An opaque cylindrical tank with an open top has a diameter of 3.00m and is completely filled with water. When the setting sun reaches an angle of 37 degrees above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. The tank will be [[1]] m deep.

With the aid of Snell's Law, we will first determine the angle of incidence and the value of$\sin r$. In order to further solve this answer, we will then determine the values of the trigonometric functions $\sin ,\cos$ and tan. Also, refer to the illustration.
Formula used: $\mu =\frac{\sin i}{\sin r},\cos \theta =\sqrt{1-{\sin }^2\theta },\tan \theta =\frac{\sin \theta }{\cos \theta }$

h= depth, angle incident to normal a
angle of reflection $=b , a=90-37=53°$
Using Snells law;
(n) air $\times \sin {53}^0=\left(n\right)$ water $\times \sin \left(b\right)$
(i)eq
n for air $=1$
n for water $=1.33$
putting in equation(i) above
$\mathrm{b}=1\times \frac{0.79}{1.33}$
$b=36.86$
Now $\tan \left(b\right)=\frac{d}{h}$

$h=\frac{3}{\tan (36.86)}$
$h=4.00\simeq 4 \mathrm{m}$

Hence, the correct answer is 4.