The side of the rhombus is 1, and one angle is 150 degrees. Find the distance from the point of intersection of its diagonals to the side.

draw a rhombus ABSD, O – the point of intersection of the diagonals DB and AS
angle BAA = BSD = 150 degrees
AC divides the angles of BAA and BSD in half, which means the angles BAO = DAo = BSO = DSo = 150: 2 = 75 degrees
Consider the ABO triangle, it is rectangular since the diagonals intersect and form a perpendicular.
The sum of the acute angles of a right-angled triangle is 90 degrees, which means
ABO = 90 – BAO = 90 – 75 = 15 degrees.
ABS angle = ADS = 15 * 2 = 30 degrees.
The distance from a point to a straight line is the perpendicular drawn from that point to a straight line.
Let’s draw perpendiculars OK to the side of blood pressure and OM to the side of BS.
Consider the KMD triangle
It is rectangular because MKD = 90 degrees (perpendicular)
Because SDA angle = 30 degrees, then the leg opposite this angle is half the hypotenuse.
KM = 1/2 MD = 0.5
Km is two distances from a point to a straight line, which means
ОМ = КМ: 2 = 0.25