Question

A vertical tower stands on horizontal plane and is 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 $\alpha$ and that of the top o flagstaff is $\beta$ . Prove that the height of the tower is $\frac{htan\alpha }{tan\beta -tan\alpha }$

Answer 1

Let height be y $\Delta OAC$

$\tan \theta =\frac{P}{B}$

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

$\tan \beta =\frac{y+h}{x}$ $(y+h)\to$ Let AB, AB$+$BC

Let OA$\to$ x

$x=\left[\frac{y+x}{\tan \beta }\right]$

Consider $\Delta OAB$

$\tan \alpha =\frac{y}{x}=\frac{perpendicular}{Base}$

$x=\frac{y}{\tan \alpha }$

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

$y\tan \beta =\tan \alpha y+\tan \alpha h$

$y\tan \beta -\tan y=\tan \alpha h$

$y(\tan \beta -\tan \alpha )=\tan \alpha h$

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

This proved.