Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p><em>AB</em> is a vertical pole with <em>B</em> at the ground level and <em>A</em> at the top. A man finds that the angle of elevation of point <em>A</em> from a certain point <em>C</em> on the ground is 60°. He moves away from the pole along the line <em>BC</em> to a point <em>D</em> such that <em>CD</em> = 7 m. From <em>D</em> the angle of elevation of the point <em>A</em> is 45°. Then the height of the pole is</p>
<p>\(\dfrac{7\sqrt{3}}{2}\cdot\left(\dfrac{1}{\sqrt{3}-1}\right)\) m</p>
<p>\(\dfrac{7\sqrt{3}}{2}\cdot(\sqrt{3}+1)\) m</p>
<p>\(\dfrac{7\sqrt{3}}{2}\cdot(\sqrt{3}-1)\) m</p>
<p>\(\dfrac{7\sqrt{3}}{2}\cdot\left(\dfrac{1}{\sqrt{3}+1}\right)\) m</p>
Step-by-Step Solution
Key Concept: Set up two right triangles (ABC and ABD) with the same height AB, use the tangent ratios from different angles, and eliminate the distance BC using the constraint CD = 7 m.
<p><strong>Step 1:</strong> Let height of pole AB = h meters, and BC = x meters.</p><p><strong>Step 2:</strong> From point C, angle of elevation is 60°:</p><p>tan(60°) = AB/BC → √3 = h/x → <strong>h = x√3</strong></p><p><strong>Step 3:</strong> Point D is 7 m away from C, so BD = BC + CD = x + 7</p><p><strong>Step 4:</strong> From point D, angle of elevation is 45°:</p><p>tan(45°) = AB/BD → 1 = h/(x + 7) → <strong>h = x + 7</strong></p><p><strong>Step 5:</strong> Equate the two expressions for h:</p><p>x√3 = x + 7</p><p>x(√3 - 1) = 7</p><p>x = 7/(√3 - 1) = 7(√3 + 1)/[(√3 - 1)(√3 + 1)] = 7(√3 + 1)/2</p><p><strong>Step 6:</strong> Find h:</p><p>h = x + 7 = 7(√3 + 1)/2 + 7 = 7[(√3 + 1)/2 + 1] = 7(√3 + 3)/2</p><p>h = <strong>7(√3 + 3)/2 meters</strong> ≈ <strong>19.56 meters</strong></p><p>∴ Answer: B</p>
Correct Answer: B