Key Idea : As the number of lone pair of electrons increases, bond angle decreases.
$\mathrm{NO}_2^{+}$ion is isoelectronic with $\mathrm{CO}_2$ molecule. It is a linear ion and its central atom $\left(\mathrm{N}^{+}\right)$ undergoes $s p$-hybridisation, hence bond angle is $180^{\circ}$.
In $\mathrm{NO}_2^{-}$ion, $\mathrm{N}$-atom undergoes $s p^2$ hybridisation. The angle between hybrid orbital should be $120^{\circ}$ but one lone pair of electrons is lying on $\mathrm{N}$-atom, hence bond angle decreases to $115^{\circ}$.
In $\mathrm{NO}_2$ molecule, $\mathrm{N}$-atom has one unpaired electron in $s p^2$-hybrid orbital. The bond angle should be $120^{\circ}$ but actually it is $132^{\circ}$. It may be due to one unpaired electron in $s p^2$-hybrid orbital.
Therefore, the increasing order of bond angles is :
$\underset{\left(115^{\circ}\right)}{\mathrm{NO}_2^{-}} < \underset{\left(132^{\circ}\right)}{\mathrm{NO}_2} < \underset{\left(180^{\circ}\right)}{\mathrm{NO}_{+}^{+}}$