Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>If the angles of elevation of the top of a tower from three collinear points \(A\), \(B\) and \(C\), on a line leading to the foot of the tower, are 30°, 45° and 60° respectively, then the ratio \(AB : BC\), is</p>
<p>\(\sqrt{3}:\sqrt{2}\)</p>
<p>\(1:\sqrt{3}\)</p>
<p>\(2:3\)</p>
<p>\(\sqrt{3}:1\)</p>
Step-by-Step Solution
Key Concept: Use the tangent relationship (tan θ = h/distance) from each point to the tower. Since all three points lie on the same line leading to the tower foot, their distances are ordered, and you can express AB and BC in terms of the tower height h.
<p><strong>Step 1:</strong> Let the tower height be <em>h</em> and foot be O. From points A, B, C with angles of elevation 30°, 45°, 60°:</p><p>tan 30° = h/OA ⟹ OA = h/tan 30° = h√3</p><p>tan 45° = h/OB ⟹ OB = h/tan 45° = h</p><p>tan 60° = h/OC ⟹ OC = h/tan 60° = h/√3</p><p><strong>Step 2:</strong> Since points are collinear on a line toward O, with A farthest, then B, then C closest:</p><p>AB = OA − OB = h√3 − h = h(√3 − 1)</p><p>BC = OB − OC = h − h/√3 = h(1 − 1/√3) = h(√3 − 1)/√3</p><p><strong>Step 3:</strong> Calculate the ratio:</p><p>AB : BC = h(√3 − 1) : h(√3 − 1)/√3 = 1 : 1/√3 = √3 : 1</p><p>∴ <strong>Answer: D</strong> (√3 : 1)</p>
Correct Answer: D