The distance from the point of intersection of the diagonals of the rhombus to one of its sides is 17

The distance from the point of intersection of the diagonals of the rhombus to one of its sides is 17, and one of the diagonals of the rhombus is 68. Find the corners of the rhombus.

Given:
Rhombus – ABCD,
O point of intersection of the diagonals of the rhombus,
AC = 68 cm,
ОН – distance from point О of side AB,
Find the degree measures: angle A, angle B, angle C, angle D -?
Solution:
1) AO = 1/2 * AC;
AO = 1/2 * 68;
AO = 34 cm;
2) consider a right-angled triangle AНO. The sine of the angle in a right-angled triangle is equal to the ratio of the opposite leg to the hypotenuse. Consequently:
sin HAO = HO / AC;
sin HAO = 17/34;
sin HAO = 1/2;
angle HAO = 30 degrees.
Then angle A = angle C = 30 * 2 = 60 degrees;
3) The sum of the degree measures of any quadrangle is 360 degrees.
angle B = angle D = (360 – 120) / 2 = 120 degrees.
Answer: 60 degrees; 60 degrees; 120 degrees; 120 degrees.