# Side lengths of triangle ABC: BC = 15; AB = 13; AC = 4. The alpha plane is drawn through AC, which makes

**Side lengths of triangle ABC: BC = 15; AB = 13; AC = 4. The alpha plane is drawn through AC, which makes an angle of 30 degrees with the ABC plane. Find the distance from the vertex B to the alpha plane.**

By Heron’s theorem, we determine the area of the triangle ABC.

The semi-perimeter of the triangle ABC is equal to: p = (AB + BC + AC) / 2 = (15 + 13 + 4) / 2 = 16 cm.

Then Savs = √p * (p – AB) * (p – BC) * (p – AC) = √16 * (16 – 15) * (16 – 13) * (16 – 4) = √576 = 24 cm2 …

Let us build the height BH of the triangle ABC, then the area of the triangle ABC is also equal to:

Sас = АС * BН / 2.

ВН = 2 * Savs / AC = 2 * 24/4 = 12 cm.

From the vertex B, we construct a perpendicular BK to the plane, then the triangle BНK is rectangular, in which the angle BНK = 30.

The leg BK lies against an angle of 30, then BK = BH / 2 = 12/2 = 6 cm.

Answer: From the top B to the plane 6 cm.