<p>The angle of elevation of the top of a vertical tower from a point \(A\), due east of it is 45°. The angle of elevation of the top of the same tower from a point \(B\), due south of \(A\) is 30°. If the distance between \(A\) and \(B\) is \(54\sqrt{2}\) m, then the height of the tower (in metres) is:</p>
Step-by-Step Solution
Key Concept: Set up a coordinate system with the tower base at origin, use angle of elevation relationships to express height in terms of distances from tower, then apply the constraint that distance AB = 54√2 to solve for height.
<p><strong>Step 1:</strong> Set up coordinates with tower base at O (origin). Point A is due east of tower, so A = (a, 0). Point B is due south of A, so B = (a, -b) where b = AB distance component southward.</p><p><strong>Step 2:</strong> From angle of elevation at A = 45°: tan(45°) = h/a, so h = a (since tan 45° = 1)</p><p><strong>Step 3:</strong> From angle of elevation at B = 30°: tan(30°) = h/OB, where OB = √(a² + b²). So 1/√3 = h/√(a² + b²), giving √(a² + b²) = h√3</p><p><strong>Step 4:</strong> The distance AB lies along the south direction from A to B, so AB = b = 54√2</p><p><strong>Step 5:</strong> Substitute into OB equation: √(a² + (54√2)²) = h√3. Since h = a: √(h² + 5832) = h√3</p><p><strong>Step 6:</strong> Square both sides: h² + 5832 = 3h². This gives 2h² = 5832, so h² = 2916, thus h = 54</p><p>∴ Answer: D (height = 54 metres)</p>
Correct Answer: D