Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression of the car to change from 30° to 45°; then after this, the time taken (in min) by the car to reach the foot of the tower, is</p>
<p>\(9(1+\sqrt{3})\)</p>
<p>\(18(1+\sqrt{3})\)</p>
<p>\(18(\sqrt{3}-1)\)</p>
<p>\(\dfrac{9}{2}(\sqrt{3}-1)\)</p>
Step-by-Step Solution
Key Concept: The angle of depression changes as the car moves closer horizontally; use tan(angle) = height/horizontal_distance to establish relationships between positions, then apply uniform speed to find the remaining time.
<p><strong>Step 1:</strong> Let the height of tower = h, and initial horizontal distance of car from tower = d₁</p><p><strong>Step 2:</strong> From angle of depression 30°: tan(30°) = h/d₁, so d₁ = h√3</p><p><strong>Step 3:</strong> From angle of depression 45°: tan(45°) = h/d₂, so d₂ = h</p><p><strong>Step 4:</strong> Distance traveled in 18 min = d₁ - d₂ = h√3 - h = h(√3 - 1)</p><p><strong>Step 5:</strong> Speed of car = h(√3 - 1)/18 per minute</p><p><strong>Step 6:</strong> Remaining distance to tower = d₂ = h</p><p><strong>Step 7:</strong> Time to reach tower = h ÷ [h(√3 - 1)/18] = 18/(√3 - 1)</p><p><strong>Step 8:</strong> Rationalize: 18/(√3 - 1) × (√3 + 1)/(√3 + 1) = 18(√3 + 1)/2 = 9(√3 + 1) = 9√3 + 9 ≈ 15.59 + 9 = 9(√3 + 1) min</p><p><strong>Step 9:</strong> Simplifying: 9(√3 + 1) ≈ 24.59 min, but exact answer = 9(√3 + 1) or approximately <strong>9 min</strong> (if options suggest simpler form) or <strong>9(√3 + 1) min</strong></p><p>∴ Answer: A</p>
Correct Answer: A