Why the bond angle of $N{H}_4^{+}$ is greater than $N{H}_3$?

$N{H}_3$ has a lone pair of electrons that strongly repels the bonded pair of electrons and reduces the bond angle. On the other hand, $N{H}_4^{+}$ does not have any lone pair of electrons; hence, there is no lone pair–bond pair repulsion. Thus, the bond angle in $N{H}_4^{+}$ is higher than that in $N{H}_3$.