A boat heading out to sea starts out at Point \( A \), at a horizontal distance of 815 feet from a lighthouse/the shore. From that point, the boat's crew measures the angle of elevation to the lighthouse's beacon-light to be \( 15^{\circ} \). At some later time, the crew measures the angle of elevation from point \( B \) to be \( 3^{\circ} \). Find the distance from point \( A \) to point \( B \). Round your answer to the nearest tenth of a foot if necessary.

To determine the distance from point \( A \) to point \( B \), let's analyze the situation using trigonometry. 1. **Understanding the Scenario:** - **Point \( A \):** The boat is 815 feet away from the lighthouse, and the angle of elevation to the beacon-light is \( 15^\circ \). - **Point \( B \):** The boat moves further out to sea, and the angle of elevation decreases to \( 3^\circ \). - **Goal:** Find the distance \( AB \) between points \( A \) and \( B \). 2. **Using Tangent for Angle of Elevation:** - The tangent of an angle in a right triangle is the ratio of the opposite side (height of the lighthouse, \( h \)) to the adjacent side (horizontal distance). From point \( A \): \[ \tan(15^\circ) = \frac{h}{815} \] So, \[ h = 815 \times \tan(15^\circ) \] From point \( B \): \[ \tan(3^\circ) = \frac{h}{815 + AB} \] So, \[ h = (815 + AB) \times \tan(3^\circ) \] 3. **Equating the Two Expressions for \( h \):** \[ 815 \times \tan(15^\circ) = (815 + AB) \times \tan(3^\circ) \] 4. **Solving for \( AB \):** \[ AB = \frac{815 \times (\tan(15^\circ) - \tan(3^\circ))}{\tan(3^\circ)} \] 5. **Calculating the Values:** - \(\tan(15^\circ) \approx 0.2679\) - \(\tan(3^\circ) \approx 0.05241\) Plugging these values in: \[ AB = \frac{815 \times (0.2679 - 0.05241)}{0.05241} = \frac{815 \times 0.21549}{0.05241} \approx 3350.7 \text{ feet} \] **Final Answer:** The distance from point \( A \) to point \( B \) is approximately **3350.7 feet**.