Based on VSEPR theory, the number of 90 degree $\mathrm{F}-\mathrm{Br}-\mathrm{F}$ angles in ${\mathrm{BrF}}_5$ is:

According to VSEPR, the valence electron pairs that surround an atom have a tendency to repel one another. As a result, they will arrange themselves in a way that minimizes this repulsion, which determines the geometry of the molecule. After lone pair-bond pair repulsion, all four planar bonds (F, Br, and F) will decrease from $90°$to ${84.8}^{\circ }$

ln ${\mathrm{BrF}}_5$, none of the bond angles are $90°$