An opaque cylindrical tank with an open top has a diameter fo`3.00m`and is completely filled with water .When the settign sun reaches an angle of `37^(@)`above the horizon,sunlight ceases to illuminate any part of the bottom of thetank .How deep is the tank?

Explanation

To find the depth of the cylindrical tank, we can use the geometry of the situation along with some trigonometric principles.

### Step-by-Step Solution:

1. **Understand the Geometry**:
The cylindrical tank has a diameter of 3.00 m, which means the radius \( r \) is:
\[ r = \frac{3.00 \, \text{m}}{2} = 1.50 \, \text{m} \]

2. **Identify the Angle of Incidence**:
The angle of the sun above the horizon is given as \( 37^\circ \). This angle will help us determine how far the sunlight can penetrate into the tank.

3. **Draw a Diagram**:
Visualize a right triangle formed by the radius of the tank, the depth of the tank \( h \), and the line of sunlight hitting the water surface. The angle at the water surface is \( 37^\circ \).

4. **Use Trigonometry**:
In the right triangle, the tangent of the angle can be expressed as:
\[ \tan(37^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{r} \]
where \( h \) is the depth of the tank and \( r \) is the radius of the tank.

5. **Substitute the Known Values**:
We can now substitute the radius \( r = 1.50 \, \text{m} \) into the equation:
\[ \tan(37^\circ) = \frac{h}{1.50 \, \text{m}} \]

6. **Calculate \( \tan(37^\circ) \)**:
Using a calculator or trigonometric table, we find:
\[ \tan(37^\circ) \approx 0.7536 \]

7. **Solve for \( h \)**:
Rearranging the equation gives:
\[ h = 1.50 \, \text{m} \cdot \tan(37^\circ) \approx 1.50 \, \text{m} \cdot 0.7536 \approx 1.1304 \, \text{m} \]

8. **Final Result**:
The depth of the tank is approximately:
\[ h \approx 1.13 \, \text{m} \]