How does VSEPR theory explain the formation of different bond angles in PCl₅? We know, that in the exited state, electrons of Phosphorus could go to higher states and form $\ce{dsp^3}$ hybridization. Therefore it will create five equal energy states of electrons. But my question is that, why are the angles of the bonds in $\ce{PCl5}$  90 and 120 degree instead of equally distributed? How does VSEPR theory explain the difference in the angles?

Why do the  five equal bonding pairs of electrons have a trigonal
 bipyramidal arrangement, with two different internal bond angles -
 120° in the plane and 90° between the plane and the axial atoms?
There are only five Platonic solids: 


tetrahedron, 
octahedron, 
cube, 
dodecahedron, and 
isocahedron.


Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane.
Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages Min-Energy Configurations of Electrons On A Sphere (Stable link via The Internet Archive).

Also there are not five equal energy states as demonstrated in the lecture notes by Marcel Schlaf and Kathryn Preuss from the University of Guelph:
Chapter 5.5 (continued 2) MO Theory ICl4, PCl5. (pdf) (Stable pdf link via The Internet Archive)