#### Answer

The apparent angle above the horizon is $7.9^{\circ}$.

#### Work Step by Step

If the angle above the horizon is $29.6^{\circ}$, then the angle $\theta_1$ from the vertical is $60.4^{\circ}$
We can use Snell's law to find $\theta_2$:
$\frac{c_1}{c_2} = \frac{sin~\theta_1}{sin~\theta_2}$
$sin~\theta_2 = \frac{c_2~sin~\theta_1}{c_1}$
$sin~\theta_2 = \frac{(2.254\times 10^8~m/s)~sin(60.4^{\circ})}{3\times 10^8~m/s}$
$\theta_2 = sin^{-1}(\frac{(2.254\times 10^8~m/s)~sin(60.4^{\circ})}{3\times 10^8~m/s})$
$\theta_2 = 40.8^{\circ}$
From Exercise 71, we know that the angle of the apparent horizon is $48.7^{\circ}$ from the vertical. To a fish, the apparent angle above the horizon is $48.7^{\circ}-40.8^{\circ}$ which is $7.9^{\circ}$.