<p>Given two poles ED = 5 m and CB = 10 m, ∠BAC = 15°. Find the distance DB (in meters).</p>
Step-by-Step Solution
Key Concept: Use the tangent ratio in right triangles formed by the poles and apply the angle condition ∠BAC = 15° to relate the horizontal distances. The key is recognizing that tan(15°) = 2 - √3 and setting up equations from two right triangles sharing point A.
<p><strong>Step 1:</strong> Set up coordinate system with B at origin. Let A be at horizontal distance x from B, with CB = 10 m (height) and ED = 5 m (height at distance x + DB from A).</p><p><strong>Step 2:</strong> From point A, tan(∠CAB) = CB/AB and tan(∠EAD) = ED/AD. Since ∠BAC = 15°, we have tan(15°) = 10/x, giving x = 10/tan(15°) = 10/(2 - √3).</p><p><strong>Step 3:</strong> Rationalize: x = 10(2 + √3)/((2 - √3)(2 + √3)) = 10(2 + √3)/(4 - 3) = 10(2 + √3) = 20 + 10√3 ≈ 37.32 m</p><p><strong>Step 4:</strong> Similarly, for the angle at E: the angle of depression/elevation to D from A relates to the geometry. Using tan(∠EAD) = 5/(x + DB), and the constraint from the angle condition, DB = distance needed such that the configuration is consistent.</p><p><strong>Step 5:</strong> From tan(15°) = (10 - 5)/(DB) when considering the angle between the lines of sight: 2 - √3 = 5/DB, so DB = 5/(2 - √3) = 5(2 + √3) = 10 + 5√3 ≈ 18.66 m</p><p>∴ Answer: <strong>18.6602 m</strong></p>
Correct Answer: 18.6602