Answer
$\theta = 23.2^{\circ}$
Work Step by Step
We can use Equation (33-45) to find the critical angle $\theta_c$ at point A:
$\theta_c = sin^{-1}~\frac{n_2}{n_1}$
$\theta_c = sin^{-1}~\frac{1.53}{1.58}$
$\theta_c = 75.55^{\circ}$
By geometry, the incident angle at point A is $90^{\circ} - \theta_r$, where $\theta_r$ is the angle of refraction when the ray enters the end of the cable.
We can find the maximum angle $\theta_r$ that will result in total internal reflection:
$\theta_r = 90^{\circ}-75.55^{\circ} = 14.45^{\circ}$
We can use Snell's law to find the maximum value of $\theta$:
$n_{air}~sin~\theta = n_1~sin~\theta_r$
$sin~\theta = \frac{n_1~sin~\theta_r}{n_{air}}$
$\theta = sin^{-1}~(\frac{n_1~sin~\theta_r}{n_{air}})$
$\theta = sin^{-1}~(\frac{1.58~sin~14.45}{1.00})$
$\theta = sin^{-1}~(0.3943)$
$\theta = 23.2^{\circ}$
