At a distance a from the foot of a tower $A B$ of known height $b$ a flagstaff $B C$ and the tower subtend equal angles, height of the flagstaff is

\frac{a^{2} + b^{2}}{a^{2} - b^{2}}

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\frac{a^{2} - b^{2}}{a^{2} + b^{2}}

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\frac{a (a^{2} - b^{2})}{a^{2} + b^{2}}

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\frac{b (a^{2} + b^{2})}{(a^{2} - b^{2})}

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Solution

The correct option is D $\frac{b (a^{2} + b^{2})}{(a^{2} - b^{2})}$
Given
$∠ A P B = ∠ B P A = θ (s a y)$
In $Δ A B P, ∠ B A P = 90 ° tan θ = \frac{b}{a}$
In $Δ C A P, ∠ C A P = 90 ° tan (2 θ) = \frac{C D + b}{a} tan θ = \frac{2 tan θ}{1 - {tan}^{2} θ} = \frac{\frac{2 b}{a}}{1 - \frac{b^{2}}{a^{2}}} = \frac{2 a b}{a^{2} - b^{2}} (∵ tan θ = \frac{b}{a}) ∴ \frac{2 a b}{a^{2} - b^{2}} = \frac{C B + b}{a} \Rightarrow \frac{2 a^{2} b}{a^{2} - b^{2}} = C B + b \Rightarrow C B = \frac{2 a^{2} b}{a^{2} - b^{2}} - b = \frac{2 a^{2} b - a^{2} b + b^{3}}{a^{2} - b^{2}} = \frac{a^{2} b + b^{3}}{a^{2} - b^{2}} \Rightarrow C B = \frac{b (a^{2} + b^{2})}{(a^{2} - b^{2})}$
Answer D

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