Physics optics question! need help Easy 10 points

The critical angle for total internal reflection at a liquid-air interface is 42.5 degrees.

A) If a ray of light traveling in the liquid has an angle of incidence at the interface of 30.5 degrees, what angle does the refracted ray in the air make with the normal

B) If a ray of light traveling in air has an angle of incidence at the interface of 30.5 degrees, what angle does the refracted ray in the liquid make with the normal?

Use: n1 sin(i) = n2 sin(r) (*)
1 = medium light ray starts in
2 = medium light ray travels into
(So when going from liquid to air, 1 = liquid, 2 = air; when going from air to liquid, 1 = air, 2 = liquid)

Refractive index of air = 1

So to find the angles you need to know the refractive index of the liquid [n(liquid)]. You can find this from the critical angle as follows. At the critical condition, where light is travelling from more dense to less dense medium, so from liquid to air:
i=C (or i = 42.5)
r = 90

so
sin(i) = sin C
sin(r) = sin 90 = 1

substitute into (*):
n(liquid) = 1/sinC

You can now use n(liquid) to find your angles using the equation (*)