Solve Snell's Law Problem: 2.1m Deep Pool

[negatifzeo]

Homework Statement

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, h = 1.6 m above the water level, onto the surface of the water at a point L = 3.0 m from his foot at the edge of the pool (Fig. 23-52). Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 2.1 m deep?

Homework Equations

$$n_{{1}}\sin \left( \theta_{{1}} \right) =n_{{2}}\sin \left( \theta_{{2<br /> }} \right)$$

The Attempt at a Solution

This problem seems straightforward, but the answer I am getting is incorrect according to my online homework submissions. I am in degree mode on my calculator. :) First I used the pythag theorem to get the hypotenuse of the first triangle and the angle with the water, which is 28.07. I subtracted this from 90 to get the angle from the normal line. This gives me 61.928 degrees. I am using 1 as the index of refraction of air and 1.33 for water, giving me $$0.7857161968= 1.33\,\sin \left( \theta_{{2}} \right)$$.
Solving this gives me $$\theta_{{2}}= 41.561$$ The depth of the pool was 2.1m. Now I just used Tan(41.561)=L/2.1 andsolved for L to get 1.86, which shows as incorrect. Can anyone see where I went wrong, or is my entire approach off?

 

[negatifzeo]

Oops. Duh, I forgot to add the original L. L=4.86. That was a waste of time but I'll leave it here just in case someone wants to see...