# The distance from the intersection of the diagonals of the rhombus to one of its sides is 19, and one of the diagonals of the rhombus is 76

**The distance from the intersection of the diagonals of the rhombus to one of its sides is 19, and one of the diagonals of the rhombus is 76. Find the angles of the rhombus**

Consider the triangle ABO.

By definition, a rhombus is a parallelogram with equal sides; therefore, all properties of a parallelogram extend to the rhombus.

Then, the diagonals of the rhombus are divided by the intersection point in half (according to the third property of the parallelogram), i.e. OB = 76/2 = 38

The triangle ABO is rectangular, since OA is the distance to the side of the rhombus, i.e. forms a right angle with the side.

sin∠ABO = AO / BO = 19/38 = 1/2 => ∠ABO = 30 ° (tabular value).

The triangles EBO and CBO are equal (on three sides).

Therefore, ∠EBO = ∠CBO = 30 °

Thus, ∠EBC = 30 ° * 2 = 60 °

By the parallelogram property, ∠EBC = ∠EDC = 60 ° and ∠BED = ∠BCD

The sum of the angles of any quadrangle is 360 °, therefore:

∠BED = ∠BCD = (360 ° – (2 * 60 °)) = (360 ° -120 °) / 2 = 120 °

Answer: ∠EBC = ∠EDC = 60 ° and ∠BED = ∠BCD = 120 °