The angle of elevation of the top of a vertic | vedclass

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Question

(A) Let the height of the tower $AB = h$ and the distance $BC = x$.In $\Delta ABC$,$	an \beta = \frac{AB}{BC} = \frac{h}{x} \Rightarrow x = \frac{h}{

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$\frac{2 \sin \alpha \sin \beta}{\sin(\beta - \alpha)}$

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(A) Let the height of the tower $AB = h$ and the distance $BC = x$.In $\Delta ABC$,$	an \beta = \frac{AB}{BC} = \frac{h}{x} \Rightarrow x = \frac{h}{	an \beta} = h \cot \beta$.In $\Delta ABP$,$	an \alpha = \frac{AB}{PB} = \frac{h}{x + 2}$.Substituting $x = h \cot \beta$,we get $	an \alpha = \frac{h}{h \cot \beta + 2}$.$\Rightarrow h \cot \beta + 2 = \frac{h}{	an \alpha} = h \cot \alpha$.$\Rightarrow 2 = h(\cot \alpha - \cot \beta) = h \left( \frac{\cos \alpha}{\sin \alpha} - \frac{\cos \beta}{\sin \beta} ight)$.$\Rightarrow 2 = h \left( \frac{\cos \alpha \sin \beta - \cos \beta \sin \alpha}{\sin \alpha \sin \beta} ight)$.$\Rightarrow 2 = h \left( \frac{\sin(\beta - \alpha)}{\sin \alpha \sin \beta} ight)$.$\Rightarrow h = \frac{2 \sin \alpha \sin \beta}{\sin(\beta - \alpha)}$.

The angle of elevation of the top of a vertical tower from a point $P$ on the horizontal ground was observed to be $\alpha$. After moving a distance $2 \ m$ from $P$ towards the foot of the tower,the angle of elevation changes to $\beta$. Then the height (in metres) of the tower is

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