A tower AB leans towards west making an angle | vedclass

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(C) Let $h$ be the height of the tower $AB$ and $x$ be the horizontal distance of $B$ from $A$. Then $x = h 	an \alpha$ and $h = h \sec \alpha$ (not

Vedclass · Accepted Answer

$3\cot \beta - \cot \gamma$

Vedclass · Answer

(C) Let $h$ be the height of the tower $AB$ and $x$ be the horizontal distance of $B$ from $A$. Then $x = h 	an \alpha$ and $h = h \sec \alpha$ (not needed here). Actually,$AB = h$ (slant height). The horizontal projection of $B$ is $B'$,where $AB' = h \sin \alpha$ and the vertical height of $B$ is $H = h \cos \alpha$.Using the $m-n$ theorem in $	riangle B'AD$ with $C$ on $AD$ such that $AC = d$ and $CD = 2d$:In $	riangle B'AC$,$\cot \beta = \frac{AC}{H} = \frac{d}{H} \Rightarrow H = d 	an \beta$.In $	riangle B'AD$,$\cot \gamma = \frac{AD}{H} = \frac{3d}{H} \Rightarrow H = 3d 	an \gamma$.Also,the horizontal distance of $B'$ from $A$ is $x = h \sin \alpha$. Since $B'$ is to the west,its coordinate is $-x$. $A$ is at $0$,$C$ is at $d$,$D$ is at $3d$.Using the cotangent rule in $	riangle B'CD$:$(d + 2d) \cot \beta = d \cot \gamma - 2d \cot(180^o - \alpha) = d \cot \gamma + 2d \cot \alpha$ is incorrect. The correct approach is:$H \cot \beta = d + x$ and $H \cot \gamma = 3d + x$.Subtracting: $H(\cot \gamma - \cot \beta) = 2d$.From $H = d 	an \beta$,we have $d = H \cot \beta$.So $H(\cot \gamma - \cot \beta) = 2H \cot \beta \Rightarrow \cot \gamma - \cot \beta = 2 \cot \beta$ (This assumes $x=0$).Given the geometry,$x = H 	an \alpha$.$H \cot \beta = d + H 	an \alpha \Rightarrow d = H(\cot \beta - 	an \alpha)$.$H \cot \gamma = 3d + H 	an \alpha \Rightarrow 3d = H(\cot \gamma - 	an \alpha)$.$3H(\cot \beta - 	an \alpha) = H(\cot \gamma - 	an \alpha)$.$3 \cot \beta - 3 	an \alpha = \cot \gamma - 	an \alpha$.$3 \cot \beta - \cot \gamma = 2 	an \alpha$.

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