In a regular triangular pyramid SABC with base ABC, the base side is 6 roots of 3, and the side edge is 10.

In a regular triangular pyramid SABC with base ABC, the base side is 6 roots of 3, and the side edge is 10. Find the angle between plane ABC and line MN, where point N is the midpoint of edge AC and point M divides edge BS so that BM: MS = 2: 1.

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.

BH = 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.

Right-angled triangles SOВ and ВМК are similar in acute angle with the similarity coefficient К = SB / ВМ = 3/2, since BM / MS = 2 / 1. Then BO / BK = 3/2, BK = 2 * BO / 3 = 2 * 6/3 = 4 cm, MK = 2 * OS / 3 = 2 * 8/3 = 16/3 cm.Then NK = BH – BK = 9 – 4 = 5 cm.

In a right-angled triangle MKN, we define the tangent of the desired angle MNK.

tgLS = MK / NC = (16/3) / 5 = 16/15 = 1 (1/15).

Angle MBK = arctan (1 (1/15).

Answer: The angle between the straight line BM and the plane of the base is equal to arctan (1 (1/15).