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A tower $A B$ leans towards west making an angle $\alpha$ with the vertical. The angular elevation of $B$, the top most point of the tower is $\beta$ as observed from a point $C$ due east of $A$ at a distance ' $d$ ' from $A$. If the angular elevation of $B$ from a point $D$ due east of $C$ at a distance $2 d$ from $C$ is $r$, then $2 \tan \alpha$ can be given as
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The correct answer is:
$3 \cot \beta-\cot \gamma$
By $\mathrm{m}-\mathrm{n}$ theorem at $\mathrm{C}$
$(\mathrm{d}+2 \mathrm{~d}) \cot \beta=\mathrm{d} \cot \gamma-2 \mathrm{~d} \cot \left(90^{\circ}+\alpha\right)$
$\begin{array}{l}
3 d \cot \beta=d \cot \gamma+2 d \tan \alpha \\
\Rightarrow 3 \cot \beta=\cot \gamma+2 \tan \alpha \\
\therefore 2 \tan \alpha=3 \cot \beta-\cot \gamma
\end{array}$
$(\mathrm{d}+2 \mathrm{~d}) \cot \beta=\mathrm{d} \cot \gamma-2 \mathrm{~d} \cot \left(90^{\circ}+\alpha\right)$
$\begin{array}{l}
3 d \cot \beta=d \cot \gamma+2 d \tan \alpha \\
\Rightarrow 3 \cot \beta=\cot \gamma+2 \tan \alpha \\
\therefore 2 \tan \alpha=3 \cot \beta-\cot \gamma
\end{array}$
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