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 O. After 1 second, the elevation of the bird from O is reduced to 30°. The speed (in m/s) of the bird is</p>
<p>(1) \(40(\sqrt{2}-1)\)</p>
<p>(2) \(20(\sqrt{3}-1)\)</p>
<p>(3) \(20\sqrt{2}\)</p>
<p>(4) \(40(\sqrt{3}-2)\)</p>
Step-by-Step Solution
Key Concept: The bird flies horizontally from the top of a 20m pole, so its height remains constant at 20m. Use elevation angles at two different times to find the horizontal distances from O, then calculate speed from the distance change in 1 second.
<p><strong>Step 1:</strong> Set up the geometry. The bird starts at height h = 20m on top of the pole. Elevation angle from O initially is 45°.</p><p><strong>Step 2:</strong> At t = 0, using tan(45°) = height/horizontal distance: tan(45°) = 20/d₀, so 1 = 20/d₀, giving d₀ = 20 m</p><p><strong>Step 3:</strong> After 1 second, the elevation angle becomes 30°. If the bird is at horizontal distance d₁ from O: tan(30°) = 20/d₁, so 1/√3 = 20/d₁, giving d₁ = 20√3 m</p><p><strong>Step 4:</strong> The horizontal distance traveled by the bird in 1 second is: Δd = d₁ - d₀ = 20√3 - 20 = 20(√3 - 1) m</p><p><strong>Step 5:</strong> Speed = Distance/Time = 20(√3 - 1)/1 = 20(√3 - 1) ≈ 20(1.732 - 1) = 20(0.732) ≈ 14.64 m/s ≈ 20(√3 - 1) m/s</p><p>∴ Answer: B</p>
Correct Answer: B