The figure shown is a rhombus in which the measure of angle A = 120°.

IMO D.

quadrilateral ABCD is a rhombus... all the sides are equal in length and opposite sides are parallel to each other.

Let the length of side of rhombus be a. So the triangle formed by joining AC is equilateral as two of its sides are equal ( sides of rhombus ) and one angle is 60 degrees. ( opposite angles of a rhombus are supplementary)

AC = a.

Drop a line perpendicular to line AB from point D and let it intersect the line AB extended out at E.

Triangle ADE is 30-60-90 triangle.

So AE is a / 2... sine rule. as the hypotenuse of this 30-60-90 triangle is AD (a)

Hence the Side BE is a + a/2 = 3a/2

This is the side of a right angled triangle DEB...
and by pythagoras theorem -> DE^2 + BE^2 = DB^2

so \(DB^2 = (\sqrt{3} a/2)^2 + (3a/2)^2\)
\(DB = \sqrt{3}a\)

Hence AC : DB = a : \sqrt{3}a

AC : DB = 1 : \sqrt{3}

Option D.

Attached is the image to explain the diagrams & calculations...

Best,
Gladi