The diagonals of the rhombus are 10 cm and 10√3 cm. Find the corners of the rhombus.

Let’s say we are given a rhombus ABCD. The diagonals of this rhombus when they intersect are divided in half at the point O. => 4 equal triangles turned out. Consider one of them Triangle ABO in which AO = 5, BO = 5√3, angle AOB = 90 degrees By the Pythagorean theorem, we find the hypotenuse AB = √AO ^ 2 + BO ^ 2 = √25 + 75 = √100 = 10 Side of the rhombus equal to 10 cm By the theorem of sines, we find the angle of the triangle AB / sinO = AO / sinB sinB = sinO * AO / AB = 1 * 5/10 = 1/2 Angle B = 30 degrees => angle BAO = 60 degrees Angle of the rhombus ABC = ADC = 60 degrees Rhombus angle DAB = DCB = 120 degrees.