A boat is heading towards a lighthouse, where Madeline is watching from a vertical
distance of 140 feet above the water. Madeline measures an angle of depression to the
boat at point to be . At some later time, Madeline takes another measurement and
finds the angle of depression to the boat (now at point ) to be . Find the distance
from point to point . Round your answer to the nearest tenth of a foot if necessary.

Ask by Cross Lawson. in the United States

Jan 27,2025

Question

A boat is heading towards a lighthouse, where Madeline is watching from a vertical
distance of 140 feet above the water. Madeline measures an angle of depression to the
boat at point to be . At some later time, Madeline takes another measurement and
finds the angle of depression to the boat (now at point ) to be . Find the distance
from point to point . Round your answer to the nearest tenth of a foot if necessary.

Ask by Cross Lawson. in the United States

Jan 27,2025

UpStudy AI · Accepted Answer

To determine the distance the boat travels from point $$ A $$ to point $$ B $$, we'll analyze the situation using trigonometry.

**Given:**
- Height of Madeline above water, $$ H = 140 $$ feet.
- Angle of depression to point $$ A $$, $$ \theta_A = 5^\circ $$.
- Angle of depression to point $$ B $$, $$ \theta_B = 50^\circ $$.

**Approach:**
1. **Understand the Angles of Depression:**
   - The angle of depression from Madeline to the boat is the same as the angle of elevation from the boat to Madeline due to alternate interior angles (assuming a horizontal line from Madeline).

2. **Set Up Right Triangles:**
   - For point $$ A $$:
     $$
     \tan(\theta_A) = \frac{H}{x_A} \implies x_A = \frac{H}{\tan(\theta_A)} = \frac{140}{\tan(5^\circ)}
     $$
   - For point $$ B $$:
     $$
     \tan(\theta_B) = \frac{H}{x_B} \implies x_B = \frac{H}{\tan(50^\circ)} = \frac{140}{\tan(50^\circ)}
     $$
   - The distance traveled by the boat, $$ AB $$, is the difference between these horizontal distances:
     $$
     AB = x_A - x_B
     $$

3. **Calculate the Distances:**
   - Calculate $$ \tan(5^\circ) $$ and $$ \tan(50^\circ) $$ using a calculator:
     $$
     \tan(5^\circ) \approx 0.0874887 \quad \text{and} \quad \tan(50^\circ) \approx 1.1917536
     $$
   - Compute $$ x_A $$ and $$ x_B $$:
     $$
     x_A \approx \frac{140}{0.0874887} \approx 1600.2 \text{ feet}
     $$
     $$
     x_B \approx \frac{140}{1.1917536} \approx 117.7 \text{ feet}
     $$
   - Calculate $$ AB $$:
     $$
     AB \approx 1600.2 \text{ feet} - 117.7 \text{ feet} = 1482.5 \text{ feet}
     $$

**Final Answer:**
The boat travels approximately **1,482.5 feet** from point $$ A $$ to point $$ B $$.

**Answer:**  
After rounding, the distance from A to B is 1 482.5 feet.
The boat travels approximately 1,482.5 feet from point A to point B.

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