Consider light travelling from a medium A to medium B separated by...

The refractive index of a medium is inversely proportional to the speed of light in that medium, given by $n = \frac{c}{v}$.
For light travelling from medium $A$ to medium $B$, the critical angle $\theta_c$ is defined when light travels from a denser to a rarer medium.
Given speeds are $v_A = 2.4 \times 10^8$ m/s and $v_B = 2.7 \times 10^8$ m/s.
The refractive indices are $n_A = \frac{c}{v_A}$ and $n_B = \frac{c}{v_B}$.
The critical angle $\theta_c$ is given by $\sin \theta_c = \frac{n_B}{n_A} = \frac{v_A}{v_B}$.
Substituting the values: $\sin \theta_c = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9}$.
To find the value in terms of $\tan^{-1}$, we use the trigonometric identity for a right-angled triangle where perpendicular $p = 8$ and hypotenuse $h = 9$.
The base $b = \sqrt{h^2 - p^2} = \sqrt{9^2 - 8^2} = \sqrt{81 - 64} = \sqrt{17}$.
Thus, $\tan \theta_c = \frac{p}{b} = \frac{8}{\sqrt{17}}$.
Therefore, $\theta_c = \tan^{-1}\left(\frac{8}{\sqrt{17}}\right)$.