The critical angle, denoted as $\theta_c$, is the angle of incidence beyond which light is totally internally reflected within a denser medium when it hits the boundary with a less dense medium. To find the critical angle for the medium in question, first, we need to understand the relationship between the speed of light in different media and their refractive indices.

Let's denote the speed of light in air as $V_1$ and in the denser medium as $V_2$. From the given information:

$V_1 = \frac{x}{t_1}$ (speed of light in air)

$$V_2 = \frac{10x}{t_2}$$ (speed of light in the denser medium)

The refractive index of a medium (n) is inversely proportional to the speed of light in that medium ($n = \frac{c}{V}$, where c is the speed of light in vacuum). Thus, the refractive index of the denser medium ($n_2$) relative to air ($n_1$) can be obtained by taking the ratio of the speed of light in air to that in the denser medium:

$$\frac{n_2}{n_1} = \frac{V_1}{V_2}$$

Substituting the expressions for $V_1$ and $V_2$:

$$\frac{n_2}{n_1} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} = \frac{t_2}{10t_1}$$

For the critical angle $\theta_c$, the light in the denser medium (index $n_2$) strikes the boundary with the less dense medium (index $n_1$) such that it refracts at 90 degrees (escapes along the boundary). Snell's Law at this boundary is:

$$n_2 \sin \theta_c = n_1 $$

Given that $n_1 = 1$ (approximating the refractive index of air), we simplify to:

$$\sin \theta_c = \frac{1}{n_2}$$

Substituting $n_2$ from the earlier relation and simplifying:

$$\sin \theta_c = \frac{10t_1}{t_2}$$

Hence, the critical angle $\theta_c$ is given by:

$$\theta_c = \sin^{-1}\left(\frac{10 t_1}{t_2}\right)$$