Trigonometry & Inverse Trigonometry
Heights and Distances
Grade None
Question:
<p>A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point \(O\) on the ground is 45°. It flies off horizontally straight away from the point \(O\). After one second, the elevation of the bird from \(O\) is reduced to 30°. Then the speed (in m/s) of the bird is</p>
<p>\(20\sqrt{2}\)</p>
<p>\(20(\sqrt{3}-1)\)</p>
<p>\(40(\sqrt{2}-1)\)</p>
<p>\(40(\sqrt{3}-\sqrt{2})\)</p>
Step-by-Step Solution
Key Concept: The bird maintains constant height (20 m) while flying horizontally. Use two elevation angles from the same observation point to find two different horizontal distances, then calculate the horizontal distance covered in 1 second.
<p><strong>Step 1:</strong> Find initial horizontal distance from point O using the 45° elevation angle.</p><p>tan(45°) = height/distance₁</p><p>1 = 20/distance₁</p><p>distance₁ = 20 m</p><p><strong>Step 2:</strong> Find horizontal distance after 1 second using the 30° elevation angle.</p><p>tan(30°) = height/distance₂</p><p>1/√3 = 20/distance₂</p><p>distance₂ = 20√3 m</p><p><strong>Step 3:</strong> Calculate the horizontal distance traveled in 1 second.</p><p>Distance flown = distance₂ - distance₁ = 20√3 - 20 = 20(√3 - 1) m</p><p><strong>Step 4:</strong> Speed = distance/time = 20(√3 - 1)/1 = 20(√3 - 1) m/s</p><p>Since √3 ≈ 1.732: Speed = 20(0.732) ≈ 14.64 m/s ≈ <strong>20(√3 - 1) m/s</strong></p><p>∴ Answer: B</p>
Correct Answer: B