# The height BM, drawn from the vertex of the corner of the rhombus ABCD, makes an angle of 30

**The height BM, drawn from the vertex of the corner of the rhombus ABCD, makes an angle of 30 degrees with side AB, the length of the diagonal AC is 6 cm. Find AM if point M lies on the extension of side AD.**

1. In the problem statement the following are given: the diagonal of the rhombus is 6 cm, the height of BM from vertex B to side AD is 30 ° with side AB.

2. It is known that the diagonals of a rhombus divide its angles in half, at the point of intersection they are perpendicular and are divided in half.

In a right-angled triangle BAM, the BAM angle is 180 ° – 30 ° – 90 ° = 60 °.

We draw a diagonal BD with an intersection with AC at point O, we obtained a triangle AOB, in which

AO = 6 cm: 2 = 3 cm, the OBA angle is half of the ABC angle, the value of which, in turn, is equal to 60 ° as it lies crosswise with the MAB angle.

Hence the angle ABO = 60 °: 2 = 30 °.

That is, in a right-angled triangle ABO AO: AB = sin 30 °,

whence AB = AO: sin 30 ° = 3: 1/2 = 6 cm.

Then AM: AB = sin 30 °, and therefore AM = AB * sin 30 ° = 6 cm * 1/2 = 3 cm.

Answer: AM = 3 cm.