Final answer:
Using the tangent function and the given angle of sunlight (23.2°), the depth of the tank can be determined by applying the trigonometric ratio to the radius of the tank and the angle at which sunlight no longer reaches the bottom.
Step-by-step explanation:
The question pertains to the principles of geometry and optics, specifically involving the angle of sunlight as it reaches the tank. Given the opaqueness of the tank and the sunlight's angle of incidence at 23.2° above the horizon, we can use trigonometric relationships to determine the depth of the tank. The fact that sunlight no longer illuminates the bottom of the tank when the sunlight is at a 23.2° angle suggests that we're looking at a right triangle formed by the sunlight's path, the depth of the tank, and the radius of the tank as the base.
Let the depth of the tank be 'd', the radius of the tank 'r' (half of the diameter, so 3.40 m / 2 = 1.70 m), and the angle of the sunlight 'θ' (23.2°). Using the tangent function which is the ratio of the opposite side (depth 'd') over the adjacent side (radius 'r'), we have tan(θ) = d/r. Hence, d = r * tan(θ) = 1.70 m * tan(23.2°). Calculating this provides the depth 'd' of the tank.
It is important to note that the Sun's rays detectable at the bottom of the tank will form an angle with the vertical that is the complement of 23.2° (90° - 23.2° = 66.8°). Thus, the angle to calculate with tangent would be 66.8°. Finally, the depth of the tank (d) would be 1.70 m * tan(66.8°).