ANSWER
EXPLANATION
From the previous part, we have that the index of refraction of the liquid is 1.37, so we have to replace this in the equation and solve,
[tex]\sin\theta_c=\frac{n_{air}}{n_{liquid}}=\frac{1.00}{1.37}\approx0.73[/tex]
And then, take the inverse of the sine to find the critical angle,
[tex]\theta_c=\sin^{-1}0.73\approx46.9\degree[/tex]
Hence, the critical angle is 46.9°, rounded to the nearest tenth.
