AOS, according to the Pythagorean theorem, we determine the length of the leg SO.

SO ^ 2 = SA ^ 2 – AO ^ 2 = 64 – 18 = 46.

SO = √46.

Then HO = SO / 2 = √46 / 2 cm.

In a right-angled triangle, ОВН, ОВ = ОА = 3 / √2.

Then tgОВН = ОН / ОВ = (√46 / 2) / (3 / √2) = √46 * √2 / 6 = √92 / 6 = 2 * √23 / 6 = √23 / 3.

OVN angle = arctan (√23 / 3).

Answer: between planes BMK and ABC is equal to arctan (√23 / 3).



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