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 angle of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is $\frac{h\tan \mathrm{\alpha }}{(\tan \mathrm{\beta }-\tan \mathrm{\alpha })}.$

Answer 1

Let $BD$ be the tower and $AD$ be the vertical flagstaff such that $AD=h$. Thus, we have:
∠$ACB=\beta$ and ∠$DCB=\alpha$
Let:
$BD=H$ and $BC=x$

In ∆$DBC$, we have:
$\frac{BD}{BC}=\tan \alpha$

⇒$\frac{H}{x}=\tan \alpha$
⇒ $x=\frac{H}{\tan \alpha }$
Or,
$H=x\tan \alpha$ ...(i)

In ∆$ABC$, we have:
$\frac{AB}{BC}=\tan \beta$

⇒ $\frac{(H+h)}{x}=\tan \beta$
⇒ $x=\frac{(H+h)}{\tan \beta }$ ...(ii)
From (i) and (ii), we get:

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

⇒ $H\tan \beta =(H+h)\tan \alpha$
⇒ $H\tan \beta =H\tan \alpha +h\tan \alpha$
⇒ $H(\tan \beta -\tan \alpha )=h\tan \alpha$
⇒ $H=\frac{h\tan \alpha }{(\tan \beta -\tan \alpha )}$
Hence, the height of the tower is $\frac{h\tan \alpha }{(\tan \beta -\tan \alpha )}$.