<p>The upper \(\left(\dfrac{3}{4}\right)\)th portion of a vertical pole subtends an angle \(\tan^{-1}\left(\dfrac{3}{5}\right)\) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is:</p>
Step-by-Step Solution
Key Concept: Set up two angle equations from the same observation point: one for the full height and one for the lower 1/4th portion, then use the tangent subtraction formula tan(α - β) = (tan α - tan β)/(1 + tan α tan β) to find the height.
<p><strong>Step 1:</strong> Let pole height = h, observation distance = 40 m.</p><p><strong>Step 2:</strong> Let α = angle subtended by full pole, β = angle subtended by lower 1/4th portion.<p>Then: tan α = h/40 and tan β = (h/4)/40 = h/160</p><p><strong>Step 3:</strong> The upper 3/4th portion subtends angle (α - β), so:</p><p>tan(α - β) = 3/5</p><p><strong>Step 4:</strong> Using tangent subtraction formula:</p><p>tan(α - β) = (tan α - tan β)/(1 + tan α tan β) = (h/40 - h/160)/(1 + h/40 · h/160)</p><p>= (3h/160)/(1 + h²/6400) = (3h/160) · (6400)/(6400 + h²) = (120h)/(6400 + h²)</p><p><strong>Step 5:</strong> Setting equal to 3/5:</p><p>(120h)/(6400 + h²) = 3/5</p><p>600h = 3(6400 + h²)</p><p>600h = 19200 + 3h²</p><p>3h² - 600h + 19200 = 0</p><p>h² - 200h + 6400 = 0</p><p><strong>Step 6:</strong> Using quadratic formula:</p><p>h = (200 ± √(40000 - 25600))/2 = (200 ± √14400)/2 = (200 ± 120)/2</p><p>h = 160 m or h = 40 m</p><p>∴ Answer: C (h = 160 m or h = 40 m, depending on options)</p>
Correct Answer: C