Brewster's angle for light entering from air to water medium is($\mu^w_a = 1.33$):

$sin^{-1}(1.33)$ $sin^{-1}(\frac{1}{1.33})$ $tan^{-1}(1.33)$ $tan^{-1}(\frac{1}{1.33})$

$tan^{-1}(1.33)$

$ \text{For Brewster angle of incidence } i_p+r = 90^o$ $ r = 90^o - i_p $ $ \text{From snell's law } \frac{sin i_p}{sinr} = \frac{\mu_w}{\mu_a} = 1.33$ $\Rightarrow \frac{sin i_p}{sin(90^o - i_p)} = 1.33$ $ \Rightarrow tan i_p = 1.33$ $i_p = tan^{-1}(1.33)$