g stAn experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a 40o angle (measured from vertical), he knows that he must release his spear at what angle

  θ = 28.9

Explanation:

For this exercise let's use the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for air and index 2 for water where the fish is

        sin θ₂ = n₁ / n₂ sin θ₁

in this case the air repair index is 1 and the water 1.33

we substitute

        sin θ₂ = 1 / 1.33 sin t 40

        sin θ = 0.4833

        θ = sin⁻¹ 0.4833

        θ = 28.9