The Brewsters angle i _ b for an interface should be

Brewster's angle is defined as the angle of incidence at which the reflection of particular polarized light is zero, and it perfectly transmitted through a transparent surface.

Brewster's angle(${i}_{b}$) is given by, $\tan \left({i}_{b}\right) = \dfrac{{\mu}_{2}}{{\mu}_{1}} = \dfrac{{\mu}_{2}}{1}$, where ${\mu}_{2}$ is the refractive index of the dielectric and ${\mu}_{1}$ is the refractive index of air.
We know that, ${\mu}_{2} > 1$. (refractive index of any dielectric will be greater than refractive index of air.)
$⇒ \infty 1$
$⇒ {90}^{o} > {i}_{b} > {45}^{o}$