## Trigonometry (10th Edition)

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