refraction: Snell's Law... details below?

Question

an aquarium filled with water has flat glass sides whose index of refraction is 1.52 A beam of light from outside the aquarium strikes the glass at a 43.5degree angle perpendicularly. What is the angle of this light ray when it enters the glass and then the water? what would be the refracted angle if the ray entered the water dircelty? Presuming the outside of the glass is air. Please help!!

Answer 1

n-air=1
n-water=1.33
n-glass=1.52

n-air*sin(43.5)=n-glass*sin(x) ==>

sin(x)=1*sin(43.5)/1.52 ==>x=26.93

n-glass*sin(26.93)=n-water*sin(y)==>

1.52/1.33*sin(26.93=sin(y) ==>
y=31.2

water-air interface

n-air*sin(43.5)=n-water*sin(z) ==>

z=31.2

Answer 2

I think that's a very hard question for anyone to answer. I thought about the question for a while, and I still couldn't answer it. I think another option would be to search similar questions on Yahoo or Google search. If you know anyone else in the subject, try calling them or e-mailing them with the question. I am sorry if I am not helping much, it's the best I could do. I hope it helps, even if it was a little.