Trigonometry Problem: Find Distance Between Boat Positions Using Angle of Elevation

Question

A boat heading out to sea starts out at Point A, at a horizontal distance of 1032 feet from a lighthouse/the shore. From that point, the boat's crew measures the angle of elevation to the lighthouse's beacon-light from that point to be 15°. At some later time, the crew measures the angle of elevation from point B to be 6°. Find the distance from point A to point B. Round your answer to the nearest tenth of a foot if necessary.

Accepted Answer

1598.5 feet (or 1599 feet when rounded to whole number)

Answer

Step 1: Calculate the height of the lighthouse using Point A. Let h = height of the lighthouse. Using the tangent function for right triangles (tan(angle) = opposite/adjacent), we get: $tan(15°) = \frac{h}{1032}$. Solving for h: $h = 1032 \times tan(15°) ≈ 1032 \times 0.2679 ≈ 276.52$ feet. Step 2: Find the horizontal distance from Point B to the lighthouse. Let x = horizontal distance from B to the lighthouse. Using $tan(6°) = \frac{h}{x}$, rearrange to solve for x: $x = \frac{h}{tan(6°)} ≈ \frac{276.52}{0.1051} ≈ 2630.5$ feet. Step 3: Compute the distance between Point A and Point B. Since both points are along the horizontal line away from the shore, $AB = x - 1032 ≈ 2630.5 - 1032 = 1598.5$ feet (rounded to 1599 feet as a whole number).