<p>We have ABC is a triangular park with AB = AC = 100 meters. M is the midpoint of BC and let the height of tower PM be \(h\) meters. If \(\cot\alpha = 3\sqrt{2}\) (where \(\alpha\) is the angle of elevation of P from A) and \(\cosec\beta = 2\sqrt{2}\) (where \(\beta\) is the angle of elevation of P from B), find \(h\) (in meters).</p>
Step-by-Step Solution
Key Concept: Use the angles of elevation from two different points (A and B) to set up two equations involving the height h and distances AM, BM. Since M is the midpoint of BC on an isosceles triangle, AM ⊥ BC, allowing you to apply trigonometry in right triangles APM and BPM separately.
<p><strong>Step 1:</strong> Set up the geometry. Since ABC is isosceles with AB = AC = 100, and M is the midpoint of BC, we have AM ⊥ BC. Let AM = d and BM = x.</p><p><strong>Step 2:</strong> From point A, angle of elevation to P is α where cot α = 3√2. In right triangle APM: cot α = AM/PM = d/h, so d/h = 3√2, giving d = 3√2·h</p><p><strong>Step 3:</strong> From point B, angle of elevation to P is β where cosec β = 2√2. In right triangle BPM (right angle at M): cosec β = BP/PM. First find BP using Pythagoras: BP² = BM² + PM² = x² + h². So cosec β = √(x² + h²)/h = 2√2, giving x² + h² = 8h²</p><p><strong>Step 4:</strong> This simplifies to x² = 7h², so x = √7·h</p><p><strong>Step 5:</strong> In right triangle ABM: AB² = AM² + BM², so 100² = (3√2·h)² + (√7·h)²</p><p><strong>Step 6:</strong> 10000 = 18h² + 7h² = 25h². Therefore h² = 400, so h = 20 meters</p><p>∴ Answer: 20</p>
Correct Answer: 20