Problem 18.9 an underwater diver sees the sun 60 ∘ above horizontal. part a how high is the sun above the horizon to a fisherman in a boat above the diver?

The angles in the equation are not the angles relative to the horizon but are relative to the "normal" which means that the line that is perpendicular to the surface.

The angle under the water is 90 – 60 = 30. 

n1 for water is 1.33, n2 for air is 1 Which you seem to understand. 

(1.33)(sin (30)) = (1.00)(sin (x2)) 

Rearranging the equation above, will give us: x^2 = sin^-1((1.33 sin 30)/1) = 41.68

But remember that that is the angle relative to the normal so you have to deduct it from 90 to get the angle relative to the horizon and you get (90 – 41.68) = 48.32 degrees.

where are the vectors ?

gas and liquid