# Point m is at a distance of √7 from the alpha plane. Two oblique MP and MK are drawn, respectively, at angles

**Point m is at a distance of √7 from the alpha plane. Two oblique MP and MK are drawn, respectively, at angles of 60 and 45 to the plane. Find РK, if the angle POK = 150, where О is the base of the perpendicular MO, MO is the perpendicular alpha.**

Segments OK and OP are inclined projections onto a plane. Since OM is perpendicular to the plane, the triangles OMP and OMK are rectangular.

In a right-angled triangle OMK OK = OM = √7 cm, since the angle OKM = 45.

In a right-angled triangle ОМР, ОP = ОМ / tg60 = √7 / √3 cm.

In the triangle PОК, by the cosine theorem:

PK ^ 2 = OP ^ 2 + OK ^ 2 – 2 * OP * OK * Cos150 = 7/3 + 7 – 2 * √ (7/3) * √7 * (-√3 / 2) = 7/3 + 7 + 7 * √3 / 3 = (7 + 21 + 8 * √3) / 3 = (28 + 8 * √3) / 3 cm.

Answer: The length of the segment PK is (28 + 8 * √3) / 3 cm.