A boat heading out to sea starts out at Point \( A \), at a horizontal distance of 1046 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 \( 6^{\circ} \). Find the distance from point \( A \) to point \( B \). Round your answer to the nearest foot if necessary.

To determine the distance between points \( A \) and \( B \) where a boat measures different angles of elevation to a lighthouse, we can use trigonometric relationships. ### Step 1: Understand the Geometry - **Point \( A \)**: The boat is 1046 feet horizontally from the lighthouse and measures a \( 15^\circ \) angle of elevation to the beacon. - **Point \( B \)**: Later, the boat measures a \( 6^\circ \) angle of elevation from a new position, farther from the lighthouse. ### Step 2: Set Up the Equations Let \( h \) be the height of the lighthouse beacon. From **Point \( A \)**: \[ \tan(15^\circ) = \frac{h}{1046} \quad \Rightarrow \quad h = 1046 \cdot \tan(15^\circ) \] From **Point \( B \)**, let the horizontal distance be \( x \): \[ \tan(6^\circ) = \frac{h}{x} \quad \Rightarrow \quad h = x \cdot \tan(6^\circ) \] ### Step 3: Equate and Solve for \( x \) \[ 1046 \cdot \tan(15^\circ) = x \cdot \tan(6^\circ) \] \[ x = \frac{1046 \cdot \tan(15^\circ)}{\tan(6^\circ)} \] ### Step 4: Calculate Using Trigonometric Values Using approximate values: \[ \tan(15^\circ) \approx 0.2679 \quad \text{and} \quad \tan(6^\circ) \approx 0.1051 \] \[ x \approx \frac{1046 \times 0.2679}{0.1051} \approx 2666.4 \text{ feet} \] ### Step 5: Find the Distance \( AB \) \[ AB = x - 1046 \approx 2666.4 - 1046 = 1620.4 \text{ feet} \] Rounding to the nearest foot: **Answer:** The distance from point \( A \) to point \( B \) is approximately \(\boxed{1620}\) feet.