60. How will you account for $104.5°$ bond angle in water?

In water, oxygen has sp3-hybridisation and the bond angle of HOH should have been $109°28\text{'}$. In H2O, the oxygen atom is surrounded by two lone pairs of electrons. From VSEPR theory, lone pair-lone pair repulsions are stronger than bond pair-bond pair repulsions.
As a result, the bond angle of HOH in water slightly decreases from the regular tetrahedral angle of $109°28\text{'}$ and $104.5°$.