In the regular triangular pyramid SABC, the edges are known: AB = 3√ 3 SC = 5. Find the angle formed

In the regular triangular pyramid SABC, the edges are known: AB = 3√ 3 SC = 5. Find the angle formed by the plane of the base and the straight line MK where K is the midpoint of the edge AC, and the point M divides the edge BS as BM: MS = 3: 1

At the base of the triangle ABC, we determine the length of the height BK. ВK = АС * √3 / 2 = 3 * √3 * √3 / 2 = 9/2.

The height ВK is also the median of the triangle ABC, then by the property of the medians, the length of the segment ОВ = ВK * 2/3 = (9/2) * (2/3) = 3 cm.

Point M divides the SB side in a ratio of 3/1, then BM = 3 * MS. MS + BM = SB = 5 cm.

4 * MS = 5, MS = 5/4 cm. BM = 5 – 5/4 = 15/4 cm

Rectangular triangles ОSB and НМВ are similar in acute angle. Then:

SB / OB = BM / BH.

BН = ОВ * BM / SB = 3 * (15/4) / 5 = 9/4 cm.

In a right-angled triangle HMB MH ^ 2 = BM ^ 2 – BH ^ 2 = (225/16) – (81/16) = 144/16 = 9.

MH = 3 cm.

КН = КB – BН = (9/2) – (9/4) = 9/4.

tgMKB = MH / KH = 9 / (9/4) = 4.

Angle MKB = arctg4.

Answer: Angle MKB = arctg4.