<p>Let a vertical tower <em>AB</em> have its end <em>a</em> on the level ground. Let <em>C</em> be the mid-point of <em>AB</em> and <em>P</em> be a point on the ground such that \(AP = 2AB\). If \(\angle BPC = \beta\), then \(\tan\beta\) is equal to</p>
Step-by-Step Solution
Key Concept: Use coordinate geometry with the tower vertical at origin, then apply the tangent subtraction formula: tan(∠BPC) = tan(∠BPA - ∠CPA) to find the relationship between angles from point P.
<p><strong>Step 1:</strong> Set up coordinates with A at origin, B at (0, h) where AB = h. Then C is at (0, h/2) and P is at (2h, 0) since AP = 2AB = 2h.</p><p><strong>Step 2:</strong> Find tan(∠BPA): In right triangle BPA, tan(∠BPA) = AB/AP = h/(2h) = 1/2</p><p><strong>Step 3:</strong> Find tan(∠CPA): In right triangle CPA, tan(∠CPA) = AC/AP = (h/2)/(2h) = 1/4</p><p><strong>Step 4:</strong> Apply tangent difference formula since β = ∠BPC = ∠BPA - ∠CPA:</p><p>tan(β) = (tan(∠BPA) - tan(∠CPA))/(1 + tan(∠BPA)·tan(∠CPA))</p><p>tan(β) = (1/2 - 1/4)/(1 + (1/2)(1/4)) = (1/4)/(1 + 1/8) = (1/4)/(9/8) = (1/4) × (8/9) = 2/9</p><p><strong>∴ Answer: tan(β) = 2/9 (Option B)</strong></p>
Correct Answer: B