Question

The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving metres towards the tower, the angle of elevation is β. Prove that the height of the tower is $\frac{\mathrm{a}\tan \mathrm{\alpha }\tan \mathrm{\beta }}{(\tan \mathrm{\beta }-\tan \mathrm{\alpha })}.$

Answer 1

Let $AB$ be the tower and $C$ be the point on the ground such that ∠$ACB=\alpha$.
On moving towards the tower at point $D$, we have $CD=a$ and ∠$ADB=\beta$.
If $AB=h$ and $BC=x$, then $BD=(BC-CD)=(x-a)$.

From ∆$ACB$, we have:
$\frac{AB}{BC}=\frac{h}{x}=\tan \alpha$

⇒ $x=\frac{h}{\tan \alpha }$ ...(i)

From ∆$ADB$, we have:
$\frac{AB}{BD}=\frac{h}{(x-a)}=\tan \beta$

⇒ $x-a=\frac{h}{\tan \beta }$

⇒ $x=a+\frac{h}{\tan \beta }$ ...(ii)
From (i) and (ii), we have:
$\frac{h}{\tan \beta }=\frac{h}{\tan \alpha }-a$

⇒ $\frac{h}{\tan \alpha }-\frac{h}{\tan \beta }=a$

⇒$\frac{h\tan \beta -h\tan \alpha }{\tan \alpha \tan \beta }=a$

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