One of the corners of the rhombus is 72 °. Find the angles that form the side of the rhombus with its diagonals.

ABCD – rhombus, AC and BD – diagonals, angle A = 72 °.
1. The opposite angles of the rhombus are equal, then angle A = angle C = 72 °, angle B = angle D = x.
By the theorem on the sum of the angles of a triangle:
angle A + angle B + angle C + angle D = 360 °;
72 ° + x + 72 ° + x = 360 °;
2x = 360 ° – 144 °;
2x = 216 °;
x = 216 ° / 2;
x = 108 °.
Then, angle B = angle D = x = 108 °.
2. The diagonals of a rhombus are the bisectors of its angles:
angle CAB = angle CAD = angle BCA = angle DCA = angle A / 2 = angle C / 2 = 72 ° / 2 = 36 °;
angle ABD = angle CBD = angle ADB = angle CDB = angle B / 2 = angle D / 2 = 108 ° / 2 = 54 °.
Answer: angle CAB = angle CAD = angle BCA = angle DCA = 36 °, angle ABD = angle CBD = angle ADB = angle CDB = 54 °.