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Let a vertical tower $\mathrm{A B}$ of height $\mathrm{2 h}$ stands on a horizontal ground. Let from a point $\mathrm{P}$ on the ground a man can see upto height $\mathrm{h}$ of the tower with an angle of elevation $\mathrm{2 \alpha.}$ When from $\mathrm{P}$ , he moves a distance $\mathrm{d}$ in the direction of $\mathrm{\overrightarrow{A P},}$ he can see the top $\mathrm{ B}$ of the tower with an angle of elevation $\mathrm{ \alpha.}$ If $\mathrm{ d=\sqrt{7} h},$ then $\mathrm{ \tan \alpha}$ is equal to
Option: 1
$\sqrt{5}-2$

Option: 2
$\sqrt{3}-1$

Option: 3
$\sqrt{7}-2$

Option: 4
$\sqrt{7}-\sqrt{3}$

Answers (1)

best_answer

$\mathrm{\tan 2 \alpha=\frac{h}{x}}\\$

$\mathrm{\text { and } \tan \alpha=\frac{2 h}{x+\sqrt{7} h}}\\$

$\mathrm{\tan \alpha=\frac{2 h}{h \cot 2 \alpha+\sqrt{7} h}}\\$

$\mathrm{\tan \alpha=\frac{2}{\frac{\left(1-\tan ^{2} \alpha\right)}{2 \tan \alpha}+\sqrt{7}}}\\$

$\mathrm{\text { Put } \tan \alpha=t\; \& \text { simplify }}\\$

$\mathrm{\Rightarrow \tan \alpha=\sqrt{7}-2}$

Hence correct option is 3

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