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.




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