Question

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A vertical tower stands on horizontal plane andis surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is a and that of the top of flagstaff is ß.
Prove that height of the tower is
$\frac{h \: tan \: \alpha }{tan \: \beta - \: tan \: \alpha }$
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Answer 1

Answer:

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Answer 2

Answer:

Let height by y ΔOAC

$\rm{ \tan θ = \frac{P}{B} }$

$\rm{ \tan \beta = \frac{CA}{OA} }$

$\rm{ \tan \beta = \frac{y + h}{x} }$

(y+h)→Let AB, AB+BC

Let OA=x

$\rm{x = [\frac{y + x}{ \tan( \beta ) }] }$

Consider ΔOAB

$\rm{ \tan α = \frac{y}{x} = \frac{perpendicular}{base} }$

$\rm{x = \frac{y}{ \tan( \alpha ) } }$

$\rm {\frac{y}{ \tan( \alpha ) } = \frac{y + h}{ \tan( \beta ) } }$

y tan β=tan αy+tan αh

y tan β–tan y=tan αh

y(tan β–tan α)=tan αh

$\rm{y = \frac{h \: tan \alpha }{ \tan \beta - \tan \alpha } }$

This proved