A tower and a flag staff on its top subtend equal angles at the observer's eye. if the heights of flagstaff, tower and the eye of the observer are respectively $a, b$ and $h$ . then the distance of the observer's eye from the base of the tower is

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$a \sqrt{(\frac{a \cdot h}{a + b})}$

None of these

$a \sqrt{\frac{a + b - 2 h}{c! + b}}$

$b \sqrt{(\frac{a + b - 2 h}{a - b})}$

answer is B.

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Detailed Solution

$\tan θ = \frac{b - h}{x}$
$\tan 2 θ = \frac{a + b - b}{x}$
$\frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{a + b - b}{x}$
$\frac{2 (\frac{b - h}{x})}{1 - {(\frac{b - h}{x})}^{2}} = \frac{a + b - h}{x}$
$\Rightarrow \frac{2 (b - h) x}{x^{2} - (b - b)^{2}} = \frac{a + b - h}{x} \Rightarrow \frac{x^{2}}{x^{2} - (b -)^{2}} = \frac{a + b - b}{2}$
$x^{2} = \frac{2 x^{2}}{a + b - b} + (b - b)^{2} \to 1 - \frac{(4 b - b)^{2}}{x^{2}} = \frac{2}{a + b - b}$
$\frac{(b - b)^{2}}{x^{2}} = \frac{a + b - h - 2}{a + b - h} \to x^{2} = \frac{a + b - b}{a + b - b - 2} (b - b)^{2}$
$x = \sqrt{\frac{a + b - 2 h}{a + b}}$