Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are $\alpha$ and $\beta ,$ respectively. Prove that the height of the tower is $\left[\frac{h\tan \alpha }{(\tan \beta -\tan \alpha )}\right]$

Answer 1

A vertical flagstaff of height $h$ is surmounted on a vertical tower of height $H$(say), such that $FP=h\&FO=H.$

The angle of elevation of the bottom and top of the flagstaff on the plane is $\angle PRO=\alpha \&\angle FRO=\beta$respectively

In $∆PRO,$ we have

$$\begin{gathered} \tan \alpha =\frac{PO}{R0}=\frac{H}{x} \\ \therefore X=\frac{H}{\tan \alpha } \end{gathered}$$

From $∆FRO,$ we have

$$\begin{gathered} \tan \beta =\frac{FO}{RO}=\frac{FP+PO}{RP}=\frac{h+H}{x} \\ \therefore X=\frac{h+H}{\tan \beta } \end{gathered}$$

On solving for $H$ we get

$$\begin{gathered} H\tan \beta =\left(h+H\right)\tan \alpha \\ H\tan \beta -H\tan \alpha =h\tan \alpha \\ H\left(\tan \beta -\tan \alpha \right)=h\tan \alpha \\ \therefore H=\frac{h\tan \alpha }{\tan \beta -\tan \alpha } \end{gathered}$$

Hence, proved.