Trigonometry & Inverse Trigonometry
Applications of Trigonometry
Grade 11
Question:
<p>Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30° and 60° and as seen from B are 60° and 45°. If AB is 30 m, the distance between the flagstaffs in metres is</p>
<p>(a) \(30 + 15\sqrt{3}\)</p>
<p>(b) \(45 + 15\sqrt{3}\)</p>
<p>(c) \(60 - 15\sqrt{3}\)</p>
<p>(d) \(60 + 15\sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Use trigonometric relations (cot) to express flagstaff heights in terms of distances from observation points, then solve the system of equations.
<p><strong>Solution:</strong> Let x and y be the heights of the flagstaffs at P and Q respectively.</p><p>Then, \(AP = x\cot 60° = \frac{x}{\sqrt{3}}\), \(AQ = y\cot 30° = y\sqrt{3}\)</p><p>\(BP = x\cot 45° = x\), \(BQ = y\cot 60° = \frac{y}{\sqrt{3}}\)</p><p>From \(BP - AP = x - \frac{x}{\sqrt{3}} = AB\):</p><p>\(x\left(1 - \frac{1}{\sqrt{3}}\right) = 30\)</p><p>\(x(\sqrt{3} - 1) = 30\sqrt{3}\)</p><p>\(x = 15(3 + \sqrt{3})\)</p><p>Similarly, \(30 = y\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) \Rightarrow y = 15\sqrt{3}\)</p><p>So that, \(PQ = BP + BQ = x + \frac{y}{\sqrt{3}} = 15(3 + \sqrt{3}) + 15 = (60 + 15\sqrt{3})\) m</p><p>∴ Answer is (d).</p>
Correct Answer: D