Point M is removed from each vertex of an acute-angled triangle ABC by 17 cm.

Point M is removed from each vertex of an acute-angled triangle ABC by 17 cm. Calculate the distance from m. M to the plane ABC, if BC = 8 cm, angle BAC = 30 degrees.

The distance from point M to the plane of the triangle ABC is perpendicular. By the condition of the problem, point M is equidistant from the vertices of the triangle, which means that the projections of the straight lines that connect point M with the vertices of the triangle are the radii of the circumscribed circle, and point M is projected to point O – the center of the circumscribed circle.
By condition, the inscribed angle BAC = 30 ° is known, respectively, the central angle BOC = 2 * ∠ BAC = 60 °.
The ВOС triangle is equilateral, OB = OС = BC = 8 cm.
In the right-angled triangle of the MOF, the hypotenuse MВ = 17 cm, the leg OB = 8 cm are known.By the Pythagorean theorem, we find the second leg MO:
MO = √ (MB² – OB²) = √ (289 – 64) = √225 = 15 (cm).
Answer: 15 cm.