Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>A tower \(T_1\) of height 60 m is located exactly opposite to a tower \(T_2\) of height 80 m on a straight road. From the top of \(T_1\), if the angle of depression of the foot of \(T_2\) is twice the angle of elevation of the top of \(T_2\), then the width (in m) of the road between the feet of the towers \(T_1\) and \(T_2\) is</p>
<p>\(10\sqrt{2}\)</p>
<p>\(10\sqrt{3}\)</p>
<p>\(20\sqrt{3}\)</p>
<p>\(20\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Set up two angle equations from the top of T₁: let α be the angle of elevation to top of T₂, then 2α is the angle of depression to foot of T₂. Use tan(α) = 20/d and tan(2α) = 60/d, where d is the road width, then apply the double angle formula tan(2α) = 2tan(α)/(1-tan²(α)).
<p><strong>Step 1:</strong> Let the road width be d meters. From the top of T₁ (at height 60m), let α = angle of elevation to top of T₂ and 2α = angle of depression to foot of T₂.</p><p><strong>Step 2:</strong> For angle of elevation to top of T₂: The vertical rise from top of T₁ to top of T₂ is (80-60) = 20m. So tan(α) = 20/d</p><p><strong>Step 3:</strong> For angle of depression to foot of T₂: The vertical drop from top of T₁ to ground is 60m. So tan(2α) = 60/d</p><p><strong>Step 4:</strong> Apply double angle formula: tan(2α) = 2tan(α)/(1-tan²(α))</p><p>60/d = 2(20/d)/(1-(20/d)²)</p><p>60/d = (40/d)/(1-400/d²)</p><p>60/d = (40/d)/((d²-400)/d²)</p><p>60/d = (40d)/(d²-400)</p><p><strong>Step 5:</strong> Cross multiply: 60(d²-400) = 40d²</p><p>60d² - 24000 = 40d²</p><p>20d² = 24000</p><p>d² = 1200</p><p>d = 20√3 m</p><p>∴ Answer: C</p>
Correct Answer: C