In a regular triangular pyramid SABC with base ABC BC = 6√3, SA = 10.

In a regular triangular pyramid SABC with base ABC BC = 6√3, SA = 10. Find the angle between the base plane and the line BK, where K is the intersection of the medians of the SAC face.

At the base of the pyramid lies a regular triangle ABC, whose point O is the center of the inscribed and circumscribed circle and the segments OH and OB, respectively, are their radii.

ОВ = R = ВС / √3 = (6 * √3) / √3 = 6 cm.

OH = r = BC / 2 * √3 = (6 * √3) / 2 * √3 = 3 cm.

ВН = OB + OH = 6 + 3 = 9 cm.

From the right-angled triangle BOS, by the Pythagorean theorem, we determine the length of the leg OS.

OS^2 = BS^2 – OB^2 = 100 – 36 = 64.

OS = 8 cm.

Rectangular triangles SOH and MCН are similar in acute angle with the similarity coefficient K = СH / MH = 3, in terms of the property of the medians. Then MK = SO / 3 = 8/3. НM = OH / 3 = 3/3 = 1 cm, then BM = BH – НM = 9 – 1 = 8 cm.

In the right-angled triangle of the MВК, we determine the tangent of the desired angle of the MВК.

tgМВК = MK / ВM = (8/3) / 8 = 1/3.

Angle MВK = arctan (1/3).

Answer: The angle between the straight line BM and the plane of the base is equal to arctan (1/3).