Find the smallest corner of a rhombus with diagonals 2 and √12.

The diagonals of the rhombus ABCD meet at point O. According to the properties of the diagonals:
∟AOВ = 90˚;
AO = OС = 1;
BO = OD = √12 / 2 = 2√3 / 2 = √3;
AO and BO are the bisectors of the angles A and B of the rhombus.
Consider ∆ AOB (∟AOB = 90˚):
tg∟BAO = ВO / AO = √3, which means
∟BAO = 60˚, then the entire angle A of the rhombus is 120˚.
∟В = 180˚ – ∟А = 180˚ – 120˚ = 60˚.
Answer. The smallest angle is 60˚.