3. WATER TOWERS A student can see a water tower from the closest point of the soccer field at San Lobos High School. The edge of the soccer field is about 110 feet from the water tower and the water tower stands at a height of 32.5 feet. What is the angle of elevation if the eye level of the student viewing the tower from the edge of the soccer field is 6 feet above the ground? Round to the nearest tenth.

Question

UpStudy AI · Accepted Answer

Given:

- Height of the water tower: $$32.5$$ feet  
- Student’s eye level: $$6$$ feet  
- Horizontal distance from student to tower: $$110$$ feet

1. **Determine the vertical distance** between the top of the water tower and the student’s eye level:
   $$
   \text{Vertical difference} = 32.5 - 6 = 26.5 \text{ feet}
   $$

2. **Set up the tangent relationship** in the right triangle formed by the vertical difference and the horizontal distance:
   $$
   \tan \theta = \frac{26.5}{110}
   $$

3. **Solve for $$\theta$$** by taking the arctan (inverse tangent):
   $$
   \theta = \arctan\left(\frac{26.5}{110}\right)
   $$

4. **Calculate the value**:
   $$
   \theta \approx \arctan(0.2409) \approx 13.6^\circ
   $$

Thus, the angle of elevation is approximately $$13.6^\circ$$.
The angle of elevation is approximately 13.6 degrees.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Trigonometry Questions