A ray of light passes from a denser medium to a rarer medium at an angle of incidence $$i$$. The reflected and refracted rays make an angle of 90$$^\circ$$ with each other. The angle of reflection and refraction are respectively $$r$$ and $$r'$$. The critical angle is given by,

A ray travels from denser medium to rarer medium at angle of incidence $$i$$. The angle of reflection equals the angle of incidence, so $$r = i$$.

The reflected and refracted rays are perpendicular to each other, so: $$r + r' = 90°$$ $$r' = 90° - r = 90° - i$$

By Snell's law at the interface (denser to rarer): $$n_1 \sin i = n_2 \sin r'$$

where $$n_1$$ is the refractive index of the denser medium and $$n_2$$ is that of the rarer medium. Assuming the rarer medium is air ($$n_2 = 1$$): $$n_1 \sin i = \sin r' = \sin(90° - i) = \cos i$$ $$n_1 = \frac{\cos i}{\sin i} = \cot i = \cot r$$

The critical angle $$\theta_c$$ is defined by $$\sin\theta_c = \frac{n_2}{n_1} = \frac{1}{n_1}$$, so: $$\sin\theta_c = \frac{1}{\cot r} = \tan r$$ $$\theta_c = \sin^{-1}(\tan r)$$